Verified Solution • 2025–26 Session moderate

QPages 198-203: Q82: HC Verma: Concepts of Physics - Volume 2 Electric Current in Conductors Pages 198-203: Q82 Solution: Consider the situation shown in figure. The switch is closed at …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently treat each capacitor in series as if it charges through the resistor with its own time constant (e.g., $R C_1$ or $R C_2$), overlooking that the series pair behaves like a single equivalent capacitor $C_{eq}=\frac{C_1C_2}{C_1+C_2}$, so the circuit’s time constant is $\tau=R C_{eq}$. Remember that the charge on series capacitors is identical at every instant, and the correct charging law is $q_1(t)=C_1\varepsilon\bigl(1-e^{-t/\tau}\bigr)$ with $\tau$ based on $C_{eq}$, not on $C_1$ or $C_2$ individually.

Consider the situation shown in figure. The switch is closed at t = 0 when the capacitors are uncharged. Find the charge on the capacitor

C_1

as a function of time t.

Step-by-Step Verified Solution

**Answer:**

q_1(t) = C_1\\varepsilon(1 - e^{-t/\\tau})

. **Step-by-Step Explanation:**

C_1

and

C_2

in series with battery.

C_{eq} = \\frac{C_1C_2}{C_1+C_2}

.

\\tau = RC_{eq}

.

q_1(t) = C_1\\varepsilon(1 - e^{-t/\\tau})

[citation:2].

All Questions in this Chapter

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