In an LCR series circuit, the potential diffe | vedclass

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(B) In an $LCR$ series circuit, the supply voltage $V$ is given by the phasor sum of the individual potential differences across the resistor $(V_R)$,

Vedclass · Accepted Answer

$50$

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(B) In an $LCR$ series circuit, the supply voltage $V$ is given by the phasor sum of the individual potential differences across the resistor $(V_R)$, inductor $(V_L)$, and capacitor $(V_C)$.
The formula is: $V = \sqrt{V_R^2 + (V_L - V_C)^2}$
Given values are:
$V_L = 60 	ext{ V}$
$V_C = 30 	ext{ V}$
$V_R = 40 	ext{ V}$
Substituting these values into the formula:
$V = \sqrt{40^2 + (60 - 30)^2}$
$V = \sqrt{1600 + (30)^2}$
$V = \sqrt{1600 + 900}$
$V = \sqrt{2500}$
$V = 50 	ext{ V}$
Therefore, the supply voltage is $50 	ext{ V}$.

In an $LCR$ series circuit, the potential difference across the terminals of the inductor, capacitor and resistor is $60 \text{ V}$, $30 \text{ V}$ and $40 \text{ V}$ respectively. Then the supply voltage will be equal to (in $\text{ V}$)

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