An alternating current I=14 sin(100+πt) A passes through a series combination . . .

Question : An alternating current I=14 sin(100+πt) A passes through a series combination of a resistor 30 Ω and an inductor of (2/5π) H. Taking √2=1.4 calculate the

(i) rms value of the voltage drops across the resistor and the inductor, and

(ii) power factor of the circuit. 

CBSE 2023 SET - 3

Solution : 

I=14sin(100πt) A

Irms=I₀/√2

Irms=14/1.4 

Irms=10 A

R=30Ω (Given)

(i) 
V_R=10×30=300V

X_L=ωL

=100π(2/5π)

=40Ω

V_L=10×XL

=10×40

=400 V

Z=√[X_L²+R²]

Z=√[(40)²+(30)²]

Z=√[1600+900]

Z=√[2500]

Z=50Ω

(ii) 

Cosφ=R/Z

=30/50

=3/5