Join thousands of students who trust us to help them ace their exams!
Multiple Choice
A 30mH inductor carries a current of 500mA. If the circuit is opened and the current stops over a period of 3.0μs, what potential develops across the inductor?
A
100V
B
380V
C
1500V
D
3400V
E
7800V
F
5000V
0 Comments
Verified step by step guidance
1
Identify the given values: Inductance \( L = 30 \text{ mH} = 30 \times 10^{-3} \text{ H} \), initial current \( I = 500 \text{ mA} = 0.5 \text{ A} \), and the time interval \( \Delta t = 3.0 \text{ } \mu\text{s} = 3.0 \times 10^{-6} \text{ s} \).
Recall the formula for the induced electromotive force (emf) across an inductor, which is given by \( \varepsilon = -L \frac{\Delta I}{\Delta t} \), where \( \Delta I \) is the change in current and \( \Delta t \) is the time over which the change occurs.
Determine the change in current \( \Delta I \). Since the current stops, \( \Delta I = I_{\text{final}} - I_{\text{initial}} = 0 - 0.5 \text{ A} = -0.5 \text{ A} \).
Substitute the known values into the formula: \( \varepsilon = - (30 \times 10^{-3} \text{ H}) \frac{-0.5 \text{ A}}{3.0 \times 10^{-6} \text{ s}} \).
Simplify the expression to find the magnitude of the potential difference across the inductor. Remember that the negative sign indicates the direction of the induced emf opposes the change in current, according to Lenz's Law.