Question

Four capacitors $4\, \mu F , 6\, \mu F , 5\, \mu F , 3\, \mu F$ are connected in series to a source of $150 \,V$. The potential difference in volt, across the $5\, \mu F$ capacitor will be

• A. $30$ volt
• B. $40$ volt
• C. $50$ volt
• D. $60$ volt

Answer 1

Answer: (A)

Equivalent capacitance $\frac{1}{ C _{\text {eq }}}=\frac{1}{ C _{1}}+\frac{1}{ C _{2}}+\frac{1}{ C _{3}}+\frac{1}{ C _{4}}$
$\frac{1}{ C _{\text {eq }}}=\frac{1}{4}+\frac{1}{6}+\frac{1}{5}+\frac{1}{3}$
$\frac{1}{ C _{\text {eq }}}=\frac{15+10+12+20}{60}=\frac{57}{60}$
$C _{ eq }=1.05 \mu F$
$q = C _{ eq } V$
$\therefore q =1.05 \times 10^{-6} \times 150=157.5 \times 10^{-6}$
$q =1.5 \times 10^{-4} C$
$\therefore V =\frac{ q }{ C }=\frac{1.5 \times 10^{-4}}{5 \times 10^{-6}}$
$=0.3 \times 10^{2}$
$V =30$ volt