Question

A capacitor of capacitance $C_1=1.0\mu F$ can with stand a maximum voltage $V_1=6.0kV$. Anothercapacitor of capacitance $C_2=2.0\mu F$ can withstand a maximum voltage $V_2=4.0kV.$ If the capacitors are connected in series, the combination can withstand a maximum voltage of

• A. 10 kV
  
• B. 9 kV
  
• C. 8 kV
  
• D. 6 kV

Answer 1

The correct option is B

9 kV

The maximum charge the first capacitor can hold is
$Q_1=C_1{V}_1=1\times {10}^{-6}\times 6000=6\times {10}^{-3}C$
The maximum charge the second capacitor can hold is
$Q_2=C_2{V}_2=2\times {10}^{-6}\times 4000=8\times {10}^{-3}C$
We know that in a series combination, the charge on each capacitor is the same. Now the first capacitor cannot hold a charge of $8\times {10}^{-3}C$; it can hold a maximum charge of $6\times {10}^{-3}C$. Therefore, the charge on the second capacitor must also be $6\times {10}^{-3}C$. Hence, the voltage across the second capacitor is
$V_2=\frac{6\times {10}^{-3}C}{2\times {10}^{-6}F}=3000volts=3kilovolts$
Thus, the maximum voltage the system can with-stand= $V_1+V_2=6kilovolts+3kilovolts=9kilovolts.$ Hence the correct choice is (b).