Question

Two capacitors having capacitances $8\mu F$ and $16\mu F$ have breaking voltages $20V$ and $80V$. They are combined in series. The maximum charge they can store individually in the combination is

• A. $160\mu C$
• B. $200\mu C$
• C. $1280\mu C$
• D. none of these

Answer 1

The correct option is A $160\mu C$

As they are combined in series so charge on both capacitor remains same.
The charge on the small capacitance $8\mu F$ is $Q_1=C_1{V}_1=8\times 20=160\mu C$
Here we can not apply charge on the capacitors greater than $Q_1=160\mu C$. Thus the maximum charge on both capacitors individually will be $160\mu C$.