Solveeit Logo

Question

Physics Question on Combination of capacitors

Three capacitors 3μF,6μF3\, \mu F , 6 \, \mu F and 6μF6\, \mu F are connected in series to a source of 120V120 \, V. The potential difference, in volt, across the 3μF3 \, \mu F capacitor will be

A

24

B

30

C

40

D

60

Answer

60

Explanation

Solution

The combination of three charges in series 1C=1C1+1C2+1C3\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}} 1C=13+16+16\frac{1}{C}=\frac{1}{3}+\frac{1}{6}+\frac{1}{6} C=64=1.5μF\Rightarrow C=\frac{6}{4}=1.5\, \mu F The charge of this circuit q=CV=1.5×120q=C V=1.5 \times 120 q=180μCq=180\, \mu C The potential difference across the 3μF3\, \mu F q=CVq=C \,V V=qC=1803=60VV=\frac{q}{C}=\frac{180}{3}=60\, V