Question

Two uncharged capacitors of capacitance $8\mu F$ and $2\mu F$ are connected in series across a $100volt$ d.c. supply. Now if the supply voltage is removed and the capacitors are connected such that similar terminals are connected together. Then the final charge on $8\mu F$ capacitor is equal to (in $\mu C)$ :

• A. $64$
• B. $256$
• C. $192$
• D. $32$

Answer 1

The correct option is B $256$

We have,
Series capacitance, $\frac{1}{C_S}=\frac{1}{C_1}+\frac{1}{C_2}$ and Parallel capacitance, $C_P=C_1+C_2$
In the first case, the effective capacitance is given by,
$\frac{1}{C_S}=\frac{1}{8\mu }+\frac{1}{2\mu }$
$⟹C_S=1.6\mu F$
Also the charge on both the capacitors is same.
i.e, $q_1=q_2=Q=C_{S}V=1.6\mu \times 100=160\mu C$
Hence the total charge due to both the capacitors is $160\mu +160\mu =320\mu C$
In the second case, both the capacitors are connected such that similar terminals are connected together. Hence, we have two capacitors in parallel.
Now the potential is given by
$V^{\prime }=\frac{Q_{Total}}{C_P}$
We have $C_P=8\mu F+2\mu F=10\mu F$
Therefore,
$V^{\prime }=\frac{320\mu C}{10\mu F}=32V$
Hence the charge on $8\mu F$ capacitor is
$q^{\prime }=8\times 32=256\mu C$