- #1
SOLVING[Series-Parallel Circuits
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- Thread starter Slashz
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In summary, the student is trying to solve a problem involving Ohm's law and the voltmeter, but is having difficulty. They have identified that I3 is the current through R3, and have provided an attempt at a solution. They have also calculated V3, which is the voltage across R4 due to the current through R3.
- #2
gneill
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Slashz said:
1. You haven't identified where I3 is (Is it the current through R3?).
2. There must be relevant equations! Otherwise, how would such problems get solved?
3. You need to provide some attempt at a solution, or at least some discussion of what you've already tried.
- #3
Slashz
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ohh sorry so this is my attempt.
1. You haven't identified where I3 is (Is it the current through R3?).Yes
2. There must be relevant equations! Otherwise, how would such problems get solved?
http://en.wikipedia.org/wiki/Ohm's_law
3. You need to provide some attempt at a solution, or at least some discussion of what you've already tried.
Rt=R1+R2+R34
Rt=10+13+2
Rt=25
I3=1
V3=I3R3
V3=1(6)
V3=6
I4=I2-I3
I4=I2-1
I2-I4=1
1. You haven't identified where I3 is (Is it the current through R3?).Yes
2. There must be relevant equations! Otherwise, how would such problems get solved?
http://en.wikipedia.org/wiki/Ohm's_law
3. You need to provide some attempt at a solution, or at least some discussion of what you've already tried.
Rt=R1+R2+R34
Rt=10+13+2
Rt=25
I3=1
V3=I3R3
V3=1(6)
V3=6
I4=I2-I3
I4=I2-1
I2-I4=1
- #4
gneill
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Have you not learned about Kirchhoff's laws? KVL? KCL?Slashz said:
Note that R1 is in parallel with the voltage source (let's call it V) . If the voltage source is ideal then the potential with respect to ground at the node where R1 and R2 meet is fixed by V regardless of the value of R1. So we can expect that the value of R1 will not play a role in determining V.
It's not clear why you calculated Rt above, or what it represents, since R1 is not in series with R2 and R3||R4.
Since you've determined the voltage that appears across R3 above, what can you say about the current through R4?I3=1
V3=I3R3
V3=1(6)
V3=6 <--- Okay, good!
I4=I2-I3
I4=I2-1
I2-I4=1 <--- Okay! (Note that you've used KCL)
- #5
Curious3141
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The quickest way to do it is to use the voltage divider principle. But doing it with Kirchoff's laws is the most instructive way and teaches basic principles well.
FAQ: SOLVING[Series-Parallel Circuits
What are series-parallel circuits?
Series-parallel circuits are electrical circuits that contain both series and parallel components. In these circuits, some components are connected in a series, while others are connected in parallel.
What is the purpose of solving series-parallel circuits?
The purpose of solving series-parallel circuits is to determine the total resistance, current, and voltage of the circuit. This information is important in designing and analyzing electrical systems.
3. How do you solve a series-parallel circuit?
To solve a series-parallel circuit, you need to first identify the series and parallel components. Then, you can use Ohm's Law and Kirchhoff's Laws to calculate the total resistance, current, and voltage of the circuit.
4. What are the advantages of using series-parallel circuits?
Series-parallel circuits offer several advantages, including the ability to handle high currents and voltages, as well as the ability to create complex circuits with different components. They are also commonly used in household wiring and electronic devices.
5. What are the common challenges in solving series-parallel circuits?
One of the common challenges in solving series-parallel circuits is accurately identifying the series and parallel components. Another challenge is dealing with non-linear components, such as diodes or transistors, which require more complex equations to solve for their behavior in the circuit.
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