Find the currents of each resistor using super position method. Where, VLeft = 160V; VRight = 180V;R₁ = 40; R₂ = 120; R3= 60; • Combined I left and I right • What is the value of 1₁?

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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**Problem Statement:**

Find the currents of each resistor using the superposition method. 

Given:
- \( V_{\text{Left}} = 160V \)
- \( V_{\text{Right}} = 180V \)
- \( R_1 = 40\Omega \)
- \( R_2 = 120\Omega \)
- \( R_3 = 60\Omega \)

**Steps:**

- Combined \( I \) left and \( I \) right
- What is the value of \( I_1 \)?

**Circuit Diagram:**

The circuit diagram consists of two voltage sources (\( V_{\text{Left}} \) and \( V_{\text{Right}} \)), each connected with resistors in the following arrangement:
- \( R_1 \) is connected in series with \( V_{\text{Left}} \).
- \( R_3 \) is connected in series with \( V_{\text{Right}} \).
- \( R_2 \) is in parallel with the combined arrangement of the two series circuits.

**Options:**
- \( \quad \) \( 0.5 \, \text{Amp} \)
- \( \quad \) \( 0.67 \, \text{Amp} \)
- \( \quad \) \( 2.0 \, \text{Amp} \)
- \( \quad \) \( 1.17 \, \text{Amp} \)
Transcribed Image Text:**Problem Statement:** Find the currents of each resistor using the superposition method. Given: - \( V_{\text{Left}} = 160V \) - \( V_{\text{Right}} = 180V \) - \( R_1 = 40\Omega \) - \( R_2 = 120\Omega \) - \( R_3 = 60\Omega \) **Steps:** - Combined \( I \) left and \( I \) right - What is the value of \( I_1 \)? **Circuit Diagram:** The circuit diagram consists of two voltage sources (\( V_{\text{Left}} \) and \( V_{\text{Right}} \)), each connected with resistors in the following arrangement: - \( R_1 \) is connected in series with \( V_{\text{Left}} \). - \( R_3 \) is connected in series with \( V_{\text{Right}} \). - \( R_2 \) is in parallel with the combined arrangement of the two series circuits. **Options:** - \( \quad \) \( 0.5 \, \text{Amp} \) - \( \quad \) \( 0.67 \, \text{Amp} \) - \( \quad \) \( 2.0 \, \text{Amp} \) - \( \quad \) \( 1.17 \, \text{Amp} \)
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