Question

In the circuit shown

• A. The charge on C₂ is greater than that of C₁
• B. The charge on C₂ is smaller than that of C₁
• C. The potential drop across C₁ is smaller than C₂
• D. The potential drop across C₁ is greater than C₂

Answer 1

Answer: (D)

The potential drop across C₁ is greater than C₂

Answer 2

Given circuit can be redrawn as follows

$C_{eq} = \frac{4 \times 8}{12} = \frac{8}{3}\mu F$

So $Q = \frac{8}{3} \times 6 = 16\mu C$

Hence potential difference $V_{1} = \frac{16}{4} = 4volt$ and

$V_{2} = \frac{16}{8} = 2volt$ i.e. V₁ > V₂