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What is the current through each resistor, using Kirchhoff's rules?
Consider the circuit shown in the figure . Suppose the four resistors in this circuit have the values R1 = 11 ohms, R2 = 6.9 ohms, R3 = 6.2 ohms, and R4 = 13 ohms, and that the emf of the battery is E = 15 V.
http://session.masteringphysics.com/problemAsset/1...
I have already found the current for each resistor using the rules of parallel/series. They were 0.95, 0.23, 0.73, 0.23, respectively. How do I find the currents through each resistor using Kirchhoff's rules?
2 Answers
- Chandler BingLv 610 years agoFavorite Answer
Use KVL for the two meshes:
1) -E + R3 ( I1-I2) + R1I1 = 0
2) R2I2 + R4I2 + R3 (I2-I1) = 0
Solve the these two equations simultaneously.
17.2 I1 -6.2 I2 = 15
13.7 I1 + 26.1 I2= 0
- Anonymous5 years ago
Kirchhoff's rules basically state conservation of energy (potential) and charge. Consider the interior loop. Let V be the potential across the battery, let I1 be the current coming out of the battery and let I2 and I3 be the current in the left and right current branches respectively. At the junction, the current in equals the current out, so: I1=I2+I3 Now consider the inner loop, starting in the lower left corner (point C). Starting at a point and returning to it means that there is no change in potential. The potential gained over the battery is V, and the potential dropped over the resistors is I1*R3 and I1*R1, and the so: V-I1*R3-I1*R1=0 V-I1*(R3+R1)=0 Considering the outer loop: V-I3*(R2+R4)-I1*R1=0 Solve this system of 3 simultaneous linear equations for I1, I2 and I3.
