In series LCR circuit voltage drop across resistance is 8V, across inductor is 6V and across capacitor is 12V. then

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Voltage of the source will be leading current in the circuit

None of these

Power factor of circuit will be 4/3

Voltage drop across each element will be less than the applied voltage

answer is D.

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Detailed Solution

Since $\cos θ = \frac{R}{Z} = \frac{I R}{I Z} = \frac{8}{10} = \frac{4}{5}$

(also, $\cos θ$ can never be greater than 1)

Hence (c) is wrong.

Also $I X_{C} > I X_{L} \Rightarrow X_{C} > X_{L}$

Current will be leading.

In an LCR circuit,

$v = \sqrt{{(v_{L} - v_{C})}^{2} + v_{R}^{2}} = \sqrt{{(6 - 12)}^{2} + 8^{2}}$

V=10; which is less than voltage drop across capacitor.

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