In the circuit shown,if the emf of the source | vedclass

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(A) In an $RC$ series circuit connected to an $AC$ source,the instantaneous $emf$ $E$ is given by the vector sum of the potential difference across th

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$3$

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(A) In an $RC$ series circuit connected to an $AC$ source,the instantaneous $emf$ $E$ is given by the vector sum of the potential difference across the resistor $(V_R)$ and the potential difference across the capacitor $(V_C)$.Since the current in the resistor and capacitor is the same,the voltage across the resistor is in phase with the current,while the voltage across the capacitor lags the current by $90^\circ$.Therefore,the relationship between the instantaneous values is $E^2 = V_R^2 + V_C^2$.Given: $E = 5 \, V$ and $V_C = 4 \, V$.Substituting these values into the equation:$5^2 = V_R^2 + 4^2$$25 = V_R^2 + 16$$V_R^2 = 25 - 16 = 9$$V_R = 3 \, V$.Thus,the potential difference across $R$ is $3 \, V$.

In the circuit shown,if the $emf$ of the source at an instant is $5 \, V$,and the potential difference across the capacitor at the same instant is $4 \, V$,then the potential difference across $R$ at that instant may be ..... $V$.

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