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9 kHz R₁ 2200 w R₂ 1800 1 mH .C 0.47 µF itions: ons: Ants do basis for analyzing most practical circuit configurations. EXAMPLE 14.15 Determine the impedance, current, and voltage values for the circuit shown in Figure 14.35a. A 300 Ω ww XL 150 Ω 400 Ω 10 V Vs R 300 0 w X₁ (90°) 2400 Xp Vs 10 V Xp Z+-384 0.38. FIGURE 14.35 Solution: (a) R 300 Xp = 240 NZ90° Xc (-90°) (b) First, combine the reactances into a single parallel-equivalent value, as follows: XLXC (150)(-400 ) Хр = 240 Ω = 240 Ω290° XL + XC 150 Ω + (-400 Ω) Since the phase angle of Xp is positive, it is inductive in nature. The circuit imped- ance has a magnitude of Z = √√ x² + R² = √(240 N)² + (300 N)² = 384 N at a phase angle of 0 = tan tan -1 = tan R 240 Ω 300 Ω = 38.7° An Alternate Approach The value of Xp can also be using 1 Xp ==== Be+B For the circuit in Fig Bc 400 02-90° BL 150 2.90° Once the circuit impedance is known, the circuit current can be found as Vs 10 V IT 26 mAZ-38.7° ZT 384 38.7° The phase angle of IT indicates that the circuit current lags the source voltage by 38.7°. Therefore, the values of circuit current and source voltage can be written as I = 26 mAZ0° and Vs = 10 VZ38.7° Once the value of circuit current is known, the voltage across the parallel LC cir- uit can be found as VLC = ITXP = (26 mAZ0°)(240 2/90°) = 6.24 VZ90° Xp Since they are 180 Bc+ B₁ = 4.1 1 4.17 mS2-9 This value of Xp m he phase angle of VLC makes sense when you remember that Xp is inductive in na- e. Using the calculated value of VLC, the values of Ic and I can be found as found in the exam VLC 6.24 VZ90° = 15.6 mAZ180° Xc 400 Z-90°

9 kHz R₁ 2200 w R₂ 1800 1 mH .C 0.47 µF itions: ons: Ants do basis for analyzing most practical circuit configurations. EXAMPLE 14.15 Determine the impedance, current, and voltage values for the circuit shown in Figure 14.35a. A 300 Ω ww XL 150 Ω 400 Ω 10 V Vs R 300 0 w X₁ (90°) 2400 Xp Vs 10 V Xp Z+-384 0.38. FIGURE 14.35 Solution: (a) R 300 Xp = 240 NZ90° Xc (-90°) (b) First, combine the reactances into a single parallel-equivalent value, as follows: XLXC (150)(-400 ) Хр = 240 Ω = 240 Ω290° XL + XC 150 Ω + (-400 Ω) Since the phase angle of Xp is positive, it is inductive in nature. The circuit imped- ance has a magnitude of Z = √√ x² + R² = √(240 N)² + (300 N)² = 384 N at a phase angle of 0 = tan tan -1 = tan R 240 Ω 300 Ω = 38.7° An Alternate Approach The value of Xp can also be using 1 Xp ==== Be+B For the circuit in Fig Bc 400 02-90° BL 150 2.90° Once the circuit impedance is known, the circuit current can be found as Vs 10 V IT 26 mAZ-38.7° ZT 384 38.7° The phase angle of IT indicates that the circuit current lags the source voltage by 38.7°. Therefore, the values of circuit current and source voltage can be written as I = 26 mAZ0° and Vs = 10 VZ38.7° Once the value of circuit current is known, the voltage across the parallel LC cir- uit can be found as VLC = ITXP = (26 mAZ0°)(240 2/90°) = 6.24 VZ90° Xp Since they are 180 Bc+ B₁ = 4.1 1 4.17 mS2-9 This value of Xp m he phase angle of VLC makes sense when you remember that Xp is inductive in na- e. Using the calculated value of VLC, the values of Ic and I can be found as found in the exam VLC 6.24 VZ90° = 15.6 mAZ180° Xc 400 Z-90°

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
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Calculate the impedance, current, and voltage values for the circuit shown in the following figure 

(The book shows how to answer but it does not show me what to do with R2) Please keep the answers similar to the form the book answers in.

9 kHz R₁ 2200 w R₂ 1800 1 mH .C 0.47 µF
Transcribed Image Text:9 kHz R₁ 2200 w R₂ 1800 1 mH .C 0.47 µF
itions: ons: Ants do basis for analyzing most practical circuit configurations. EXAMPLE 14.15 Determine the impedance, current, and voltage values for the circuit shown in Figure 14.35a. A 300 Ω ww XL 150 Ω 400 Ω 10 V Vs R 300 0 w X₁ (90°) 2400 Xp Vs 10 V Xp Z+-384 0.38. FIGURE 14.35 Solution: (a) R 300 Xp = 240 NZ90° Xc (-90°) (b) First, combine the reactances into a single parallel-equivalent value, as follows: XLXC (150)(-400 ) Хр = 240 Ω = 240 Ω290° XL + XC 150 Ω + (-400 Ω) Since the phase angle of Xp is positive, it is inductive in nature. The circuit imped- ance has a magnitude of Z = √√ x² + R² = √(240 N)² + (300 N)² = 384 N at a phase angle of 0 = tan tan -1 = tan R 240 Ω 300 Ω = 38.7° An Alternate Approach The value of Xp can also be using 1 Xp ==== Be+B For the circuit in Fig Bc 400 02-90° BL 150 2.90° Once the circuit impedance is known, the circuit current can be found as Vs 10 V IT 26 mAZ-38.7° ZT 384 38.7° The phase angle of IT indicates that the circuit current lags the source voltage by 38.7°. Therefore, the values of circuit current and source voltage can be written as I = 26 mAZ0° and Vs = 10 VZ38.7° Once the value of circuit current is known, the voltage across the parallel LC cir- uit can be found as VLC = ITXP = (26 mAZ0°)(240 2/90°) = 6.24 VZ90° Xp Since they are 180 Bc+ B₁ = 4.1 1 4.17 mS2-9 This value of Xp m he phase angle of VLC makes sense when you remember that Xp is inductive in na- e. Using the calculated value of VLC, the values of Ic and I can be found as found in the exam VLC 6.24 VZ90° = 15.6 mAZ180° Xc 400 Z-90°
Transcribed Image Text:itions: ons: Ants do basis for analyzing most practical circuit configurations. EXAMPLE 14.15 Determine the impedance, current, and voltage values for the circuit shown in Figure 14.35a. A 300 Ω ww XL 150 Ω 400 Ω 10 V Vs R 300 0 w X₁ (90°) 2400 Xp Vs 10 V Xp Z+-384 0.38. FIGURE 14.35 Solution: (a) R 300 Xp = 240 NZ90° Xc (-90°) (b) First, combine the reactances into a single parallel-equivalent value, as follows: XLXC (150)(-400 ) Хр = 240 Ω = 240 Ω290° XL + XC 150 Ω + (-400 Ω) Since the phase angle of Xp is positive, it is inductive in nature. The circuit imped- ance has a magnitude of Z = √√ x² + R² = √(240 N)² + (300 N)² = 384 N at a phase angle of 0 = tan tan -1 = tan R 240 Ω 300 Ω = 38.7° An Alternate Approach The value of Xp can also be using 1 Xp ==== Be+B For the circuit in Fig Bc 400 02-90° BL 150 2.90° Once the circuit impedance is known, the circuit current can be found as Vs 10 V IT 26 mAZ-38.7° ZT 384 38.7° The phase angle of IT indicates that the circuit current lags the source voltage by 38.7°. Therefore, the values of circuit current and source voltage can be written as I = 26 mAZ0° and Vs = 10 VZ38.7° Once the value of circuit current is known, the voltage across the parallel LC cir- uit can be found as VLC = ITXP = (26 mAZ0°)(240 2/90°) = 6.24 VZ90° Xp Since they are 180 Bc+ B₁ = 4.1 1 4.17 mS2-9 This value of Xp m he phase angle of VLC makes sense when you remember that Xp is inductive in na- e. Using the calculated value of VLC, the values of Ic and I can be found as found in the exam VLC 6.24 VZ90° = 15.6 mAZ180° Xc 400 Z-90°
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