Question

A capacitor $C_1=4\mu F$ is connected in series with another capacitor $C_2=1\mu F$. The combination is connected across $d.c$. source of $200V$. 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
$q=C_s\times V=\frac{4}{5}\times 200=160V$
$\therefore V_1\text{ (potential across }{R}_1)=\frac{160}{C_1}=\frac{160}{4}=40V$
and $V_2$ (potential across $R_2)=\frac{160}{1}=160V$
So, $\frac{V_2}{V_1}=\frac{160}{40}=\frac{4}{1}\Rightarrow V_2:V_1=4:1$