Question

Two capacitors A and B of capacitance $6\mu F$ and $10\mu F$ respectively are connected in parallel and this combination is connected in series with a third capacitors C of $4\mu F$. A potential difference of 100 volt is applied across the entire combination. Find the charge and potential difference across $6\mu F$ capacitor.

• A. $120\mu C;20V.$
• B. $200\mu C;20V.$
• C. $320\mu C;80V.$
• D. $320\mu C;60V.$

Answer 1

The correct option is A $120\mu C;20V.$

Since A and B are connected and Parallel,

Hence,

Equivalence capacitance of capacitor$A$ and $B=6\mu F+10\mu F=16\mu F$

Now $16\mu F$ and capacitor C of $4\mu F$ are connected in series

Hence,

the equivalence capacitance $\left(EC\right)$ will be given by

$\frac{1}{EC}=\frac{1}{16}+\frac{1}{4}$

we get,

Equivalence capacitance $\left(EC\right)=\frac{16}{5}\mu F$

$Charge=EC\times P.D.$

$Charge=\frac{16}{5}\times 100$

$Charge\left(Q\right)=320\mu F$

Now,

$Q_A=320\times \frac{6}{16}\mu F$

$Q_A=120\mu F$

$Q_B=320\times \frac{10}{16}\mu F$

$Q_B=200\mu F$

Now proceeding for P.D. across each capacitor,

$V_A=\frac{Q_A}{C_A}$

$V_A=\frac{120}{6}=20V$ ,

$V_B=\frac{Q_B}{C_B}$

$V_B=\frac{200}{10}=20V$ ,

$V_C=\frac{Q_C}{C_C}$

$V_C=\frac{320}{4}=80V$

this is the required solution.