Resistance - Physics 152 (or Phys 2) question

Question

A generator is connected to a resistor and a 0.038-H inductor in series. The rms voltage across the generator is 7.4 V. When the generator frequency is set to 120 Hz, the rms voltage across the inductor is 3.8 V. Determine the resistance of the resistor in this circuit.

Answer 1

The rms current through an inductor is I = V/X_L where X_L is equal to 2πfL. The rms current is therefore I = 3.8V / (2 x π x 120 x 0.038 = 0.133 A rms.

As this is a series circuit, the same current flows through the resistor. The impedance, Z, is obtained by Z = V/I = 7.4 V / 0.133A = 55.6 ohms.

But Z² = R² + X_L² so R = sqrt ( Z² – X_L²) = sqrt (55.6² – X_L²) where X_{L =} 2πfL = 28.6 ohms. Substitute this into the equation for R, and you should get 47.7 ohms.

The vector sum of the voltage drops across R and X_L should equal the applied voltage.

V_applied = sqrt ( V_R² + V_L²). As the voltage drop across R = 0.133A x 47.7 ohms = 6.33 V, and the voltage across the inductor is given as 3.8V, the vector sum of the two voltage drops is

Sqrt ( 3.8² + 6.33²) = 7.38 which is approximately 7.4 V.