An AC voltmeter connected between points A an | vedclass

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(A) Given is a series $L-C-R$ circuit.Let $V_L$,$V_R$,and $V_C$ be the voltages across the inductor,resistor,and capacitor,respectively.From the circu

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$\sqrt{481} \,V$

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(A) Given is a series $L-C-R$ circuit.Let $V_L$,$V_R$,and $V_C$ be the voltages across the inductor,resistor,and capacitor,respectively.From the circuit,the voltmeter readings are:$1$. Between $A$ and $B$: $V_L = 36 \,V$$2$. Between $A$ and $C$: $\sqrt{V_L^2 + V_R^2} = 39 \,V$$3$. Between $B$ and $D$: $\sqrt{V_R^2 + V_C^2} = 25 \,V$From $(1)$ and $(2)$:$V_L^2 + V_R^2 = 39^2$$36^2 + V_R^2 = 1521$$1296 + V_R^2 = 1521$$V_R^2 = 1521 - 1296 = 225$$V_R = 15 \,V$From $(3)$ and $V_R = 15 \,V$:$V_R^2 + V_C^2 = 25^2$$15^2 + V_C^2 = 625$$225 + V_C^2 = 625$$V_C^2 = 400$$V_C = 20 \,V$The voltage between $A$ and $D$ is the total voltage $V_{AD} = \sqrt{V_R^2 + (V_L - V_C)^2}$:$V_{AD} = \sqrt{15^2 + (36 - 20)^2}$$V_{AD} = \sqrt{225 + 16^2} = \sqrt{225 + 256} = \sqrt{481} \,V$

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