Question

Two capacitors of $10PF$ and $20PF$ are connected to $200V$ and $100V$ 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)

Given, $C_1=10pF=10\times 10^{-12}F$
$C_2=20pF=20\times 10^{-12}F$
$V_1=200V,V_2=100V$
$C_1=$ 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
$V=\frac{q_1+q_2}{C_1+C_2}=\frac{V_1{C}_1+V_2{C}_2}{C_1+C_2}$
$=\frac{200\times 10\times 10^{-12}+100\times 20\times 10^{-12}}{10\times 10^{-12}+20\times 10^{-12}}$
$=\frac{200\times 10+100\times 20}{10+20}$
$=\frac{2000+2000}{30}=\frac{4000}{30}=133.3V$