Question

Three capacitors of capacitance 1 muF, 2muF and 3 muF are connected in series and a potential difference of 11 V is applied across the combination. Then, the potential difference across the plates of 1 muF capacitor is

Answer 1

Answer:

2V

Explanation:

Ceq=6/11 muF

q=CV=(2/11)×11=2muC

for series combination charge is same for all

so V1=q/C1=2/1=2V

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Answer 2

SoluTion :-

$\sf {Equivalent\ capacitance\ \frac{1}{C_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3}}\\\\\sf {\longmapsto C_{eq}=\frac{6}{11}\ \mu F }\\\\\\\sf {Charge\ supplied\ from\ battery\ Q=\frac{6}{11}\times 11 = 6\ \mu C }\\\\\\\sf {Hence\ potential\ difference\ across\ 2\mu F\ capacitor}\\\\\sf {=\frac{6}{1} = 6V }$