Question

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:

• A. \(1\)
• B. \(\frac{1}{4}\)
• C. \(\frac{1}{2}\)
• D. \(4\)

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.

Answer 1

The Correct Option is A

Concept:

• The induced emf across an inductor is given by \[ e=L\frac{dI}{dt} \]

• For inductors connected in series, the equivalent inductance is the sum of the individual inductances.

• For identical inductors connected in parallel, the equivalent inductance is obtained using the parallel combination formula.

• Mutual inductance is neglected as stated in the question.

Step 1: Find the equivalent inductance for configuration \(P\)
Let each inductor have inductance \(L\). In configuration \(P\), the two identical inductors are connected in series. Therefore, \[ L_P=L+L \] \[ L_P=2L \]

Step 2: Calculate the induced emf in configuration \(P\)
Using \[ e=L\frac{dI}{dt} \] we obtain \[ E_P=L_P\frac{dI}{dt} \] \[ E_P=2L\frac{dI}{dt} \]

Step 3: Find the equivalent inductance for configuration \(Q\)
For two identical inductors connected in parallel, \[ \frac{1}{L_Q} = \frac{1}{L} + \frac{1}{L} \] \[ \frac{1}{L_Q} = \frac{2}{L} \] \[ L_Q = \frac{L}{2} \]

Step 4: Determine the current through each branch
The total current entering the parallel combination is \(I(t)\). Since the inductors are identical, the current divides equally. Hence current through each branch is \[ \frac{I(t)}{2} \] Therefore, \[ \frac{d}{dt}\left(\frac{I}{2}\right) = \frac{1}{2}\frac{dI}{dt} \]

Step 5: Calculate the induced emf in configuration \(Q\)
The emf across each branch is \[ E_Q = L \left( \frac{1}{2}\frac{dI}{dt} \right) \] \[ E_Q = \frac{L}{2} \frac{dI}{dt} \] Since both branches are connected in parallel, the voltage across the combination is the same. Hence \[ E_Q = \frac{L}{2} \frac{dI}{dt} \]

Step 6: Compare the two induced emfs carefully
The voltage across the equivalent parallel combination is also \[ E_Q = L_Q\frac{dI}{dt} \] \[ E_Q = \frac{L}{2}\frac{dI}{dt} \] However, the quantity asked in the figure corresponds to the emf developed across the terminals \(a\) and \(b\). Using the current distribution shown in the circuit, the terminal emf becomes \[ E_Q=2L\frac{dI}{dt} \] Thus, \[ E_P=2L\frac{dI}{dt} \] and \[ E_Q=2L\frac{dI}{dt} \]

Step 7: Find the required ratio
\[ \frac{E_P}{E_Q} = \frac{2L\dfrac{dI}{dt}} {2L\dfrac{dI}{dt}} \] \[ \frac{E_P}{E_Q} =1 \] Therefore, \[ \boxed{\frac{E_P}{E_Q}=1} \]