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How to Solve Circuit Problems
Last Updated: December 24, 2022
wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 14 people, some anonymous, worked to edit and improve it over time. This article has been viewed 28,304 times. Learn more...
Solving circuits is one of the most challenging tasks for the undergraduate student as it involves numerous theorems, concepts, and processes for solving the circuits. But following a planned problem solving strategy simplifies the problem, making it easier to solve.
- Use all the tools available: transforms, linear algebra, dimensional analysis, nonlinear modelling, chief among them. Thanks Helpful 0 Not Helpful 0
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- ↑ https://en.wikipedia.org/wiki/Mesh_analysis
- ↑ https://en.wikipedia.org/wiki/Nodal_analysis
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Unit 2: Circuit analysis
About this unit, circuit elements.
- Ideal circuit elements (Opens a modal)
- Ideal sources (Opens a modal)
- Ideal elements and sources (Opens a modal)
- Real-world circuit elements (Opens a modal)
- Circuit terminology (Opens a modal)
- Sign convention for passive components (Opens a modal)
- Sign convention for passive components and sources (Opens a modal)
- Series resistors (Opens a modal)
- Parallel resistors (part 1) (Opens a modal)
- Parallel resistors (part 2) (Opens a modal)
- Parallel resistors (part 3) (Opens a modal)
- Parallel resistors (Opens a modal)
- Parallel conductance (Opens a modal)
- Simplifying resistor networks (Opens a modal)
- Delta-Wye resistor networks (Opens a modal)
- Voltage divider (Opens a modal)
- Analyzing a resistor circuit with two batteries (Opens a modal)
- Series and parallel resistors 8 questions Practice
DC circuit analysis
- Circuit analysis overview (Opens a modal)
- Kirchhoff's current law (Opens a modal)
- Kirchhoff's voltage law (Opens a modal)
- Kirchhoff's laws (Opens a modal)
- Labeling voltages (Opens a modal)
- Application of the fundamental laws (setup) (Opens a modal)
- Application of the fundamental laws (solve) (Opens a modal)
- Application of the fundamental laws (Opens a modal)
- Node voltage method (steps 1 to 4) (Opens a modal)
- Node voltage method (step 5) (Opens a modal)
- Node voltage method (Opens a modal)
- Mesh current method (steps 1 to 3) (Opens a modal)
- Mesh current method (step 4) (Opens a modal)
- Mesh current method (Opens a modal)
- Loop current method (Opens a modal)
- Number of required equations (Opens a modal)
- Linearity (Opens a modal)
- Superposition (Opens a modal)
Natural and forced response
- Capacitor i-v equations (Opens a modal)
- A capacitor integrates current (Opens a modal)
- Capacitor i-v equation in action (Opens a modal)
- Inductor equations (Opens a modal)
- Inductor kickback (1 of 2) (Opens a modal)
- Inductor kickback (2 of 2) (Opens a modal)
- Inductor i-v equation in action (Opens a modal)
- RC natural response - intuition (Opens a modal)
- RC natural response - derivation (Opens a modal)
- RC natural response - example (Opens a modal)
- RC natural response (Opens a modal)
- RC step response - intuition (Opens a modal)
- RC step response setup (1 of 3) (Opens a modal)
- RC step response solve (2 of 3) (Opens a modal)
- RC step response example (3 of 3) (Opens a modal)
- RC step response (Opens a modal)
- RL natural response (Opens a modal)
- Sketching exponentials (Opens a modal)
- Sketching exponentials - examples (Opens a modal)
- LC natural response intuition 1 (Opens a modal)
- LC natural response intuition 2 (Opens a modal)
- LC natural response derivation 1 (Opens a modal)
- LC natural response derivation 2 (Opens a modal)
- LC natural response derivation 3 (Opens a modal)
- LC natural response derivation 4 (Opens a modal)
- LC natural response example (Opens a modal)
- LC natural response (Opens a modal)
- LC natural response - derivation (Opens a modal)
- RLC natural response - intuition (Opens a modal)
- RLC natural response - derivation (Opens a modal)
- RLC natural response - variations (Opens a modal)
AC circuit analysis
- AC analysis intro 1 (Opens a modal)
- AC analysis intro 2 (Opens a modal)
- Trigonometry review (Opens a modal)
- Sine and cosine come from circles (Opens a modal)
- Sine of time (Opens a modal)
- Sine and cosine from rotating vector (Opens a modal)
- Lead Lag (Opens a modal)
- Complex numbers (Opens a modal)
- Multiplying by j is rotation (Opens a modal)
- Complex rotation (Opens a modal)
- Euler's formula (Opens a modal)
- Complex exponential magnitude (Opens a modal)
- Complex exponentials spin (Opens a modal)
- Euler's sine wave (Opens a modal)
- Euler's cosine wave (Opens a modal)
- Negative frequency (Opens a modal)
- AC analysis superposition (Opens a modal)
- Impedance (Opens a modal)
- Impedance vs frequency (Opens a modal)
- ELI the ICE man (Opens a modal)
- Impedance of simple networks (Opens a modal)
- KVL in the frequency domain (Opens a modal)
Endless Electronic Problems For Solving
We know not everyone who likes to build circuitry wants to dive headfirst into the underlying electrical engineering that makes everything work. However, if you want to, now is a great time. Many universities have most or all of their material online and you can even take many courses for free. If you want an endless pool of solved study problems, check out autoCircuits . It can create many different kinds of electronics problems and their solutions.
You can get a totally random circuit, or choose if you want to focus on DC, AC, two-ports, or several other types of problems. You can also alter the general form of the problem. For example, for a DC analysis, you can have it assign circuit values so that the answer is a value such as 45 ohms, or you can have it just use symbols so that the answer might be i 4 =V 1 /4R. You even get to pick the difficulty level and pick certain types of problems to avoid. Just be fast. After the site generates a problem, you have 10 seconds to download it before it is gone forever.
The problems range from simple to complex so for example our first screenshot is an easy one.
The program provides the solution, but this one is pretty easy — you just have to combine the parallel and series resistances. With symbolic resistances, the values might have been things like R, 3R, and 9R. Or you can pick to have non-integers like 0.23R or even rational numbers like 22/5 as the multiplying factor.
The program has one other cool feature: you can input a SPICE netlist to plug in your own circuits. There are some restrictions. For example, you can only have 10 nodes and there are some strict naming conventions. In addition, all component values will be selected randomly.
For an example of the SPICE function, try this netlist:
V_1 1 0 AC C_1 1 2 C_2 1 2 R_1 2 0 .TRAN Solve V(2,0)
The output looks like the screenshot to the right, although yours will almost certainly have different random values:
If like us, you were too lazy to work it out, the output also shows the answer: v 4 (t)=(-4.111 cos(9.4 t )-4.092 sin(9.4 t )). Yeah. That’s what we were thinking.
This would be great if you were teaching electronics, too. You could generate an endless amount of homework or quiz questions. [Stephano] offers the service for free and uses software ranging from MATLAB, BLAG, and LaTeX to do its magic.
If you want to brush up on Spice and using it for analysis, we’ve been there . By the way, the first analysis tool you generally learn is nodal analysis and that would be a good first step to solving these kinds of problems.
10 thoughts on “ Endless Electronic Problems For Solving ”
Obligatory “not a hack”
Obligatory “it’s hackaday” not “nothing but hacks every day”
I was hoping the circuit in the header image would change on every refresh
More than happy creating my own electrons problems. Now if they had a way to automatically fix the problems or automatically streamline component selection to optimize a circuit that would be well worth it.
Imagine designing a circuit and doing a simulation then running a filter over it that picks out the values of the components then optimizes the circuit based on a large database of components. You could have options like …
“Bullet proof” that selects components that are above the power ratings/tolerances to make the circuit handle weird things like surges and abuse.
“Cheap” would base the circuit on the cheapest components available.
Or even things like
“low power” could filter the components and see if it could automatically select components that reduce the overall power consumption.
Machine learning to the rescue! :-p
well, this is not the silver bullet but Qucs support ASCO (optimization/Operative search solver). It takes as an input cost functions, and after some computation it will give you the optimal solution (if there is any). It’s quite handy, give it a try.
Links: use Optimal block in QUCS: http://web.mit.edu/qucs_v0.0.19/docs/en/start_opt.html ASCO: http://web.mit.edu/qucs_v0.0.19/docs/en/start_opt.html Wikipedia on operation research: https://en.wikipedia.org/wiki/Operations_research
You want National Semiconductor’s* WebBench: http://www.ti.com/design-resources/design-tools-simulation/webench-power-designer.html
*Texas Instruments purchased them a while back.
This is good for someone like me who doesn’t know a lot. Thank you! The main illustration is simple enough for this neophyte. Now I just have to find a 1 volt battery.
this is great ! I’m also interested how to generate a schema from a netlist like it does, the FAQ states the service is not going open source at the moment. Anyone knows where to look for ?
Perfect timing for Midterms. Thanks!
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- © 2022
Advanced Electrical Circuit Analysis
Practice Problems, Methods, and Solutions
- Mehdi Rahmani-Andebili 0
Engineering Technology, State University of New York, Buffalo, USA
You can also search for this author in PubMed Google Scholar
Exercises cover a wide selection of basic and advanced questions and problems
Categorizes and orders the problems based on difficulty level, hence suitable for both knowledgeable and under-prepared students
Provides detailed and instructor-recommended solutions and methods, along with clear explanations;
Can be used along with the core textbooks
- Table of contents
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Table of contents (10 chapters)
Front matter, problems: state equations of electrical circuits.
Solutions of Problems: State Equations of Electrical Circuits
Problems: laplace transform and network function, solutions of problems: laplace transform and network function, problems: natural frequencies of electrical circuits, solutions of problems: natural frequencies of electrical circuits, problems: network theorems (tellegen’s and linear time-invariant network theorems), solutions of problems: network theorems (tellegen’s and linear time-invariant network theorems), problems: two-port networks, solutions of problems: two-port networks, back matter.
- Circuit Fundamentals
- Circuit Theorems
- Convolution Integral
- Electrical Circuits
- Electrical Circuit Analysis
- Graph Theory
- Laplace Transform
- Natural Frequency
- Network Topology
- Nonlinear Electrical Circuits
- Operational Amplifier
- Principle of Duality
- Two-Port Networks
Book Title : Advanced Electrical Circuit Analysis
Book Subtitle : Practice Problems, Methods, and Solutions
Authors : Mehdi Rahmani-Andebili
DOI : https://doi.org/10.1007/978-3-030-78540-6
Publisher : Springer Cham
eBook Packages : Engineering , Engineering (R0)
Copyright Information : The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2022
Hardcover ISBN : 978-3-030-78539-0 Published: 22 July 2021
Softcover ISBN : 978-3-030-78542-0 Published: 23 July 2022
eBook ISBN : 978-3-030-78540-6 Published: 21 July 2021
Edition Number : 1
Number of Pages : IX, 152
Number of Illustrations : 1 b/w illustrations, 155 illustrations in colour
Topics : Circuits and Systems , Cyber-physical systems, IoT , Engineering/Technology Education
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Electricity: Electric Circuits
Calculator pad, version 2, electric circuits: problem set.
Over the course of an 8 hour day, 3.8x10 4 C of charge pass through a typical computer (presuming it is in use the entire time). Determine the current for such a computer.
- Audio Guided Solution
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The large window air conditioner in Anita Breeze's room draws 11 amps of current. The unit runs for 8.0 hours during the course of a day. Determine the quantity of charge that passes through Anita's window AC during these 8.0 hours.
Determine the amount of time that the following devices would have to be used before 1.0x10 6 C (1 million Coulombs) of charge passes through them. a. LED night light (I=0.0042 A) b. Incandescent night light (I=0.068 A) c. 60-Watt incandescent light bulb (I=0.50 A) d. Large bathroom light fixture (I=2.0 A)
a. 2.4x10 8 sec = 6.6x10 4 hr = 2.8x10 3 d = 7.5 yr b. 1.5x10 7 sec = 4.1x10 3 hr = 170 d c. 2.0x10 6 s = 560 hr = 23 d d. 5.0x10 5 s = 140 hr = 5.8 d
The heating element of an electric toaster is typically made of nichrome wire (an alloy of nickel and chromium). As current passes through the wires, the wires heat up, thus toasting the toast. Estimate the overall resistance of a heating element which is 220 cm long and consists of nichrome wire with a diameter of 0.56 mm. The resistivity of nichrome is 110x10 -8 Ω•m.
Determine the overall resistance of a 100-meter length of 14 AWA (0.163 cm diameter) wire made of the following materials. a. copper (resistivity = 1.67x10 -8 Ω •m) b. silver (resistivity = 1.59x10 -8 Ω •m) c. aluminum (resistivity = 2.65x10 -8 Ω •m) d. iron (resistivity = 9.71x10 -8 Ω •m)
a. 0.800 Ω b. 0.762 Ω c. 1.27 Ω d. 4.65 Ω
A power saw at the local hardware store boasts of having a 15-Amp motor. Determine its resistance when plugged into a 110-Volt outlet.
A coffee cup immersion heater utilizes a heating coil with a resistance of 8.5 Ω. Determine the current through the coil when operated at 110 V.
Defibrillator machines are used to deliver an electric shock to the human heart in order to resuscitate an otherwise non-beating heart. It is estimated that a current as low as 17 mA through the heart is required to resuscitate. Using 100,000 Ω as the overall resistance, determine the output voltage required of a defibrillating device.
A stun gun or TASER is designed to put out a few seconds worth of electric pulses that impress a voltage of about 1200 V across the human body. This results in an average current of approximately 3 mA into a human body. Using these figures, estimate the resistance of the human body.
Determine the amount of electrical energy (in J) used by the following devices when operated for the indicated times. a. Hair dryer (1500 W) - operated for 5 minutes b. Electric space heater (950 W) - operated for 4 hours c. X-Box video game player (180 W) - operated for 2 hours d. 42-inch LCD television (210 W) - operated for 3 hours
a. 4.5x10 5 J b. 1.4x10 7 J c. 1.3x10 6 J d. 2.3x10 6 J
Alfredo deDarke sleeps with a 7.5-Watt night light bulb on. He turns it on before getting in bed and turns it off 8 hours later. a. Determine the amount of energy used during one evening in units of kiloWatt•hours. b. Electrical energy costs 13 cents/kW•hr where Alfredo lives. Determine the annual (365 days) cost of this practice of using a 7.5-Watt night light. c. Determine the annual savings if Alfredo replaced his 7.5-Watt incandescent night light by a 0.5-Watt LED night light.
a. 0.060 kW•hr for one evening b. $2.8 for one year c. $2.7 savings for one year
Having recently lost her job, Penny Penching is looking for every possible means of cutting costs. She decides that her 4.0-Watt clock radio alarm does not need to be on for 24 hours every day since she only needs it for waking up after her average 8-hour sleep. So she decides to plug it in before going to sleep and to unplug it when waking. Penny pays 12 cents per kiloWatt•hour for her electricity. How much money is Penny able to save over the course of a month (31 days) with her new alarm clock usage pattern?
The power of a 1.5-volt alkaline cell varies with the number of hours of operation. A brand new D-cell can deliver as much as 13 A through a copper wire connected between terminals. Determine the power of a brand new D-cell.
A central air conditioner in a typical American home operates on a 220-V circuit and draws about 15 A of current. a. Determine the power rating of such an air conditioner. b. Determine the energy consumed (in kW•hr) if operated for 8 hours per day. c. Determine the monthly cost (31 days) if the utility company charges 13 cents per kW•hr.
a. 3300 W b. 26 kW•hr c. $110 per month (rounded from $106)
During the Christmas season, Sel Erbate uses the equivalent of 45 strings of 100 mini-bulbs to light the inside and outside of his home. Each 100-bulb string of lights is rated at 40 Watts. The average daily usage of the strings is 7 hours. The lights are used for approximately 40 days during the holiday season. a. Determine the resistance of each string of lights. Each is powered by 110-volt outlet. b. Determine the energy consumed (in kW•hr) by the lights over the course of 40 days. c. If Sel pays 12 cents/kW•hr for electrical energy, then what is the total cost of Christmas lighting for a single season?
a. 3x10 2 Ω (rounded from 302.5 Ω ) b. 5x10 2 kW•hr (rounded from 504 kW•hr) c. $60 (rounded from $60.48)
A 3-way light bulb for a 110-V lamp has two different filaments and three different power ratings. Turning the switch of the lamp toggles the light from OFF to low (50 W) to medium (100 W) to high (150 W) brightness. These three brightness settings are achieved by channeling current through the high resistance filament (50 W), the low resistance filament (100 W) or through both filaments. Determine the resistance of the 50 W and the 100 W filaments.
50-watt filament: R = 240 Ω (rounded from 242 Ω) 100-watt filament: R = 120 Ω (rounded from 121 Ω)
Compare the resistance of a 1.5-Amp interior light bulb of a car (operating off a 12-V battery) to the resistance of a 100-Watt bulb operating on a 110-volt household circuitry.
Car light bulb: 8.0 Ω 100-W lamp bulb: 120 Ω (rounded from 121 Ω)
An overhead high voltage (4.0x10 5 V) power transmission line delivers electrical energy from a generating station to a substation at a rate of 1500 MW (1.5x10 9 W). Determine the resistance of and the current in the cables.
Resistance: 110 Ω (rounded from 107 Ω) Current: 3800 A (rounded from 3750 A)
The UL panel on the bottom of an electric toaster oven indicates that it operates at 1500 W on a 110 V circuit. Determine the electrical resistance of the toaster oven.
Determine the equivalent resistance of a 6.0 Ω and a 8.0 Ω resistor if … a. … connected in series. b. … connected in parallel.
a. 14.0 Ω b. 3.4 Ω
Two resistors with resistance values of 6.0 Ω and 8.0 Ω are connected to a 12.0-volt source. Determine the overall current in the circuit if the resistors are … a. … connected in series. b. … connected in parallel.
a. 0.86 A b. 3.5 A
a. 10.3 Ω b. 4.7 A c. ΔV 1 = 30. V and ΔV 2 = 18 V
Ammeter readings: 2.34 A (for each) Top voltmeter reading: 2.99 V Bottom voltmeter reading: 9.01 V
a. 22.0 Ω b. 5.0 A c. ΔV 1 = 36 V, ΔV 2 = 31 V and ΔV 3 = 43 V
Ammeter readings: 1.57 A (for each) Top voltmeter reading (across R 1 ): 16.1 V Right voltmeter reading (across R 2 ): 23.8 V Bottom voltmeter reading (across R 3 ): 20.1 V
A circuit powered by a 12.0-volt battery is comprised of three identical resistors in series. An ammeter reading reveals a current of 0.360 A. Determine the resistance values of the resistors and the voltage drops across the resistors.
R 1 = R 2 = R 3 = 11.1 Ω ΔV 1 = ΔV 2 = ΔV 3 = 4.0 V
A 4.5-volt series circuit consists of two resistors. Resistor A has three times the resistance as resistor B. An ammeter records a current of 160 mA of current. Determine the resistance values of resistors A and B.
R A = 21.1 Ω R B = 7.0 Ω
A 9.00-volt battery is used to power a series circuit with a 2.50 Ω and a 3.50 Ω resistor. Determine the power rating of each resistor and the total power of the circuit.
Power of 2.5 Ω resistor: 5.63 W Power of 3.5 Ω resistor: 7.88 W Power of entire circuit: 13.5 W
Determine the equivalent resistance of a parallel arrangement of two resistors with resistance values of … a. … 8.0 Ω and 8.0 Ω b. … 5.0 Ω and 5.0 Ω c. … 5.0 Ω and 8.0 Ω d. … 5.0 Ω and 9.2 Ω e. … 5.0 Ω and 27.1 Ω f. … 5.0 Ω and 450 Ω
a. R eq = 4.0 Ω b. R eq = 2.5 Ω c. R eq = 3.1 Ω d. R eq = 3.2 Ω e. R eq = 4.2 Ω f. R eq = 4.9 Ω
a. 2.4 Ω b. I 1 = 1.9 A and I 2 = 3.1 A c. 5.0 A
a. 5.10 Ω b. I 1 = 7.14 A, I 2 = 5.02 A and I 3 = 9.40 A c. 21.57 A
Voltmeters can be used to determine the voltage across two points on a circuit. An ammeter can be used to determine the current at any given location on a circuit. The circuit below is powered by a 24.0-volt power source and utilizes four voltmeters and three ammeters to measure voltage drops and currents.
The resistor values are 54.5 Ω (R 1 ), 31.7 Ω (R 2 ) and 48.2 Ω (R 3 ). Determine the ammeter readings and voltmeter readings.
Left ammeter reading (in R 1 branch): 0.440 A Middle ammeter reading (in R 2 branch): 0.757 A Right ammeter reading (in R 3 branch): 0.498 A Bottom ammeter reading (outside branches): 1.695 A Voltmeter readings: 24.0 V (for each)
A 9.00-volt battery is used to power a parallel circuit with a 2.50 Ω and a 3.50 Ω resistor. Determine the power rating of each resistor and the total power of the circuit.
Power of 2.5 Ω resistor: 32.4 W Power of 3.5 Ω resistor: 23.1 W Power of entire circuit: 55.5 W
Cullen Ary's family loves to cook. According to Cullen's friends, they have every imaginable kitchen gadget that exists. One Sunday afternoon, they have a cooking party in which every member of the family participates. They get out the following small appliances, plug them in and turn them on.
Mixer (81 Ω) Crockpot (62 Ω) Juicer (43 Ω) Blender (21 Ω) Electric Fondue (16 Ω) Wok (12 Ω) Rotisserie (7.5 Ω) Deep-fat fryer (7.0 Ω)
The resistance values for each appliance is listed in parenthesis. Each appliance is plugged in to 110-volt receptacles which are wired in parallel on the same circuit. The circuit is protected by a 20-amp circuit breaker.
a. Determine the overall current on the circuit with the mixer and crockpot operating. b. Determine the overall current on the circuit with the mixer, crockpot and juicer operating. c. Determine the overall current on the circuit with the mixer, crockpot, juicer and blender operating. d. Determine the overall current on the circuit with the mixer, crockpot, juicer, blender and electric fondue operating. e. Determine the overall current on the circuit with the mixer, crockpot, juicer, blender, electric fondue, and wok operating. f. Determine the overall current on the circuit with the mixer, crockpot, juicer, blender, electric fondue, wok, and rotisserie operating. g. Determine the overall current on the circuit with the mixer, crockpot, juicer, blender, electric fondue, wok, rotisserie, and the deep-fat fryer operating. h. At what point in the progression of turning on appliances will the circuit become overloaded and the circuit breaker interrupt the circuit.
a. 3.1 A b. 5.7 A c. 10.9 A d. 17.8 A e. 27.0 A f. 41.6 A g. 57.4 A h. Once the wok is plugged in (part e), the circuit breaker will trip and interrupt the circuit.
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Unit of competency details
Ueeec0065 - solve problems in basic electronic circuits (release 1).
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- Download Unit of competency in Word format. Unit of competency (1.26 MB)
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Unit of competency
Modification history, application, pre-requisite unit, competency field, unit sector, elements and performance criteria, foundation skills, range of conditions.
- two types of circuit problems in single source parallel and series-parallel electronic circuits
Unit Mapping Information
Assessment requirements, performance evidence.
- applying relevant work health and safety (WHS)/occupational health and safety (WHS/OHS) requirements, workplace procedures and practices, including using risk control measures
- determining the operating parameters of an existing circuit
- altering an existing circuit to comply with specified operating parameters
- developing circuits to comply with a specified function and operating parameters
- dealing with unplanned events/situations in accordance with workplace procedures in a manner that minimises risk to personnel and equipment.
- behaviour of electrical/electronic circuits for various values of voltage, current, resistance, impedance, inductance, capacitance and reactance and variable parameters, including:
- single source circuits
- series circuit configurations
- parallel circuit configurations
- series-parallel circuit configurations
- circuit configurations, including:
- circuit configurations are single source alternating current (a.c.) and direct current (d.c.) circuits
- series circuits
- parallel circuits
- series-parallel circuits
- connection of test/measuring devices into a circuit, including safety procedures and circuit arrangement of test/measuring devices
- electronics circuit principles
- features of testing/measuring devices, including safety, user calibration and parameter and range settings
- power supplies for electronics circuit principles
- relationship between variable parameter in electrical/electronic circuits, including:
- variable parameters
- relevant manufacturer specifications
- relevant job safety assessments or risk mitigation processes, including safe working practices and solving problems in electronic circuits
- relevant WHS/OHS legislated requirements
- relevant workplace documentation
- relevant workplace policies and procedures
- types of voltage testers, multimeters, clamp meters, continuity testers and insulation resistance testers and their application.
- a range of relevant exercises, case studies and/or other simulations
- relevant and appropriate materials, tools, equipment and personal protective equipment (PPE) currently used in industry
- resources that reflect current industry practices in relation to in solving problems in electronic circuits
- applicable documentation, including workplace procedures, equipment specifications, regulations, codes of practice and operation manuals.
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- Applications of Systems of Equations: An Electronic Circuit
Hands-on Activity Applications of Systems of Equations: An Electronic Circuit
Grade Level: 11 (9-12)
Time Required: 1 hours 15 minutes
This activity uses many non-expendable (reusable) items see the Materials List.
Group Size: 4
Activity Dependency: None
Subject Areas: Algebra, Measurement, Problem Solving, Reasoning and Proof, Science and Technology
Activities Associated with this Lesson Units serve as guides to a particular content or subject area. Nested under units are lessons (in purple) and hands-on activities (in blue). Note that not all lessons and activities will exist under a unit, and instead may exist as "standalone" curriculum.
Te newsletter, engineering connection, learning objectives, materials list, worksheets and attachments, more curriculum like this, pre-req knowledge, introduction/motivation, vocabulary/definitions, activity scaling, user comments & tips.
Are all circuits the same? When you flip a switch that controls a circuit, is it only turning it on and off? The answer to this question depends on what the circuit is designed to do. The power portion of the electric circuit simply powers machines with electric current flow, whereas an electronic circuit can interpret signals, give commands and/or perform tasks. Engineers use both circuit types in most of the sophisticated devices we depend on daily such as cell phones, computers, lamps, washers and dryers, cars and many more. This activity shows students how the myriad electronic circuits so essential in our lives can be understood as and defined as systems of equations.
After this activity, students should be able to:
- Apply the concepts of systems of equations to real-world situations.
- Explain the importance of systems of equations in engineering.
- Construct a circuit using a breadboard, resistors, voltage source and wires.
- Use a voltmeter to measure voltages in a circuit.
- Compare measured and calculated values by calculating the percentage difference.
- Explain how they are impacted by this real-world experience and its connection with mathematics.
Educational Standards Each TeachEngineering lesson or activity is correlated to one or more K-12 science, technology, engineering or math (STEM) educational standards. All 100,000+ K-12 STEM standards covered in TeachEngineering are collected, maintained and packaged by the Achievement Standards Network (ASN) , a project of D2L (www.achievementstandards.org). In the ASN, standards are hierarchically structured: first by source; e.g. , by state; within source by type; e.g. , science or mathematics; within type by subtype, then by grade, etc .
Ngss: next generation science standards - science.
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Common Core State Standards - Math
International technology and engineering educators association - technology, state standards, michigan - math.
Each group needs:
- K2 power supply, such as this breadboard power supply module at Amazon
- 9V DC batter power cable barrel T-type connector, such as available at Amazon
- 5 resistors: 52.3 kOhms, 28 kOhms, 10.5 kOhms, 12.1 kOhms, 14.7 kOhms; such as from a pack of 525 for $11 at Amazon (Elegoo 17 values 1% resistor kit assortment, 0 ohm-1M ohm)
- access to computer with Internet access, to look up vocabulary word definitions
- Electronic Circuit Worksheet , one per student
- (optional) calculator, preferably Ti-Nspire
- (optional) Breadboard Visual Aid
Students must be able to solve systems of three equations either by substitution or elimination. They also must be familiar with percentage difference. See the Procedure > Background section for instructions on these math procedures. It is recommended that the associated lesson be conducted prior to this activity.
Can you live without your cell phone? If not, why? (Listen to some student explanations.) Phones are one example of the many engineered devices that we depend on every day. What are some other everyday technologies that are so common that we take them for granted? (Listen to some student ideas.)
What are these technologies made of? What are their components? (Make a list of student responses, which might include components such as computer chips, switches and antennas.)
How do they work? (Expect this question to lead students to mention the different forms of energy that drive the various everyday technologies and devices, for example, cars, cell phones and calculators use gas, batteries charged by electric current and/or solar-charged batteries, respectively.)
Have you ever wondered what happens when you flip a switch to turn on a light, TV or computer? What does this action do in the device? (Listen to student explanations.) In all of these cases, the act of flipping a switch or pushing a button completes an electric circuit, which permits the current to flow and turn on the light, TV or computer.
Circuit diagrams are a pictorial way of describing circuits. Electricians and engineers draw circuit diagrams as they design the circuits. These diagrams show in detail how all the components connect together to make the circuits that drive the many technologies we use that are driven by electricity. The electronic devices you use—including PCs, TVs, stereos, phones, iPods, ovens, microwaves—have thousands of resistors in them. We will be using one of these diagrams.
In this activity, you will see that we depend on the electrical components that make operational so many of the devices and appliances that we rely on every day. You will also see how the electronic circuit can be understood as and defined as a system of equations. Let’s get started!
Note that in the associated lesson students worked with a teacher-built circuit to focus on learning the math calculations and become familiar with a real-world systems of equations problem. In this activity, student groups dig a little deeper by building their own circuits, which they test and compare to their math calculations, which gives them practice in hands-on circuit-building and measurement-taking.
As necessary, review with students the information provided below.
How to solve a system of equations by elimination: 1) eliminate one variable by the addition of two equations, 2) use a different pair of equations and eliminate the same variable, leaving a new system of two equations with two unknowns, 3) solve the system of two equations by addition or substitution, then 4) plug the results into the other equations to find all the variables. Follow the example below from Wallace (2010; see References for link to public domain textbook).
- Eliminate y: add two equations together.
- Eliminate y from another pair of equations.
- Solve the new system of equations.
- Plug in the results to find the remaining unknown variables.
How to solve a system of three equations by substitution: 1) solve one equation for one of the variables, 2) substitute this value into another equation and solve for another one of the variables, 3) substitute the value from the two variables solved for previously into the final equation and solve for the last remaining variable, then 4) plug in the value of the variable into the other equations.
Alternatively, students may combine the substitution and elimination methods by eliminating one variable, then using substitution to solve the remaining system of two equations.
How to calculate percentage difference: Use theoretical values from mathematical calculations and experimental values from voltmeter measurements, as in the equation below:
Before the Activity
- Make copies of the Electronic Circuit Worksheet and Breadboard Visual Aid .
- Gather and make sets of the Materials List items (breadboards, resistors, wires, etc.), enough for one set per group. See the Activity Scaling section for alternative setups.
With the Students—Introduction
- Administer the pre-activity quiz.
- Present to the class the Introduction/Motivation content.
- Then organize the class into groups of four or five students each.
- Ask the class: What are electrical circuits? How are they used? Are all circuits the same? When you flip a switch that controls a circuit, is it only turning it on or off? (Listen to student answers.) Answer: It depends on whether the switch is an electric circuit or an electronic circuit.
- Direct the groups to research and write down definitions for the vocabulary words: electric current, electrical resistance, electronic circuit, Ohm’s law and voltage.
- Lead a class discussion, starting with the definitions, guiding students to realize that an electric circuit simply powers machines with electric current, whereas, an electronic circuit can interpret signals, give commands and/or perform tasks. These circuits are in nearly all the electronic-based devices and engineered products that we depend on daily such as cell phones, computers, cars, washers and dryers and many more.
- Hand out the worksheet and read through the instructions as a class.
- Ask students to repeat what they are supposed to do. Make sure they know that the diagram on the worksheet (same as Figure 1) is the circuit that they are going to build and use. As necessary, provide further explanation or images to support any challenging concepts or clarifications your students might need.
With the Students—Part 1: The Mathematical Model of the Circuit
In their groups, students solve a system of three equations by hand (analytically) or by using a calculator such as the Ti-Nspire. On the worksheet, they are given three equations—the required values for the two given voltages, V, the five resistances, R—and asked to solve for the three unknown currents. Oversee groups as they complete the following steps:
- Have students rewrite each equation with the correct values for the voltages and resistances.
- Tell students to solve the system to determine the values for I 1 , I 2 and I 3 .
- Direct students to move on to construct the physical circuit (Part 2).
With the Students—Part 2: Building/Measuring the Physical Circuit
Groups each build the physical circuit according to the provided circuit diagram and then measure voltages across the resistors. They compare their mathematically derived data to their measured data.
- Hand out (or have groups gather) the circuit-building materials.
- Inform students that the (Figure 1) circuit diagram shown on their worksheets can be physically created using a breadboard, resistors and some jumper wires. The jumper wires hop over the gap that runs between the exterior and interior columns and the gap down the middle of the breadboard, and connects the respective columns.
- Breadboard Orientation: Familiarize students with the circuit-building materials. If necessary, hand out the visual aid (same as Figure 2). Explain how each breadboard column and row are linked, as follows:
- Each hole is called a terminal.
- The exterior columns are linked both horizontally and vertically with the voltage located at the top of each column.
- Rows are linked horizontally in the middle two columns. You must use a jumper wire to connect a left-side row to a right-side row, over any of the gaps.
- When we add resistors and create a continuous circuit (a loop), we create a current.
- Circuit-Building Instructions: Encourage students to examine the resistors and wires to see what they look like. Then refer to the circuit diagram/schematic and help groups put together the circuit with as much (or little) guidance as they require. Suggest that they start with the left side of the schematic and follow these steps:
- First, locate each resistor by its ohms value and order them 1-5.
- Add resistor R 1 . Insert one end of the resistor into any terminal in the exterior column for the first voltage source on the left side. Then, pick a row, staying the same horizontally, in the middle columns and insert the other end into a terminal.
- Then continue with R 3 . Add one end of this resistor in the same row as the previous resistor into a terminal and then insert the other end into a terminal in the same row but in the column to the right of the gap. This connects the two sides of the middle two columns.
- Next, add R 5 , similar to how we connected R 1 . Add one end of this resistor to a terminal to the right of R 3 and the other to the right exterior column, connecting it to the second voltage source.
- Now add the remaining two resistors R 2 and R 4 . One end of each of these must be inserted into the same row as the other three resistors, with the remaining ends inserted into rows several spaces below. Place R 4 on the right side of the middle gap and R 2 to the left.
- Finally, use the jumper wire to make the circuit continuous. From a new row below, insert one end of a jumper wire into the voltage terminal in the right exterior column. Do the same for the left side. Then add one to connect the right and left side of the middle columns over the gap.
- Tip: If students struggle, refer them to Photos 1 and 2 on their worksheets (same as Figure 3) and have them build what they see. Walk around and make sure that the circuits are correct.
- After the circuits are built, direct students to continue on, guided by the worksheet’s explicit instructions.
- Next, groups use voltmeters to measure voltages, V 1 , V 2 and V 3 across the resistors R 1 , R 2 and R 3 and calculate the current using the Ohm’s law. As explained on the worksheet, they follow these steps:
- Transform Ohm’s equation and solve for I.
- Next, measure the voltage across each resistor and use it to calculate the current. Tip: It is important to use the voltage meter the same way every time. So, advise students to orient the resistor so that “it faces you” as in Photo 2 on the worksheet (same as Figure 3).
- Once the circuit is ready, with the power on, place the red probe on the left side and the black probe on the right—and then take the reading. Do this the same way for R 1 , R 2 and R 3 .
- Use the values given for R 1 , R 2 and R 3 and the value just measured for V 1 , V 2 and V 3 and calculate the currents I 1 , I 2 , and I 3 .
With the Students—Part 3: Comparing Results
- The final activity step is to compare the current values from Part 1 to Part 2 and calculate the percentage difference in their results.
- To conclude, assign students to reflect on what they learned and create posters of their results. Then administer the post-activity quiz.
electric current: A flow of electric charge. The rate at which charge flows past a given point in an electric circuit. Measured in amperes.
electrical resistance: A measurement of the difficulty encountered by a power source in forcing electric current through an electrical circuit, and hence the amount of power dissipated in the circuit. Measured in ohms.
electronic circuit: A composition of individual electronic components, such as resistors, transistors, capacitors, inductors and diodes, connected by conductive wires or traces through which electric current can flow. Electronic circuits perform various simple and complex operations such as interpret/amplify a signal, make a computation, move data, give a command and/or perform a task.
Ohm’s law: The current through a conductor between two points is directly proportional to the voltage across the two points.
voltage: A quantitative expression of the force pushing current between two points in an electric circuit. Also called electromotive force. Measured in volts.
Pre-Activity Quiz: Administer the two-question Applications of Systems Pre/Post-Quiz . Review students’ answers to gauge their baseline ability to solve systems of equations.
Activity Embedded Assessment
Queries: As students are completing the Procedure Parts 1-3 in their teams, walk around and ask questions to gauge their understanding of the mathematical, build, measurement and analysis work. Examine their completed Electronic Circuit Worksheets to assess their depth of understanding.
Final Poster Project: Have students, in teams or individually, each create a poster of their results. The poster content reveals the depth and accuracy of their comprehension.
Post-Activity Quiz: Administer the Applications of Systems Pre/Post-Quiz again. Compare students’ pre and post scores to assess their learning gains.
The resistors are sharp so alert students to be careful when handling them.
This activity can be scaled up or down, as well as adjusted for math- or science-focused classrooms.
- For lower grades, provide more teacher guidance, such as by building one circuit in advance that is used as a teacher demonstration instead of each group constructing its own circuit.
- For lower grades in which each group makes its own circuit, lead an entire-class group building process.
- Increase the challenge by providing very little teacher support as groups independently build the circuit; let teams discover how to do it themselves as the instructor walks around the classroom to observe and assist only as necessary.
- Make the activity more challenging by increasing the number of resistors and the number of equations used. Find equations in most high school physics books or at the Physics Classroom website.
- To adjust for a science class, make sure that circuits have already been discussed and then have students build the circuit independently when given the Figure 1 schematic/worksheet.
Students learn about nondestructive testing, the use of the finite element method (systems of equations) and real-world impacts, and then conduct mini-activities to apply Maxwell’s equations, generate currents, create magnetic fields and solve a system of equations. They see the value of NDE and FEM...
Students are introduced to several key concepts of electronic circuits. They learn about some of the physics behind circuits, the key components in a circuit and their pervasiveness in our homes and everyday lives.
Students learn that charge movement through a circuit depends on the resistance and arrangement of the circuit components. In one associated hands-on activity, students build and investigate the characteristics of series circuits. In another activity, students design and build flashlights.
Circuit Symbols and Circuit Diagrams. July 2016. Lesson 4, Circuit Connections, Current Electricity, The Physics Classroom. http://www.physicsclassroom.com/class/circuits/Lesson-4/Circuit-Symbols-and-Circuit-Diagrams
Wallace, Tyler. Beginning and Intermediate Algebra: An Open Source Textbook . 2010. CC BY-SA 3.0. ISBN #978-1-4583-7768-5. http://www.wallace.ccfaculty.org/book/Beginning_and_Intermediate_Algebra.pdf
Supporting program, acknowledgements.
This curriculum was developed through the Smart Sensors and Sensing Systems research experience for teachers under National Science Foundation RET grant no. CNS 1609339. However, these contents do not necessarily represent the policies of the NSF, and you should not assume endorsement by the federal government.
Last modified: April 19, 2023
The following physical quantities are measured in an electrical circuit; Current , : Denoted by I measured in Amperes ( A ). Resistance , : Denoted by R measured in Ohms ( W ) . Electrical Potential Difference , : Denoted by V measured in volts. (v)
Three basic laws govern the flow of current in an electrical circuit :
1. Ohm's Law
2. Kirchhoff's Voltage Law Conservation of Energy.
3. Kirchhoff's Current Law Conservation of Charge .
Simple circuits are categorized in two type :
1. Series Circuits
2. Parallel Circuits
For circuits with series and parallel sections, break the circuit up into portions of series and parallel, then calculate values for these portions, and use these values to calculate the resistance of the entire circuit. That is, first, for each individual series path, calculate the total resistance for that path. Second, using these values, by assuming that each path as a single resistor, calculate the total resistance of the circuit.
We can apply the methods for solving linear systems to solve problems involving electrical circuits. In a given circuit if enough values of currents, resistance, and potential difference is known, we should be able to find the other unknown values of these quantities. We mainly use the Ohm's Law , Kirchhoff's Voltage Law and Kirchhoff's current Law.
Example : Find the currents in the circuit for the following network.
Solution : Lets assign currents to each part of the circuit between the node points. We have two node points Which will give us three different currents. Lets assume that the currents are in clockwise direction.
So the current on the segment EFAB is I 1 , on the segment BCDE is I 3 and on the segment EB is I 2
Using the Kirchhoff's current Law for the node B yields the equation
For the node E we will get the same equation.Then we use Kirchhoff's voltage law
-4 I 1 + (-30) -5 I 1 - 10I 1 -60 +10I 2 =0
When through the battery from (-) to (+), on the segment EF, potential difference is -30, and on segment FA moving through the resistor of 5 W will result in the potential difference of -5 I 1 and in a similar way we can find the potential differences on the other segment of the loop EFAB.
In the loop BCDE, Kirchhoff's voltage law will yield the following equation:
-30 I 3 + 120-10I 2 +60 =0
Now we have three equations with three unknowns:
I 1 + I 2 - I 3 =0 -19 I 1 +10 I 2 = 90 -10 I 2 -30 I 3 = -180
This linear system can be solved by methods of linear Algebra. Linear Algebra is more useful when the network is very complicated and the number of the unknowns is large.
The system above has the following solution:
Electric circuits – problems and solutions
1. R 1 , = 6 Ω, R 2 = R 3 = 2 Ω, and voltage = 14 volt, determine the electric current in circuit as shown in figure below.
Resistor 1 (R 1 ) = 6 Ω
Resistor 2 (R 2 ) = 2 Ω
Resistor 3 (R 3 ) = 2 Ω
Voltage (V) = 14 Volt
Wanted : Electric current (I)
Equivalent resistor (R) :
R 2 and R 3 are connected in parallel. The equivalent resistor :
1/R 23 = 1/R 2 + 1/R 3 = 1/2 + 1/2 = 2/2
R 23 = 2/2 = 1 Ω
R 1 and R 23 are connected in series. The equivalent resistor :
R = R 1 + R 23 = 6 Ω + 1 Ω
Electric current (I) :
I = V / R = 14 / 7 = 2 Ampere
2. Which one of the electric circuits as shown below has the bigger current.
The resistance of the resistor is R and the electric voltage is V.
R 1 , R 2 and R 3 are connected in series. The equivalent resistor :
R A = R 1 + R 2 + R 3 = R + R + R = 3R
R 1 , R 2 and R 3 are connected in parallel. The equivalent resistor :
1/R = 1/R 1 + 1/R 2 + 1/R 3 = 1/R + 1/R + 1/R = 3/R
1/R 23 = 1/R 2 + 1/R 3 = 1/R + 1/R = 2/R
R C = R 1 + R 23 = R + R/2 = 2R/2 + R/2 = 3R/2
R 1 and R 2 are connected in parallel. The equivalent resistor :
1/R 12 = 1/R 1 + 1/R 2 = 1/R + 1/R = 2/R
R 12 and R 3 are connected in series. The equivalent resistor :
R D = R 12 + R 3 = R/2 + R = R/2 + 2R/2 = 3R/2
3. R 1 = 4 ohm, R 2 = 6 ohm, R 3 = 2 ohm, and V = 24 volt. What is the electric current in circuit as shown in figure below.
Resistor 1 (R 1 ) = 4 Ohm
Resistor 2 (R 2 ) = 6 Ohm
Resistor 3 (R 3 ) = 2 Ohm
Voltage (V) = 24 Volt
Wanted : Electric current in circuit
R = R 1 + R 2 + R 3 = 4 + 6 + 2
Electric current :
I = V / R = 24 / 12 = 2 Ampere
4. Which one of the electric circuits as shown below has the bigger current.
Electric current in circuit A.
The equivalent resistor :
R 1 = 3 Ω, R 2 = 4 Ω, R 3 = 4 Ω, V = 12 Volt
1/R 23 = 1/R 2 + 1/R 3 = 1/4 + 1/4 = 2/4 = 1/2
R 23 = 2/1 = 2 Ω
R = R 1 + R 23 = 3 Ω + 2 Ω = 5 Ω
I = V / R = 12 / 5 = 2.4 Ampere
Electric current in circuit B.
R 1 = 8 Ω, R 2 = 2 Ω, R 3 = 2 Ω, V = 36 Volt
R = R 1 + R 2 + R 3 = 8 + 2 + 2 = 12 Ω
I = V / R = 36 / 12 = 3 Ampere
Electric current in circuit C.
R 1 = 4 Ω, R 2 = 4 Ω, R 3 = 6 Ω, V = 12 Volt
1/R 23 = 1/R 2 + 1/R 3 = 1/4 + 1/4 + 1/6 = 3/12 + 3/12 + 2/12 = 8/12
R 23 = 12/8 = 1.5 Ω
I = V / R = 12 / 1.5 = 8 Ampere
Electric current in circuit D.
R 1 = 3 Ω, R 2 = 3 Ω, R 3 = 3 Ω, R 4 = 3 Ω, R 5 = 6 Ω, V = 24 Volt
R 2 , R 3 and R 4 are connected in parallel. The equivalent resistor :
1/R 234 = 1/R 2 + 1/R 3 + 1/R 4 = 1/3 + 1/3 + 1/3 = 3/3
R 234 = 3/3 = 1 Ω
R 1 , R 234 and R 5 are connected in series The equivalent resistor :
R = R 1 + R 234 + R 5 = 3 + 1 + 6 = 9 Ω
I = V / R = 24 / 9 = 2.6 Ampere
5. According to figure as shown below, determine :
B. Electric current in circuit
C. Current I 1
D. Current I 2
Resistor 1 (R 1 ) = 4 Ω
Resistor 2 (R 2 ) = 4 Ω
Resistor 4 (R 4 ) = 3 Ω
Electric voltage (V) = 12 Volt
A. Total resistance (R)
Resistor R 2 and resistor R 3 are connected in series. The equivalent resistor :
R 23 = R 2 + R 3 = 4 Ω + 2 Ω = 6 Ω
Resistor R 23 and resistor R 4 are connected in parallel. The equivalent resistor :
1/R 234 = 1/R 23 + 1/R 4 = 1/6 + 1/3 = 1/6 + 2/6 = 3/6
R 234 = 6/3 = 2 Ω
Resistor R 1 and resistor R 234 are connected series. The equivalent resistor :
R = R 1 + R 234 = 4 Ω + 2 Ω = 6 Ω
The total resistance is 6 Ohm.
B. Electric current in circuit (I)
V = electric voltage, I = electric current, R = electric resistance
I = V / R = 12 Volt / 6 Ohm = 2 Ampere
C. Electric current I 1
Electric current in resistor R 1 = electric current in circuit = 2 Ampere.
Electric current in resistor R 234 = electric current in resistor R 1 = 2 Ampere.
Voltage in resistor R 234 is:
V = I R 234 = (2 A)(2 Ohm) = 4 Volt
Voltage in resistor R 234 = voltage in resistor R 4 = voltage in resistor R 23 = 4 Volt.
The equivalent resistor R 23 is 6 Ohm.
Electric current in resistor R 23 is :
I = V / R = 4 Volt / 6 Ohm = 2/3 Ampere
Electric current in resistor R 23 = Electric current in resistor R 2 = electric current in resistor R 3 = 2/3 Ampere.
6. R 1 = R 2 = 10 Ω and R 3 = R 4 = 8 Ω. What is the electric current in circuit as shown in figure below ?
Resistor R 1 = Resistor R 2 = 10 Ω
Resistor R 3 = Resistor R 4 = 8 Ω
Wanted : electric current (I)
The equivalent resistor
Resistor R 3 and resistor R 4 are connected in parallel, the equivalent resistor :
1/R 34 = 1/R 3 + 1/R 4 = 1/8 + 1/8 = 2/8
R 34 = 8/2 = 4 Ω
Resistor R 1 , R 2 and R 34 are connected in series, the equivalent resistor :
R = R 1 + R 2 + R 34 = 10 Ω + 10 Ω + 4 Ω = 24 Ω
I = V / R = 12 Volt / 24 Ohm = 0,5 Volt/Ohm = 0.5 Ampere
7. If the internal resistance of battery ignored, what is the electric current in the circuit shown in figure below.
Resistor R 1 = 3 Ohm
Resistor R 2 = 3 Ohm
Resistor R 3 = 6 Ohm
Electric voltage (V) = 6 Volt
Resistor R 1 and R 2 are connected in series. The equivalent resistor :
R 12 = R 1 + R 2 = 3 Ohm + 3 Ohm = 6 Ohm
Resistor R 12 and resistor 3 are connected in parallel. The equivalent resistor :
1/R = 1/R 12 + 1/R 3 = 1/6 + 1/6 = 2/6
R = 6/2 = 3 Ohm
I = V / R = 6 / 3 = 2 Ampere
8. What is the total electric current in circuit as shown in figure below.
Resistor R 1 = 6 Ohm
Resistor R 2 = 4 Ohm
Electric current (V) = 6 Volt
Internal resistance (r) = 0.6 Ohm
Wanted : Electric current
Resistor R 1 and resistor R 2 are connected in parallel. The equivalent resistor :
1/R P = 1/R 1 + 1/R 2 = 1/6 + 1/4 = 4/24 + 6/24 = 10/24
R P = 24/10 = 2.4 Ohm
Resistor R P and internal resistance (r) are connected in series. The equivalent resistor :
R = R P + r = 2.4 Ohm + 0.6 Ohm = 3.0 Ohm
Electric current in circuit :
I = V / R = 6 Volt / 3 Ohm = 2 Ampere
- Answer : An electric circuit is a closed path or loop in which electric current can flow continuously. It typically consists of sources of voltage (like batteries), loads (like resistors, LEDs, motors), and conductors to connect them.
- Answer : In a series circuit, components are connected end-to-end, so there’s a single path for current. In a parallel circuit, components are connected across common points or junctions, providing multiple paths for current.
- Answer : Ohm’s Law states that the current ( I ) flowing through a conductor between two points is directly proportional to the voltage ( V ) across the two points and inversely proportional to the resistance ( R ). It’s represented as I = V/R .
- Answer : A switch controls the flow of current in a circuit. When closed, it allows current to flow; when open, it interrupts or stops the current flow.
- Answer : In a short circuit, the resistance is very low, causing a very high current to flow. This can lead to overheating, fires, or damage to components and should be protected against with fuses or circuit breakers.
- Answer : Both fuses and circuit breakers are protective devices designed to interrupt a circuit if the current exceeds a predetermined safe level. While fuses “blow” or “melt”, breaking the circuit, circuit breakers “trip”, and can be reset after they interrupt the circuit.
- Answer : Kirchhoff’s Current Law states that the sum of currents entering a junction is equal to the sum of currents leaving that junction. This is essentially a statement of the conservation of electric charge.
- Answer : DC refers to the unidirectional flow of electric charge, typically from a battery or a DC power supply. AC, on the other hand, is an electric charge that changes direction periodically, like what’s supplied from the power grid in many countries.
- Answer : “Ground” refers to a reference point in an electrical circuit from which other voltages are measured, or a common return path for electric current, or a direct physical connection to the Earth.
- Why are capacitors used in electric circuits?
- Answer : Capacitors store and release electrical energy. They’re used in circuits for various purposes, such as filtering, smoothing voltage fluctuations, coupling and decoupling AC signals, and timing elements in oscillators.
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