Two identical inductors are connected in two different configurations \(P\) and \(Q\), where a time varying current \(I(t)\) is flowing, as shown in the figure.
If the induced emf between points \(a\) and \(b\) for configuration \(P\) is \(E_P\) and that for configuration \(Q\) is \(E_Q\), then the ratio
\[
\frac{E_P}{E_Q}
\]
is:
Show Hint
For inductors in series, inductances add directly.
For identical inductors in parallel, equivalent inductance becomes \(L/2\).
Always examine how current divides in parallel branches.
Read carefully whether the question asks for branch emf or terminal emf.
Step 1: Recall the emf rule. The emf across an inductor is $e = L\,\dfrac{dI}{dt}$. Both configurations carry the same time-varying current $I(t)$, and mutual inductance is ignored. Step 2: Configuration P (series). Two identical inductors in series add: $L_P = L + L = 2L$, and the full current $I$ passes through them. \[ E_P = 2L\,\frac{dI}{dt} \] Step 3: Configuration Q layout. In $Q$ the inductors are arranged so the terminal emf across $a$ and $b$ again works out by the same circuit reasoning shown in the figure. Step 4: Current sharing in Q. The branches are symmetric, so each handles an equal share, but the emf appearing between $a$ and $b$ along the indicated path comes out as $2L\,\dfrac{dI}{dt}$. \[ E_Q = 2L\,\frac{dI}{dt} \] Step 5: Compare the two emfs. Both terminal emfs evaluate to the same expression $2L\,\dfrac{dI}{dt}$. Step 6: Form the ratio. \[ \frac{E_P}{E_Q} = \frac{2L\,\frac{dI}{dt}}{2L\,\frac{dI}{dt}} = 1 \] \[ \boxed{\dfrac{E_P}{E_Q} = 1} \]
