A series combination of three capacitors of capacities 1 μ F , 2 μ F and 8 μ F is connected to a battery of e.m.f. 13 volt. The potential difference across the plates of 2 μ F capacitor will be

Question

A series combination of three capacitors of capacities 1 μ F , 2 μ F and 8 μ F is connected to a battery of e.m.f. 13 volt. The potential difference across the plates of 2 μ F capacitor will be (A) 1 V (B) 8 V (C) 4 V (D) (13/3) V

Tardigrade Admin · Accepted Answer

(1/Ce q)=(1/1)+(1/2)+(1/8) ⇒ Ce q=(8/13) μ F Total charge, Q=Ce q V=(8/13) × 13=8 μ C Potential difference across 2 μ F capacitor =(8/2)=4 V

Q. A series combination of three capacitors of capacities $1 μ F, 2 μ F$ and $8 μ F$ is connected to a battery of e.m.f. $13$ volt. The potential difference across the plates of $2 μ F$ capacitor will be

1 V

8 V

4 V

$\frac{13}{3} V$

$\frac{1}{C _{e q}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{8}$
$\Rightarrow C_{e q} = \frac{8}{13} μ F$
Total charge, $Q = C_{e q} V = \frac{8}{13} \times 13 = 8 μ C$
Potential difference across $2 μ F$ capacitor
$= \frac{8}{2} = 4 V$

Q. A series combination of three capacitors of capacities 1μF,2μF and 8μF is connected to a battery of e.m.f. 13 volt. The potential difference across the plates of 2μF capacitor will be

1 V

8 V

4 V

313​V

Ceq​1​=11​+21​+81​ ⇒Ceq​=138​μF Total charge, Q=Ceq​V=138​×13=8μC Potential difference across 2μF capacitor =28​=4V

Q. A series combination of three capacitors of capacities $1 μ F, 2 μ F$ and $8 μ F$ is connected to a battery of e.m.f. $13$ volt. The potential difference across the plates of $2 μ F$ capacitor will be

$\frac{13}{3} V$

$\frac{1}{C _{e q}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{8}$
$\Rightarrow C_{e q} = \frac{8}{13} μ F$
Total charge, $Q = C_{e q} V = \frac{8}{13} \times 13 = 8 μ C$
Potential difference across $2 μ F$ capacitor
$= \frac{8}{2} = 4 V$