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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

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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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} =
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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