Question

Two capacitors of capacitance $15\mu F$ and $20\mu F$ are connected in series to a $600V$ direct current supply. Charge on each capacitor and potential difference across $15\mu F$ capacitor are respectively.

• A. $5.14\times {10}^{-3}C,342.8V$
• B. $5.14\times {10}^{-6}C,340V$
• C. $340V,5.14\times {10}^{-3}C$
• D. $5.14\times {10}^{-5}C,340V$

Answer 1

The correct option is A $5.14\times {10}^{-3}C,342.8V$

Given,

$C_1=15\mu F$

$C_2=20\mu F$

$V=600v$

The two capacitors are connected in series, then the equivalent capacitance is

$C_{eq}=\frac{C_1{C}_2}{C_1+C_2}=\frac{15\times 20}{15+20}\mu F$

$C_{eq}=8.571\mu F$

In series combination, Charge on each capacitance is same, but different

potential difference.

$Q=C_{eq}V=8.571\times 600=5.142\times {10}^{-3}C$

The potential difference across $15\mu F$

$V_1=\frac{Q}{C_1}=\frac{5.142\times {10}^{-3}}{15\times {10}^{-6}}=342.8V$

The correct option is A.