In a simple L-R circuit with an A.C. source,t | vedclass

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Question

(A) In an $L-R$ series circuit,the current $I$ is the same through both the inductor and the resistor.However,the voltage across the resistor $(V_R)$

Vedclass · Accepted Answer

$V_L^2 + V_R^2 = V_{AC}^2$

Vedclass · Answer

(A) In an $L-R$ series circuit,the current $I$ is the same through both the inductor and the resistor.However,the voltage across the resistor $(V_R)$ is in phase with the current,while the voltage across the inductor $(V_L)$ leads the current by a phase angle of $90^{\circ}$.Therefore,the phasor sum of the voltages must be equal to the source voltage $V_{AC}$.According to the phasor diagram,the relationship between the root-mean-square $(RMS)$ values is $V_{AC}^2 = V_L^2 + V_R^2$.Since the question asks for the relationship at any instant,and the phase difference between $V_L$ and $V_R$ is $90^{\circ}$,the instantaneous source voltage is the vector sum of the instantaneous voltages across the components,which follows the Pythagorean relationship $V_{AC}^2 = V_L^2 + V_R^2$.

In a simple $L-R$ circuit with an $A.C.$ source,the potential difference at any instant across the inductor and the resistor are $V_L$ and $V_R$ respectively,and the $A.C.$ source has a potential difference $V_{AC}$ at the same instant. Then:

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