Questions: What is the total Z of a 600-Ω R in parallel with a 300-Ω × L? Assume 600 V for the applied voltage 671 Ω 268 Ω 450 Ω 200 Ω

Solution Steps

We have two components:

1. A resistor \( R \) with resistance \( R = 600 \, \Omega \).
2. An inductor \( L \) with inductive reactance \( X_L = 300 \, \Omega \).

The impedance of the resistor is purely real: \[ Z_R = 600 \, \Omega \]

The impedance of the inductor is purely imaginary: \[ Z_L = j300 \, \Omega \]

For components in parallel, the total impedance \( Z \) is given by: \[ \frac{1}{Z} = \frac{1}{Z_R} + \frac{1}{Z_L} \]

Substitute the values: \[ \frac{1}{Z} = \frac{1}{600} + \frac{1}{j300} \]

Convert the imaginary part to a common denominator: \[ \frac{1}{Z} = \frac{1}{600} - \frac{j}{300} \]

Combine the fractions: \[ \frac{1}{Z} = \frac{1}{600} - \frac{j}{300} = \frac{1 - 2j}{600} \]

To find \( Z \), take the reciprocal of the complex number: \[ Z = \frac{600}{1 - 2j} \]

Multiply the numerator and the denominator by the complex conjugate of the denominator: \[ Z = \frac{600(1 + 2j)}{(1 - 2j)(1 + 2j)} = \frac{600(1 + 2j)}{1 + 4} = \frac{600(1 + 2j)}{5} = 120 + 240j \]

The magnitude of \( Z \) is: \[ |Z| = \sqrt{(120)^2 + (240)^2} = \sqrt{14400 + 57600} = \sqrt{72000} = 268.3282 \]

Final Answer