KCET · Physics · Capacitance
A capacitor of capacitance \( C \) charged by an amount \( Q \) is connected in parallel with an
uncharged capacitor of capacitance 2C. The final charges on the capacitors are
- A \( \frac{Q}{3}, \frac{2 Q}{3} \)
- B \( \frac{Q}{2}, \frac{Q}{2} \)
- C \( \frac{Q}{5}, \frac{4 Q}{5} \)
- D \( \frac{Q}{4}, \frac{3 Q}{4} \)
Answer & Solution
Correct Answer
(A) \( \frac{Q}{3}, \frac{2 Q}{3} \)
Step-by-step Solution
Detailed explanation
(A)
The two capacitors attain common potential \( \left(V_{C}\right) \) given by the relation
\( V_{c}=\frac{\text { Total charge }}{\text { Total Capacitance }}=\frac{Q+0}{C+2 C}=\frac{Q}{3 C} \)
The final charges on two capacitors are
\( Q_{1}=C V_{c}=\frac{C Q}{3 C}=\frac{Q}{3} \) and \( Q_{2}=2 C V_{c}=\frac{2 Q}{3} \)
The two capacitors attain common potential \( \left(V_{C}\right) \) given by the relation
\( V_{c}=\frac{\text { Total charge }}{\text { Total Capacitance }}=\frac{Q+0}{C+2 C}=\frac{Q}{3 C} \)
The final charges on two capacitors are
\( Q_{1}=C V_{c}=\frac{C Q}{3 C}=\frac{Q}{3} \) and \( Q_{2}=2 C V_{c}=\frac{2 Q}{3} \)
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