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A test voltage is applied to the input of an electrical network. v(t) = 1V · sin(2π · 5kHz. t) The input current is measured. i(t) = 268mA sin(2à · 5kHz · t – 46°) What is the circuit's input impedance, Zin? v(t) Zin Electrical Network

A test voltage is applied to the input of an electrical network. v(t) = 1V · sin(2π · 5kHz. t) The input current is measured. i(t) = 268mA sin(2à · 5kHz · t – 46°) What is the circuit's input impedance, Zin? v(t) Zin Electrical Network

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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### Determining Input Impedance of an Electrical Network In this example, a test voltage is applied to the input of an electrical network. The details are as follows: #### Applied Voltage \[ v(t) = 1V \cdot \sin(2\pi \cdot 5kHz \cdot t) \] #### Measured Input Current \[ i(t) = 268mA \cdot \sin(2\pi \cdot 5kHz \cdot t - 46^\circ) \] The objective is to determine the circuit's input impedance, \( Z_{in} \). ### Diagram Explanation To the right of the text is a diagram representing the setup. - The voltage source \( v(t) \) is connected in series with the electrical network. - The voltage \( v(t) \) is applied to the input of the electrical network. - The resulting current \( i(t) \) flows through the network, and \( Z_{in} \) represents the input impedance of the electrical network. ### Determining the Input Impedance By analyzing the given voltage and current, the input impedance \( Z_{in} \) of the circuit can be calculated. The voltage is: \[ v(t) = 1V \cdot \sin(2\pi \cdot 5kHz \cdot t) \] And the current is: \[ i(t) = 268mA \cdot \sin(2\pi \cdot 5kHz \cdot t - 46^\circ) \] Using the impedance formula: \[ Z_{in} = \frac{V_{max}}{I_{max}} \cdot e^{j\theta} \] Where \( V_{max} \) is the peak voltage, \( I_{max} \) is the peak current, and \( \theta \) is the phase angle. Given: \[ V_{max} = 1V \] \[ I_{max} = 268mA = 0.268A \] \[ \theta = -46^\circ \] \[ Z_{in} = \frac{1V}{0.268A} \cdot e^{-j46^\circ} \] \[ Z_{in} = 3.73 \Omega \cdot e^{-j46^\circ} \] Expressed in rectangular form: \[
Transcribed Image Text:### Determining Input Impedance of an Electrical Network In this example, a test voltage is applied to the input of an electrical network. The details are as follows: #### Applied Voltage \[ v(t) = 1V \cdot \sin(2\pi \cdot 5kHz \cdot t) \] #### Measured Input Current \[ i(t) = 268mA \cdot \sin(2\pi \cdot 5kHz \cdot t - 46^\circ) \] The objective is to determine the circuit's input impedance, \( Z_{in} \). ### Diagram Explanation To the right of the text is a diagram representing the setup. - The voltage source \( v(t) \) is connected in series with the electrical network. - The voltage \( v(t) \) is applied to the input of the electrical network. - The resulting current \( i(t) \) flows through the network, and \( Z_{in} \) represents the input impedance of the electrical network. ### Determining the Input Impedance By analyzing the given voltage and current, the input impedance \( Z_{in} \) of the circuit can be calculated. The voltage is: \[ v(t) = 1V \cdot \sin(2\pi \cdot 5kHz \cdot t) \] And the current is: \[ i(t) = 268mA \cdot \sin(2\pi \cdot 5kHz \cdot t - 46^\circ) \] Using the impedance formula: \[ Z_{in} = \frac{V_{max}}{I_{max}} \cdot e^{j\theta} \] Where \( V_{max} \) is the peak voltage, \( I_{max} \) is the peak current, and \( \theta \) is the phase angle. Given: \[ V_{max} = 1V \] \[ I_{max} = 268mA = 0.268A \] \[ \theta = -46^\circ \] \[ Z_{in} = \frac{1V}{0.268A} \cdot e^{-j46^\circ} \] \[ Z_{in} = 3.73 \Omega \cdot e^{-j46^\circ} \] Expressed in rectangular form: \[
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