Question

A series RL circuit with R = 10 Ω and L = `(100/pi)` mH is connected to an ac source of voltage V = 141 sin (100 πt), where V is in volts and t is in seconds. Calculate

1. the impedance of the circuit
2. phase angle, and
3. the voltage drop across the inductor.

Answer 1

Given:

R = 10 Ω

L = `(100/pi)` mH

V = 141 sin (100 π)t

From this equation, we get the value of ω = 100π and V = 141 volt

1. To find: Impedance (Z)
   Z = `sqrt(R^2 + X_L^2)`
   Where Z is the impedance, R is the resistance, and X_L is the impedance,
   X_L = ωL
   X_L = `(100pi xx 100)/(pi xx 10^-3)`
   X_L = 10Ω
   Z = `sqrt(R^2 + X_L^2)`
   Z = `sqrt((10)^2 + (10)^2)`
   Z = `sqrt200`
   Z = `10sqrt2`Ω
2. Phase Angle (Φ):
   We can calculate the phase angle by the following formula,
   `cosphi = R/Z`
   `cosphi = 10/(10sqrt2)`
   `cosphi = 1/sqrt2`
   `phi = 45^circ`
3. Voltage drop:
   `V_L = IX_L`
   = `V/Z xx X_L`
   `V_L = 141/(10sqrt2) xx 10`
   `V_L = 141/sqrt2`
   `V_L ≅ 100` volt