An electricity and magnetism problem by Manmeswar Patnaik
Electricity and Magnetism
Level
2
The current through 60 ohm resistor in the following network is
The answer is 0.04.
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Taking the 40-ohm resistor and the shorted wire in parallel on the far right, we obtain an equivalent resistance of:
R_eq = (0*40) / (0 + 40) = 0-ohms
which is just an ordinary wire in this circuit. Now taking two clockwise currents in each circuit loop, call them I1 (left) and I2 (right) respectively, we now just apply Kirchhoff's Voltage Law in each branch and obtain the following system of equations in I1 and I2:
30 I1 + 60 (I1 - I2) + 30*I1 = 6 (i)
70 I2 + 50 I2 + 60*(I2 - I1) = 0 (ii)
which yields I1 = 3/50 A and I2 = 1/50 A. The current through the central 60-ohm resistor is just |I1 - I2| = |3/50 - 1/50| = |1/25| = 0.04 A.