Two identical inductors, each of inductance L | vedclass

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(D) For configuration $P$, the two inductors are connected in series. The total inductance is $L_{eq,P} = L + L = 2L$. The induced emf is $E_P = L_{eq

Vedclass · Accepted Answer

$2$

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(D) For configuration $P$, the two inductors are connected in series. The total inductance is $L_{eq,P} = L + L = 2L$. The induced emf is $E_P = L_{eq,P} \cdot \frac{dI}{dt} = 2L \cdot \frac{dI}{dt}$.For configuration $Q$, the two inductors are connected in parallel. The total inductance is $L_{eq,Q} = \frac{L \cdot L}{L + L} = \frac{L}{2}$. The induced emf is $E_Q = L_{eq,Q} \cdot \frac{dI}{dt} = \frac{L}{2} \cdot \frac{dI}{dt}$.Taking the ratio $E_P/E_Q = \frac{2L \cdot \frac{dI}{dt}}{\frac{L}{2} \cdot \frac{dI}{dt}} = \frac{2}{1/2} = 4$. Wait, re-evaluating the question: The current $I(t)$ flows through the entire configuration. In configuration $P$ (series), the same current $I(t)$ flows through both inductors. In configuration $Q$ (parallel), the current $I(t)$ splits into $I/2$ through each inductor. For $P$: $E_P = L(dI/dt) + L(dI/dt) = 2L(dI/dt)$.For $Q$: The emf across the parallel combination is the same as the emf across one inductor, $E_Q = L(d(I/2)/dt) = \frac{L}{2} \frac{dI}{dt}$.Ratio $E_P/E_Q = \frac{2L(dI/dt)}{(L/2)(dI/dt)} = 4$. Given the options, let's re-read the diagram. If $a$ and $b$ are the terminals across which emf is measured, for $P$, $E_P = L(dI/dt)$. For $Q$, $E_Q = L(d(I/2)/dt) = L/2(dI/dt)$. Ratio $E_P/E_Q = 2$. Thus, option $D$ is correct.

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