Fig P1 show a resistor R its ac voltage vR and current iR in

Fig. P1 show a resistor R, its ac voltage v_R and current i_R in an ac circuit. v_R leads i_R by 90 degree v_R lags i_R by 90 degree v_R and i_R are in phase In Fig. P2, an ac current i = I_m cos(omega t) flows through a load consisting of resistor R and an inductor L in series. The voltages across the resistor and the inductor are v_R and v_L. respectively. v_R leads v_L v_R lags v_L v_R and v_L are in phase In Fig. 2. if R = 100 ohm, L = 0.1 H, the angular frequency of ac current is omega = 1000 rad/s, the total impedance of the load is 100 ohm 100 + j100 ohm 100 - j0.01 il ohm In Fig. 2 and problem 3. the power factor of the RL load is pf = 1 pf = 0.707 leading pf = 0.707 lagging

Solution

1.Ans) C- Vr and Ir are in phase   

Reason : In resistive circuit Resistance is a pure real quantity so Ir=Vr/R has phase of Voltage Vr because Resistance doesn\'t add any phase

2.Ans) B---- vR lags vL

Reason: Given i=im*cos(wt)

Vr =i*R=R*im*cos(wt)

VL=Ldi/dt=w*L*im*-sin(wt)=w*L*im*cos(wt+90)

From the above equations we can see voltage VL has phase advance of 90 deg over Vr

so VL leads Vr by 90 deg or you can say Vr lags VL by 90 deg

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3.Ans) B- 100+j100 ohms

Given R=100; L=0.1 H and w=100o rad/sec

Impeadance Z= R+jwL for series RL circuit

Z=100+j(1000*0.1)= 100+j100 ohms

4 Ans) C-0.707 lag

   Power Factor cos(phi)=R/|Z|=100/(sqrt(100²+100²))=100/141.42=0.707 lagging because circuit is inductive

PF=0.707 lag