Question

Three capacitors \(C_1=1\mu F\), \(C_2=2\mu F\) and \(C_3=3\mu F\) are connected in series across a \(10V\) battery. Find the potential difference across \(C_2\).

• A. \(2.73\,V\)
• B. \(3.33\,V\)
• C. \(10\,V\)
• D. \(5\,V\)

Hint:

In a series capacitor combination, charge remains identical on every capacitor while voltage divides inversely with capacitance.

Answer 1

The Correct Option is A

Concept: In series combination of capacitors: \[ \frac{1}{C_{eq}} = \frac{1}{C_1} +\frac{1}{C_2} +\frac{1}{C_3} \] and charge remains the same on every capacitor.

Step 1: Calculate equivalent capacitance. \[ \frac{1}{C_{eq}} = 1+\frac12+\frac13 \] \[ = \frac{11}{6} \] \[ C_{eq} = \frac{6}{11}\mu F \]

Step 2: Calculate common charge. \[ Q=C_{eq}V \] \[ Q= \frac{6}{11}\times10 \] \[ Q=\frac{60}{11}\mu C \]

Step 3: Potential across \(C_2\). \[ V_2=\frac{Q}{C_2} \] \[ V_2= \frac{60/11}{2} \] \[ V_2=\frac{30}{11} \] \[ V_2=2.73V \] Final Answer: \[ \boxed{2.73V} \] Hence option \[ \boxed{(A)} \]