**Problem: Circuit Impedance Calculation**A resistor with a resistance of 50 Ω is connected in parallel with an inductor with an inductance of 0.175 henry. The power source is 60 Hz. What is the impedance of the circuit?**Explanation:**To solve for the impedance of the circuit, follow these steps:1. **Calculate Inductive Reactance (\(X_L\)):** - Use the formula: \(X_L = 2\pi fL\) - Where \(f\) is the frequency (60 Hz) and \(L\) is the inductance (0.175 H). 2. **Determine the Impedance of Each Component:** - The impedance of the resistor (\(Z_R\)) is purely real and equal to 50 Ω. - The impedance of the inductor (\(Z_L\)) is purely imaginary and given by \(jX_L\).3. **Combine the Impedances in Parallel:** - Use the formula for combined impedance (\(Z\)) of parallel components: \[ \frac{1}{Z} = \frac{1}{Z_R} + \frac{1}{Z_L} \]4. **Calculate the Resultant Impedance (\(Z\)):** - Compute the real and imaginary components separately to find the total impedance.By solving the above, you will find the combined impedance of the parallel resistor-inductor circuit.

Inductive reactance of inductor = 2×pi×frequency ×inductance. If two impedances Z1 , Z2 are connected in parallel than equivalant impedance is (Z1×Z2)/(Z1 +Z2).

Answered: A resistor with a resistance of 50 Q is…

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A resistor with a resistance of 50 Q is connected in parallel with an inductor with an inductance of 0.175 henry. The power source is 60 Hz. What is the impedance of the cir- cuit?

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

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Question

**Problem: Circuit Impedance Calculation**

A resistor with a resistance of 50 Ω is connected in parallel with an inductor with an inductance of 0.175 henry. The power source is 60 Hz. What is the impedance of the circuit?

**Explanation:**

To solve for the impedance of the circuit, follow these steps:

1. **Calculate Inductive Reactance (\(X_L\)):**
- Use the formula: \(X_L = 2\pi fL\)
- Where \(f\) is the frequency (60 Hz) and \(L\) is the inductance (0.175 H).

2. **Determine the Impedance of Each Component:**
- The impedance of the resistor (\(Z_R\)) is purely real and equal to 50 Ω.
- The impedance of the inductor (\(Z_L\)) is purely imaginary and given by \(jX_L\).

3. **Combine the Impedances in Parallel:**
- Use the formula for combined impedance (\(Z\)) of parallel components:
\[
\frac{1}{Z} = \frac{1}{Z_R} + \frac{1}{Z_L}
\]

4. **Calculate the Resultant Impedance (\(Z\)):**
- Compute the real and imaginary components separately to find the total impedance.

By solving the above, you will find the combined impedance of the parallel resistor-inductor circuit.

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