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Standard XII

Physics

Alternating Current Applied to an Inductor

A 50Hz AC...

Question

A $50 H z$ $A C$ source of $20 V$ is connected across $R$ and $C$ as shown in figure. The voltage across $R$ is $12 V$ . The voltage across $C$ is

8 V

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16 V

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10 V

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Not possible to determine unless value of

R

and

C

are given.

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Solution

The correct option is B $16 V$
$V_{C} = \sqrt{V^{2} - V_{R}^{2}} = \sqrt{(20)^{2} - (12)^{2}}$
$= 16 V$

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A 50Hz AC source of 20V is connected across R and C as shown in figure. The voltage across R is 12V. The voltage across C is

The correct option is B 16VVC=√V2−V2R=√(20)2−(12)2=16V

A $50 H z$ $A C$ source of $20 V$ is connected across $R$ and $C$ as shown in figure. The voltage across $R$ is $12 V$ . The voltage across $C$ is

The correct option is B $16 V$
$V_{C} = \sqrt{V^{2} - V_{R}^{2}} = \sqrt{(20)^{2} - (12)^{2}}$
$= 16 V$