A 3.5 mH inductor and a 4.5 mH inductor are connected in parallel.When the total emf of the combination is 16 V, the rate of change of the current in the larger inductor is: A)2.0 * 10³ A/s B)3.6 * 10³ A/s C)4.6*10³ A/s D)7.0 * 10³ A/s E)8.1 *10³ A/s

Question

Quizplus · Accepted Answer

The rate of change of current in an inductor is given by $V = L \frac{dI}{dt}$, where $V$ is the voltage across the inductor, $L$ is the inductance, and $\frac{dI}{dt}$ is the rate of change of current. For the larger inductor (4.5 mH), rearranging the formula gives $\frac{dI}{dt} = \frac{V}{L} = \frac{16}{4.5 \times 10^{-3}} = 3.56 \times 10^{3} A/s$, which is approximately $3.6 \times 10^{3} A/s$.

A 35 MH Inductor and a 4