Question

A capacitor of 10 \(\mu\text{F}\) charged up to 200 V is connected in parallel with another capacitor of 20 \(\mu\text{F}\) charged up to 50 V. The common potential is:

• A. \( 400 \text{ V} \)
• B. \( 300 \text{ V} \)
• C. \( 200 \text{ V} \)
• D. \( 100 \text{ V} \)

Hint:

The common potential is a weighted average of the initial potentials, weighted by the capacitance of each capacitor.

Answer 1

The Correct Option is D

Step 1: Charge is conserved.
Connecting two charged capacitors in parallel only lets charge move around; the total charge before equals the total after. They settle to one common voltage.
Step 2: Initial charge on the first capacitor.
$Q_1 = C_1V_1 = 10\,\mu\text{F} \times 200\,\text{V} = 2000\,\mu\text{C}$.
Step 3: Initial charge on the second.
$Q_2 = C_2V_2 = 20\,\mu\text{F} \times 50\,\text{V} = 1000\,\mu\text{C}$.
Step 4: Total charge.
$Q = Q_1 + Q_2 = 3000\,\mu\text{C}$.
Step 5: Total capacitance in parallel.
Parallel capacitances add: $C = C_1 + C_2 = 30\,\mu\text{F}$.
Step 6: Common potential.
\[ V = \frac{Q}{C} = \frac{3000\,\mu\text{C}}{30\,\mu\text{F}} = 100\,\text{V}. \]
\[ \boxed{100\ \text{V}} \]