A 200ohm resistor a 400mu F capacitor and a 0200H inductor a

A 20.0-ohm resistor, a 40.0-mu F capacitor, and a 0.200-H inductor are connected in series with a 60.0-Hz, 70.0-V source. The inductor is found to have a resistance of 15.0 ohm. Find the current in the circuit and the voltage drop across the inductor.

Solution

Inductive reactance

X_L=2pifL =2pi*60*0.2 =75.4 ohms

Capacitive reactance

X_C=1/2pifC =1/2pi*60*(40*10^-6) =66.3 ohms

Equivalent resistance

R_eq =20+15=35 ohms

Impedance

Z=sqrt[R²+(X_L-X_C)²] =sqrt[35²+(75.4-66.3)²]

Z=36.16 ohms

a)

Current in the circuit is

I=V/Z =70/36.16

I=1.9356 A

b)

Impedance across inductor

Z_L=sqrt[15²+75.4²]=76.88 ohms

Voltage drop across inductor

V_L=IZ_L =1.9356*76.88

V_L=148.8 Volts