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Using superposition, determine the current through the inductance XL for the network below I = 0.3 A Z 60° L'with E shorted = L"with I opened IL = XL ell IL 8 Ω Xc 5Ω + E = 10 V 0° units units units

Using superposition, determine the current through the inductance XL for the network below I = 0.3 A Z 60° L'with E shorted = L"with I opened IL = XL ell IL 8 Ω Xc 5Ω + E = 10 V 0° units units units

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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Using superposition, determine the current through the inductance XL for the network below

**Using superposition, determine the current through the inductance \( X_L \) for the network below** The circuit diagram includes the following components: 1. **Current Source (I):** - Value: \( 0.3 \, A \, \angle 60^\circ \) 2. **Inductance (\( X_L \)):** - Inductive Reactance: \( 8 \, \Omega \) 3. **Capacitance (\( X_C \)):** - Capacitive Reactance: \( 5 \, \Omega \) 4. **Voltage Source (E):** - Voltage: \( 10 \, V \, \angle 0^\circ \) ### Calculations to Perform - \( I_L' \) with \( E \) shorted = [ ] \( \angle \) [ ] units - \( I_L'' \) with \( I \) opened = [ ] \( \angle \) [ ] units - \( I_L \) = [ ] \( \angle \) [ ] units **Explanation:** - **Superposition Method:** This technique involves analyzing the circuit multiple times, each with only one independent source active, and then combining the effects to find the total current through a specific component (in this case, the inductance \( X_L \)). - **\( I_L' \):** Current through \( X_L \) when the voltage source \( E \) is shorted. - **\( I_L'' \):** Current through \( X_L \) when the current source \( I \) is opened. - **\( I_L \):** Total current through \( X_L \), calculated by combining \( I_L' \) and \( I_L'' \) using vector addition.
Transcribed Image Text:**Using superposition, determine the current through the inductance \( X_L \) for the network below** The circuit diagram includes the following components: 1. **Current Source (I):** - Value: \( 0.3 \, A \, \angle 60^\circ \) 2. **Inductance (\( X_L \)):** - Inductive Reactance: \( 8 \, \Omega \) 3. **Capacitance (\( X_C \)):** - Capacitive Reactance: \( 5 \, \Omega \) 4. **Voltage Source (E):** - Voltage: \( 10 \, V \, \angle 0^\circ \) ### Calculations to Perform - \( I_L' \) with \( E \) shorted = [ ] \( \angle \) [ ] units - \( I_L'' \) with \( I \) opened = [ ] \( \angle \) [ ] units - \( I_L \) = [ ] \( \angle \) [ ] units **Explanation:** - **Superposition Method:** This technique involves analyzing the circuit multiple times, each with only one independent source active, and then combining the effects to find the total current through a specific component (in this case, the inductance \( X_L \)). - **\( I_L' \):** Current through \( X_L \) when the voltage source \( E \) is shorted. - **\( I_L'' \):** Current through \( X_L \) when the current source \( I \) is opened. - **\( I_L \):** Total current through \( X_L \), calculated by combining \( I_L' \) and \( I_L'' \) using vector addition.
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