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
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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:
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1069a960-ab44-4829-baef-0330e3e99a25%2F3b330541-bb7d-42c7-9613-ee1850acbc4d%2Fc9hfta_processed.jpeg&w=3840&q=75)
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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