Question

A capacitor $C_1 = 4\,μF$ is connected in series with another capacitor $C_2 = 1\,μF$. The combination is connected across $d.c$. source of $200 \,V$. The ratio of potential across $C_2$ to $C_1$ is

• A. $2 : 1$
• B. $4 : 1$
• C. $8 : 1$
• D. $16 : 1$

Answer 1

Answer: (B)

In series combination, $\frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{4}+\frac{1}{1}=\frac{5}{4}$
$ \therefore \,\,C_{s}=\frac{4}{5} \mu F$
Now, total charge in series combination is given by
$q=C_{s} \times V=\frac{4}{5} \times 200=160 \,V $
$\therefore V_{1} \text { (potential across } \left.R_{1}\right)=\frac{160}{C_{1}}=\frac{160}{4}=40 \,V$
and $V_{2}$ (potential across $\left.R_{2}\right)=\frac{160}{1}=160\, V$
So, $\frac{V_{2}}{V_{1}}=\frac{160}{40}=\frac{4}{1} \Rightarrow V_{2}: V_{1}=4: 1$