In a series RLC circuit,the potential differe | vedclass

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(C) In a series $RLC$ circuit,the total voltage (source $e.m.f.$) is given by the phasor sum of the individual potential differences:$V_{source} = \sq

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$50$

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(C) In a series $RLC$ circuit,the total voltage (source $e.m.f.$) is given by the phasor sum of the individual potential differences:$V_{source} = \sqrt{V_R^2 + (V_L - V_C)^2}$Given:$V_R = 30 \,V$$V_L = 60 \,V$$V_C = 100 \,V$Substituting these values into the formula:$V_{source} = \sqrt{(30)^2 + (60 - 100)^2}$$V_{source} = \sqrt{(30)^2 + (-40)^2}$$V_{source} = \sqrt{900 + 1600}$$V_{source} = \sqrt{2500} = 50 \,V$Thus,the $e.m.f.$ of the source is $50 \,V$.

In a series $RLC$ circuit,the potential differences across $R, L$ and $C$ are $30 \,V, 60 \,V$ and $100 \,V$ respectively,as shown in the figure. The $e.m.f.$ of the source (in volts) is:

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