Question

Two capacitors of $10\, PF$ and $20\, PF$ are connected to $200 \,V$ and $100\, V$ sources respectively. If they are connected by the wire, what is the common potential of the capacitors ?

• A. 133.3 volt
• B. 150 volt
• C. 300 volt
• D. 400 volt

Answer 1

Answer: (A)

133.3 volt

Answer 2

Given, $C_{1}=10 pF =10 \times 10^{-12} F$ $C_{2}=20 pF =20 \times 10^{-12} F$ $V_{1}=200\, V ,\, V_{2}=100\, V$ $C_{1}=$ Capacitance of Ist capacitor $C _{2}=$ Capacitance of IInd capacitor $V_{1}=$ Voltage across Ist capacitor $V_{2}=$ Voltage across IInd capacitor We know that $V_{1}=\frac{q_{1}}{C_{1}}$ and $V_{2}=\frac{q_{2}}{C_{2}}$ $\Rightarrow q_{1} =V_{1} C_{1}$...(i) $q_{2} = V_{2} C_{2}$...(ii) So, common potential of capacitors $V=\frac{q_{1}+q_{2}}{C_{1}+C_{2}}=\frac{V_{1} C_{1}+V_{2} C_{2}}{C_{1}+C_{2}}$ $=\frac{200 \times 10 \times 10^{-12}+100 \times 20 \times 10^{-12}}{10 \times 10^{-12}+20 \times 10^{-12}}$ $=\frac{200 \times 10+100 \times 20}{10+20}$ $=\frac{2000+2000}{30}=\frac{4000}{30}=133.3\, V$