First slide

Capacitance

Question

When two capacitors one of 3 $μ$ F and the other of 6 $μ$ F arc connected in series and the combination is charged to a potential difference of 120 volt, the potential difference across 3 $μ$ F capacitor is

Easy

A

40 V
B

60 V
C

80 V
D

180 V

Solution

$V_{1} + V_{2} = 120 volt and \frac{V_{1}}{V_{2}} = \frac{C_{2}}{C_{1}} = 2 ∴ V_{1} = 2 V_{2}$
Now $2 V_{3} + V_{2} = 120 or V_{2} = 40 volt$
and $V_{1} = 2 \times 40 = 80 volt .$

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