If in a series L-C-R ac circuit, the voltages across R, L, C are $V_{1}, V_{2}, V_{3}$ respectively. Then the voltage of applied AC source is always equal to :

V_{1} + V_{2} + V_{3}

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\sqrt{V_{1}^{2} + (V_{2} + V_{3})^{2}}

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V_{1} - V_{2} - V_{3}

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\sqrt{V_{1}^{2} + (V_{2} - V_{3})^{2}}

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Solution

The correct option is B $\sqrt{V_{1}^{2} + (V_{2} - V_{3})^{2}}$
$V_{2}$ and $V_{3}$ will be $180^{0}$ out of phase with each other.

Hence, resultant is $V_{2} - V_{3}$ .
Each of $V_{2}$ and $V_{3}$ is having a phase diff of $90^{0}$ with $V_{1}$ .
Hence, resultant AC voltage which is equal to the applied AC voltage is $\sqrt{V_{1}^{2} + (V_{2} - V_{3})^{2}}$ .

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