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I. Multiple Choice Questions · Q10

Q.In a series RL circuit, the resistance and inductive reactance are the same. Then the phase difference between the voltage and current in the circuit is

(a) π4\dfrac{\pi}{4}
(b) π2\dfrac{\pi}{2}
(c) π6\dfrac{\pi}{6}
(d) zero
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Concept understanding — Series RLC Circuit and Impedance

Setting up. For a resistor R, inductor L and capacitor C all in series across v=Vmsin⁡ωtv=V_m\sin\omega t, the SAME current i flows through every element, so VRV_R is in phase with i, VLV_L leads i by 90∘90^{\circ}, and VCV_C lags i by 90∘90^{\circ} -- meaning VLV_L and VCV_C point in exactly OPPOSITE directions on a phasor diagram. Combining VRV_R with the net (VL−VC)(V_L-V_C) by the parallelogram law gives the applied voltage's magnitude,

Vm2=VR2+(VL−VC)2V_m^2 = V_R^2+(V_L-V_C)^2

Dividing through by Im2I_m^2 (since each voltage is current times its own R or reactance) gives

Im=VmZ,Z=R2+(XL−XC)2I_m=\dfrac{V_m}{Z}, \qquad Z = \sqrt{R^2+(X_L-X_C)^2}

where Z, the impedance, is the circuit's total effective opposition (in ohms) to AC, combining resistance and net reactance.

Phase angle -- from the similar voltage/impedance triangles. tan⁡ϕ=(VL−VC)/VR=(XL−XC)/R\tan\phi = (V_L-V_C)/V_R = (X_L-X_C)/R. …

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