Question

In the given circuit C₁ = C, C₂ = 2C, V₁ = V₂ = V & R₁ = R₂ = R₃ = R and all switches-are closed at t = 0 then choose the incorrect statement at time t = RC (initially all capacitors are uncharged)

• A. Current through V₁ is less than $\frac{V}{2Re}[e-1]$
• B. Power delivered by V₂ is less than $\frac{V^2}{3eR}$
• C. Current through V₁ is greater than $\frac{V}{2Re}[e-1]$

Answer 1

Answer: (C)

Incorrect statements are A and B.

Answer 2

We will show that if we view the circuit as a single‐loop network so that the two sources “assist” each other (i.e. their EMFs add up to 2V), then the three resistors R₁, R₂ and R₃ (all equal to R) lie in series and the two capacitors (of capacitances C and 2C in series) have an equivalent capacitance

$C_{eq} = \frac{C \cdot 2C}{C+2C} = \frac{2C}{3}$.

Thus the overall time‐constant is

$\tau = (R₁ + R₂ + R₃) \cdot C_{eq} = (3R) \cdot \frac{2C}{3} = 2RC$.

For a standard RC–circuit the current (if the net emf is 2V) is $I(t) = \frac{2V}{3R} \cdot e^{\frac{-t}{\tau}}$.

At t = RC we have $I(RC) = \frac{2V}{3R} \cdot e^{\frac{-RC}{2RC}} = \frac{2V}{3R} \cdot e^{\frac{-1}{2}}$.

Numerically, since $e^{\frac{-1}{2}} \approx 0.6065$, $I(RC) \approx \frac{2V}{3R} \cdot 0.6065 \approx 0.4043 \cdot \frac{V}{R}$.

Now notice that the “benchmark” given in options A and C is

$\frac{V}{2eR} \cdot (e - 1)$.

Since e ≈ 2.718, e – 1 ≈ 1.718 so that $\frac{V}{2eR} \cdot (e - 1) \approx \frac{V}{5.436R} \cdot 1.718 \approx 0.316 \cdot \frac{V}{R}$.

Thus we have

$I(RC) \approx 0.4043 \frac{V}{R} > 0.316 \frac{V}{R}$.

In other words, the current coming from V₁ is greater than the given benchmark value. Therefore:

• Statement A (“Current through V₁ is less than …”) is false.
• Statement C (“Current through V₁ is greater than …”) is correct.

Next, the power supplied by source V₂ (which—since the circuit is in series—has the same current I flowing through it) is

$P₂ = V \cdot I(RC) \approx V \cdot (0.4043 \cdot \frac{V}{R}) = 0.4043 \cdot \frac{V^2}{R}$.

But the option B claims that $P₂ < \frac{V^2}{3eR}$.

Since 3e ≈ 8.154, we have $\frac{V^2}{3eR} \approx 0.1226 \cdot \frac{V^2}{R}$. Clearly $0.4043 \cdot \frac{V^2}{R}$ is not less than $0.1226 \cdot \frac{V^2}{R}$. Thus statement B is also false.

So, the incorrect statements (the “wrong” ones) are A and B.

Summary of Core Steps:

1. The loop has three resistors (3R) and the series combination of C and 2C gives $C_{eq} = \frac{2C}{3}$ so that τ = 2RC.
2. With net emf 2V, the transient current is $I(t) = \frac{2V}{3R} \cdot e^{\frac{-t}{2RC}}$. At t = RC, $I \approx 0.4043 \cdot \frac{V}{R}$.
3. The given benchmark current is $\frac{V}{2eR}(e – 1) \approx 0.316 \cdot \frac{V}{R}$. Thus statement A is false and C true.
4. Also, power from V₂ is $V \cdot I \approx 0.4043 \cdot \frac{V^2}{R}$ which is not less than $\frac{V^2}{3eR} \approx 0.1226 \frac{V^2}{R}$. So B is false.