In an R-L-C series circuit, the potential dif | vedclass

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(A) Initially, the potential difference across each element is $V_R = V_L = V_C = 20 \, V$. Since $V_L = V_C$, the circuit is at resonance, meaning th

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$20 \, V, 10 \, V, 10 \, V$

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(A) Initially, the potential difference across each element is $V_R = V_L = V_C = 20 \, V$. Since $V_L = V_C$, the circuit is at resonance, meaning the impedance $Z = R$. The source voltage is $V = V_R = 20 \, V$.When the resistance $R$ is doubled to $2R$, the new impedance becomes $Z' = \sqrt{(2R)^2 + (X_L - X_C)^2}$. Since $X_L = X_C$ at resonance, $Z' = 2R$.The new current in the circuit is $I' = V / Z' = V / (2R) = I / 2$, where $I$ is the original current.The new potential difference across the resistor is $V_R' = I' 	imes (2R) = (I/2) 	imes (2R) = IR = 20 \, V$.The new potential difference across the inductor is $V_L' = I' X_L = (I/2) X_L = V_L / 2 = 20 / 2 = 10 \, V$.The new potential difference across the capacitor is $V_C' = I' X_C = (I/2) X_C = V_C / 2 = 20 / 2 = 10 \, V$.Thus, the new potential differences are $20 \, V, 10 \, V, 10 \, V$.

In an $R-L-C$ series circuit, the potential difference across each element is $20 \, V$. If the value of the resistance $R$ is doubled, what will be the potential difference across $R, L$, and $C$ respectively?

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