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  4. In a series LCR circuit, the potential drop across L, C and R respectively are 40 V, 120 V and 60 V. Then the source voltage is

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Q. In a series $LCR$ circuit, the potential drop across $L, C$ and $R$ respectively are $40\, V$, $120\, V$ and $60\, V$. Then the source voltage is

KCETKCET 2016Alternating Current

Solution:

In series $L-C-R$ circuit, the source voltage
Here, $V =\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}} $
$V_{R} =60 V \Rightarrow V_{L}=40\, V $
$V_{C} =120\, V $
$V =\sqrt{(60)^{2}+(40-120)^{2}} $
$=\sqrt{(60)^{2}+(-80)^{2}}=\sqrt{3600+6400} $
$=\sqrt{10000}=100 \,V$