9. For the circuit of Figure 4.8, determine voltages across R, L and C if the source is 7 volts RMS. j 200 Figure 4.8 -j 300 100
9. For the circuit of Figure 4.8, determine voltages across R, L and C if the source is 7 volts RMS. j 200 Figure 4.8 -j 300 100
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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For the circuit of Figure 4.8, determine voltages across R, L and C if the source is 7 volts RMS.
![### Problem Statement
Determine the voltages across R, L, and C in the circuit of Figure 4.8, if the source voltage is 7 volts RMS.
### Circuit Description
The circuit in Figure 4.8 comprises:
- An AC voltage source E with a root mean square (RMS) value of 7 volts.
- A resistor (R) with a resistance of 100 ohms.
- An inductor (L) with an inductive reactance of \( j200 \) ohms.
- A capacitor (C) with a capacitive reactance of \( -j300 \) ohms.
### Circuit Diagram
The provided diagram (Figure 4.8) shows a simple AC circuit where:
- The voltage source \( E \) is connected to a ground (referenced as zero potential).
- The components are connected in series, starting with the voltage source, followed by the inductor \( j200 \), then the resistor (R) with a value of 100 ohms, and finally, the capacitor with a reactance of \( -j300 \).
### Solution Explanation
To solve for the voltages across each component (R, L, and C), we will follow these steps:
1. **Determine Total Impedance (Z):**
- The impedance \( Z \) of the circuit is the sum of the individual impedances:
\[
Z = R + j\omega L + \frac{1}{j\omega C}
\]
- Substituting the given values: \( R = 100 \) ohms, \( j\omega L = j200 \) ohms, \( \frac{1}{j\omega C} = -j300 \) ohms,
\[
Z = 100 + j200 - j300 = 100 - j100 \text{ ohms}
\]
2. **Calculate the Current (I):**
- Using Ohm’s Law, \( V = IZ \), the current \( I \) is given by:
\[
I = \frac{V}{Z} = \frac{7}{100 - j100} \text{ A}
\]
- Converting \( Z \) to polar form for easier calculation:
\[
|Z| = \sqrt{(100)^2 + (-100)^2} = \sqrt{20000} =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5842debd-c7e0-436d-83d9-66a4971dddbf%2Ffbcb41f0-c4f5-4fb7-8042-44c4fc0ca104%2Fl9qd91g_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Determine the voltages across R, L, and C in the circuit of Figure 4.8, if the source voltage is 7 volts RMS.
### Circuit Description
The circuit in Figure 4.8 comprises:
- An AC voltage source E with a root mean square (RMS) value of 7 volts.
- A resistor (R) with a resistance of 100 ohms.
- An inductor (L) with an inductive reactance of \( j200 \) ohms.
- A capacitor (C) with a capacitive reactance of \( -j300 \) ohms.
### Circuit Diagram
The provided diagram (Figure 4.8) shows a simple AC circuit where:
- The voltage source \( E \) is connected to a ground (referenced as zero potential).
- The components are connected in series, starting with the voltage source, followed by the inductor \( j200 \), then the resistor (R) with a value of 100 ohms, and finally, the capacitor with a reactance of \( -j300 \).
### Solution Explanation
To solve for the voltages across each component (R, L, and C), we will follow these steps:
1. **Determine Total Impedance (Z):**
- The impedance \( Z \) of the circuit is the sum of the individual impedances:
\[
Z = R + j\omega L + \frac{1}{j\omega C}
\]
- Substituting the given values: \( R = 100 \) ohms, \( j\omega L = j200 \) ohms, \( \frac{1}{j\omega C} = -j300 \) ohms,
\[
Z = 100 + j200 - j300 = 100 - j100 \text{ ohms}
\]
2. **Calculate the Current (I):**
- Using Ohm’s Law, \( V = IZ \), the current \( I \) is given by:
\[
I = \frac{V}{Z} = \frac{7}{100 - j100} \text{ A}
\]
- Converting \( Z \) to polar form for easier calculation:
\[
|Z| = \sqrt{(100)^2 + (-100)^2} = \sqrt{20000} =
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