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A network consisting of a 1 ohm resistor, a 5 V cell and a 5 mH inductor (in series) is part of a complete circuit between points A and B. At a certain instant the current I = 5 A flows from A towards B and is decreasing at the rate dI/dt = 10³ A/s. Find V_B - V_A.

1. 20 V
2. 15 V
3. 10 V
4. 5 V

Correct answer: 15 V

Solution

Walking from A to B through the series elements: drop across the 1 ohm resistor = IR = 5 V, plus the 5 V cell, plus the inductor term L|dI/dt| = (5x10⁻³)(10³) = 5 V (the decreasing current makes the inductor behave so as to add 5 V). Summing: 5 + 5 + 5 = 15 V, giving V_B - V_A = 15 V.

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