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 DC source 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)

4 : 1

Given: $C_1=4\mu F$

$C_2=1\mu F$

$V=200$ volts

Equivalent capacitance of the series combination, $C_{eq}=\frac{C_1\times C_2}{C_1+C_2}$

$\Rightarrow C_{eq}=\frac{4\times 1}{4+1}=0.8\mu F$

Charge flowing through the circuit, $q=C_{eq}V=0.8\mu F\times 200$ volts $=160\mu C$

Potential across $C_1$: $V_1=\frac{q}{C_1}=\frac{160\mu C}{4\mu C}=40$ volts

Potential across $C_2$: $V_2=\frac{q}{C_2}=\frac{160\mu C}{1\mu C}=160$ volts

$⟹$ $V_2:V_1=4:1$