Question

Following is a circuit containing an ideal battery of emf E in which the elements X & Y are unknown & may be a capacitor C, a resistor R, a battery of emf E, or may be their combination as mentioned in each case. The connecting wires are perfectly conducting.

The switch S is initially open & it is closed at t = 0.

In which case the charge drawn from the battery (s) during t = 0 to t = RC is the least?

• A. II, i, S
• B. I, iii, P
• C. III, ii, Q
• D. IV, i, Q

Answer 1

Answer: (C)

III, ii, Q

Answer 2

The question asks in which case the charge drawn from the battery during $t = 0$ to $t = RC$ is the least. We analyze each case in Column-1.

Case (I): Element X is R, Element Y is C. The circuit is an RC charging circuit. The current is $i(t) = \frac{E}{R} e^{-t/RC}$. The charge drawn from the battery from $t=0$ to $t=RC$ is $Q_I = \int_0^{RC} \frac{E}{R} e^{-t/RC} dt = \frac{E}{R} [-RC e^{-t/RC}]_0^{RC} = \frac{E}{R} (-RC e^{-1} - (-RC e^0)) = CE(1 - e^{-1}) \approx 0.632 CE$.

Case (II): Both elements X & Y are R each. The circuit is a simple resistive circuit with total resistance $2R$. The current is constant $i(t) = \frac{E}{2R}$. The charge drawn from the battery from $t=0$ to $t=RC$ is $Q_{II} = \int_0^{RC} \frac{E}{2R} dt = \frac{E}{2R} [t]_0^{RC} = \frac{E}{2R} (RC) = \frac{EC}{2} = 0.5 CE$.

Case (III): Element X is R, element Y is a parallel combination of R and C. The circuit has X=R in series with a parallel combination of R and C. The current from the battery is $i(t) = \frac{E}{2R}(1 + e^{-2t/RC})$. The charge drawn from the battery from $t=0$ to $t=RC$ is $Q_{III} = \int_0^{RC} \frac{E}{2R}(1 + e^{-2t/RC}) dt = \frac{E}{2R} [t - \frac{RC}{2} e^{-2t/RC} \frac{1}{-2/RC}]_0^{RC} = \frac{E}{2R} [t + \frac{RC}{2} e^{-2t/RC}]_0^{RC} = \frac{E}{2R} [(RC + \frac{RC}{2} e^{-2}) - (0 + \frac{RC}{2} e^{0})] = \frac{E}{2R} [RC + \frac{RC}{2} e^{-2} - \frac{RC}{2}] = \frac{E}{2R} [\frac{RC}{2} + \frac{RC}{2} e^{-2}] = \frac{EC}{4}(1 + e^{-2}) \approx \frac{EC}{4}(1 + 0.135) = 0.28375 CE$.

Case (IV): Element X is R element Y is battery E joined with same polarity. The circuit has X=R in series with two batteries of emf E connected in series with the same polarity. The total emf is $2E$. The total resistance is R. The current is constant $i(t) = \frac{2E}{R}$. The charge drawn from the battery from $t=0$ to $t=RC$ is $Q_{IV} = \int_0^{RC} \frac{2E}{R} dt = \frac{2E}{R} [t]_0^{RC} = \frac{2E}{R} (RC) = 2EC$.

Comparing the calculated charges: $Q_I \approx 0.632 CE$ $Q_{II} = 0.5 CE$ $Q_{III} \approx 0.28375 CE$ $Q_{IV} = 2 CE$ The least charge drawn is in Case (III).

The question asks for the option corresponding to the case with the least charge drawn. The options are given as combinations of entries from Column-1, Column-2, and Column-3. We found that Case (III) has the least charge drawn. The options are:

• II, i, S
• I, iii, P
• III, ii, Q
• IV, i, Q The option corresponding to Case (III) is "III, ii, Q".

Let's briefly check the validity of the statements in option "III, ii, Q". Case (III): Element X is R, element Y is a parallel combination of R and C. Statement (ii): Total charge drawn from the battery(s) after long time is $\frac{CE}{2}$ (Q). In Case (III), the current is $i(t) = \frac{E}{2R}(1 + e^{-2t/RC})$. As $t \to \infty$, $i(t) \to \frac{E}{2R}$. The total charge drawn from the battery after a long time is $Q(\infty) = \int_0^\infty i(t) dt = \int_0^\infty \frac{E}{2R}(1 + e^{-2t/RC}) dt$. This integral diverges to infinity because the current approaches a non-zero constant value. So statement (ii) is incorrect for Case (III).

However, since the question asks "In which case the charge drawn from the battery (s) during t = 0 to t = RC is the least?", and our calculation clearly shows that Case (III) has the least charge drawn among the four cases, the answer should correspond to Case (III). The options provided are in the format of matching entries from the columns. Given that the option "III, ii, Q" is provided, it is likely that the question intends for this option to be the correct answer based on some intended matching, even if the statement (ii) is not correct for Case (III) as stated. Assuming the question is primarily about identifying the case with the least charge drawn from $t=0$ to $t=RC$ and then finding the corresponding option, Case (III) is the correct case. The option starting with III is the third option.

Let's re-examine statement (ii). Perhaps it refers to the charge stored on the capacitor after a long time. In Case (III), the capacitor is in parallel with a resistor R. The voltage across the parallel combination is $V_Y(t) = \frac{E}{2}(1 - e^{-2t/RC})$. After a long time, $V_Y(\infty) = \frac{E}{2}$. The charge on the capacitor after a long time is $q_C(\infty) = C V_Y(\infty) = C \frac{E}{2} = \frac{CE}{2}$. So, statement (ii) is the charge stored on the capacitor after a long time, not the total charge drawn from the battery. If statement (ii) refers to the charge stored on the capacitor, then it is correct for Case (III).

Let's assume the question implicitly asks for the option that correctly matches the case with the least charge drawn from $t=0$ to $t=RC$ and the corresponding statements from Column-2 and Column-3, even if the statements themselves might be part of a separate matching problem. Based on our calculation, Case (III) has the least charge drawn during $t=0$ to $t=RC$. The option corresponding to Case (III) is "III, ii, Q". We have verified that statement (ii) is the charge stored on the capacitor after a long time in Case (III). The figure in Column-3 is not labeled with letters P, Q, R, S, so we cannot check which figure corresponds to Q. However, based on the structure of the options, it seems the question is asking to select the option that starts with the case having the least charge drawn from $t=0$ to $t=RC$.