Explain current divider rule in ac circuits with two parallel impedances Z1 and Z2. Calculate the currents I1 and I2 in phasor form for the circuit given in figure. ### Current Divider Rule in AC CircuitsThe problem requires us to explain the current divider rule for AC circuits where there are two parallel impedances \( Z_1 \) and \( Z_2 \). We are also asked to calculate the currents \( I_1 \) and \( I_2 \) in phasor form for the given circuit.#### Circuit Description- **Total Current \( I \):** 30 A at an angle of 40°.- **Components in Parallel:** - Resistor \( R \) with a resistance of 22 Ω. - Inductor \( X_L \) with an inductive reactance of 60 Ω.- **Currents:** - \( I_1 \) flows through the resistor \( R \). - \( I_2 \) flows through the inductor \( X_L \).For the given configuration, the current divider rule can be used to find the individual branch currents \( I_1 \) and \( I_2 \), which are the currents through \( Z_1 \) (resistor) and \( Z_2 \) (inductor), respectively.

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Explain current divider rule in ac circuits with two parallel impedances Z1 and Z2. Calculate the currents I1 and I2 in phasor form for the circuit given in figure.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Question

Explain current divider rule in ac circuits with two parallel impedances Z₁ and Z₂. Calculate the currents I₁and I₂ in phasor form for the circuit given in figure.

### Current Divider Rule in AC Circuits

The problem requires us to explain the current divider rule for AC circuits where there are two parallel impedances \( Z_1 \) and \( Z_2 \). We are also asked to calculate the currents \( I_1 \) and \( I_2 \) in phasor form for the given circuit.

#### Circuit Description
- **Total Current \( I \):** 30 A at an angle of 40°.
- **Components in Parallel:**
- Resistor \( R \) with a resistance of 22 Ω.
- Inductor \( X_L \) with an inductive reactance of 60 Ω.
- **Currents:**
- \( I_1 \) flows through the resistor \( R \).
- \( I_2 \) flows through the inductor \( X_L \).

For the given configuration, the current divider rule can be used to find the individual branch currents \( I_1 \) and \( I_2 \), which are the currents through \( Z_1 \) (resistor) and \( Z_2 \) (inductor), respectively.

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