Question

Two capacitors of \( 100 \mu\text{F} \) and \( 50 \mu\text{F} \) are connected in parallel. If the potential difference across \( 100 \mu\text{F} \) is 20 V and across \( 50 \mu\text{F} \) is 40 V, then the common potential of the parallel combination will be (same polarities of the capacitor connected together)

• A. 20 V
• B. 60 V
• C. \( \frac{3}{80} \) V
• D. \( \frac{80}{3} \) V

Hint:

Charge is conserved: Total Charge $Q = Q_1 + Q_2$ before and after connection.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the final steady-state potential when two charged capacitors are connected in parallel.
Step 2: Key Formula or Approach:
Common Potential \( V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} \).
Step 3: Detailed Explanation:
Given: \( C_1 = 100 \mu\text{F}, V_1 = 20 \text{ V} \) and \( C_2 = 50 \mu\text{F}, V_2 = 40 \text{ V} \).
Common potential \( V \):
\[ V = \frac{(100 \times 20) + (50 \times 40)}{100 + 50} \] \[ V = \frac{2000 + 2000}{150} = \frac{4000}{150} = \frac{400}{15} \] \[ V = \frac{80}{3} \text{ V} \] Step 4: Final Answer:
The common potential is \( \frac{80}{3} \) V.