Question

The circuit below has an ideal inductor and a switch as shown in figure. The switch can be closed for a long time and then can be opened at t = 0 or it can be closed at t = 0. List-I gives the different circuit elements and list-II gives the value.

	List-I		List-II
(I)	Current in inductor	(P)	$\frac{6V}{7}$
(II)	Voltage across inductor	(Q)	$\frac{3V}{2}$
(III)	Current supplied by battery	(R)	$\frac{V}{4R}$
(IV)	Energy stored in the inductor	(S)	$\frac{V}{R}$
		(T)	Zero
		(U)	$\frac{LV^2}{32R^2}$

If the switch was closed at t = 0 then which of the following options are correct match of list I and II for that instant?

• A. $\frac{6V}{7}$
• B. $\frac{3V}{2}$
• C. $\frac{V}{4R}$
• D. $\frac{V}{R}$
• E. Zero
• F. $\frac{LV^2}{32R^2}$

Answer 1

Answer: (E)

The correct matches are (I)-T, (II)-P, (IV)-T. The value for (III) is not in List-II.

Answer 2

Here's a breakdown of the solution:

Understanding the Circuit at t = 0+

When the switch is closed at t = 0, we analyze the circuit's behavior immediately after the switch closure (t = 0+). Crucially, the current through an inductor cannot change instantaneously.

Step-by-step Analysis

1. (I) Current in Inductor:

   • At t = 0+, the inductor acts as an open circuit because the current through it was zero before the switch closed.
   • Therefore, the current in the inductor, $I_L(0^+) = 0$. This matches option (T).

2. (II) Voltage across Inductor:

   • The voltage across the inductor at t = 0+ can be found by analyzing the equivalent circuit. The inductor branch (L and 3R) is in parallel with the 3R resistor (2R + R). This parallel combination is in series with the R/2 resistor, and the battery V.
   • The current supplied by the battery is $I = \frac{V}{R/2 + 3R} = \frac{2V}{7R}$. This current flows through the 3R resistor.
   • The voltage across the 3R resistor (and therefore across the inductor branch) is $V_{3R} = \frac{2V}{7R} \cdot 3R = \frac{6V}{7}$. Since the inductor current is zero at t=0+, the voltage across the inductor is simply the voltage across the parallel branch, which is $\frac{6V}{7}$. This matches option (P).

3. (III) Current Supplied by Battery:

   • As calculated above, the current supplied by the battery at t = 0+ is $\frac{2V}{7R}$. However, this value is not present in List-II.

4. (IV) Energy Stored in the Inductor:

   • The energy stored in an inductor is given by $E = \frac{1}{2}LI^2$.
   • Since the current through the inductor at t = 0+ is zero, the energy stored in the inductor is also zero. This matches option (T).

Conclusion

Based on the analysis, the correct matches are (I)-T, (II)-P, and (IV)-T. The value for (III) is calculated to be $\frac{2V}{7R}$, which is not found in List-II. Therefore, we can only definitively match (I), (II), and (IV).