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R₂ 0 www U ww www R₁ U 후 Note OBJECTIVE 2 Although each circuit in this ection contains two voltage ources, the superposition corem can be applied to cuits containing any number voltage and/or current The following example demonstrates the superposition analysis of a multisource circun EXAMPLE 7.1 Determine the values of V1, V2, and V₁ for the circuit shown in Figure 7.6a. 12 V A₁ 30 Ω ww www R3 R2 15 Ω ww 10 Ω e At this point, we have cale Using the same appro R VEQ Since this voltage is ac and (a) R₁ VA 12 V R₁₁ 30 Ω W V1A V3A R3 R2 15 Ω W 10 Ω V2A VA 12 V 30 Ω W V₁A VEQ R₁ 30 Ω ww V₁B R2 15 Ω W 2B + (b) 1B V3B R3 10 Ω www For the circuit in Fig Finally, we dete results. If you com ities. Therefore, REQ 60 Figure 7.6 also s Finally, V3A and R₂ 15 Ω W + V2B These are the c Practice Prob A circuit like VB 9 V VEQ REQ 7.5 Ω VB = 7V, R₁ 9V and V3 for the FIGURE 7.6 Solution: ow) diw bobal (c) i The first step is to split the circuit into two single source circuits. These circuits (along with their series equivalents) are shown in Figures 7.6b and 7.6c. For Figure 7.6b, REQ = R2|| R3 = 15 || 10 N = 6 N and burn The results Figure 7.7a sho ple. Note the greater voltage P ■ Since V ■ Since W Polar REQ 60 VEQ = VA = (12 V) =2 V ✓ VA 12 V REQ + R₁ 36 Ω Since this voltage is across the parallel combination of R2 and R3, and V2A = V3A = 2 V E=12 V-2 V10 V₁ = VA - VEQ = 12 V 2V = 10 V Circuit Analysis Techniques * ✓ FIGURE 7

R₂ 0 www U ww www R₁ U 후 Note OBJECTIVE 2 Although each circuit in this ection contains two voltage ources, the superposition corem can be applied to cuits containing any number voltage and/or current The following example demonstrates the superposition analysis of a multisource circun EXAMPLE 7.1 Determine the values of V1, V2, and V₁ for the circuit shown in Figure 7.6a. 12 V A₁ 30 Ω ww www R3 R2 15 Ω ww 10 Ω e At this point, we have cale Using the same appro R VEQ Since this voltage is ac and (a) R₁ VA 12 V R₁₁ 30 Ω W V1A V3A R3 R2 15 Ω W 10 Ω V2A VA 12 V 30 Ω W V₁A VEQ R₁ 30 Ω ww V₁B R2 15 Ω W 2B + (b) 1B V3B R3 10 Ω www For the circuit in Fig Finally, we dete results. If you com ities. Therefore, REQ 60 Figure 7.6 also s Finally, V3A and R₂ 15 Ω W + V2B These are the c Practice Prob A circuit like VB 9 V VEQ REQ 7.5 Ω VB = 7V, R₁ 9V and V3 for the FIGURE 7.6 Solution: ow) diw bobal (c) i The first step is to split the circuit into two single source circuits. These circuits (along with their series equivalents) are shown in Figures 7.6b and 7.6c. For Figure 7.6b, REQ = R2|| R3 = 15 || 10 N = 6 N and burn The results Figure 7.7a sho ple. Note the greater voltage P ■ Since V ■ Since W Polar REQ 60 VEQ = VA = (12 V) =2 V ✓ VA 12 V REQ + R₁ 36 Ω Since this voltage is across the parallel combination of R2 and R3, and V2A = V3A = 2 V E=12 V-2 V10 V₁ = VA - VEQ = 12 V 2V = 10 V Circuit Analysis Techniques * ✓ FIGURE 7

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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My book shows me how to calculate V1 through V3 but I need to answer for a circuit that has a V4 under V3.

The book shows how too calculate the equavilat rnesistance with V1,V2 and V3 but I am not sure how to calculate resistance for V4 in Superposition Theorem. Can you help me to calculate for R4 to get V4? I just need to know what to do with the additional resistor R4.

R₂ 0 www U ww www R₁ U 후
Transcribed Image Text:R₂ 0 www U ww www R₁ U 후
Note OBJECTIVE 2 Although each circuit in this ection contains two voltage ources, the superposition corem can be applied to cuits containing any number voltage and/or current The following example demonstrates the superposition analysis of a multisource circun EXAMPLE 7.1 Determine the values of V1, V2, and V₁ for the circuit shown in Figure 7.6a. 12 V A₁ 30 Ω ww www R3 R2 15 Ω ww 10 Ω e At this point, we have cale Using the same appro R VEQ Since this voltage is ac and (a) R₁ VA 12 V R₁₁ 30 Ω W V1A V3A R3 R2 15 Ω W 10 Ω V2A VA 12 V 30 Ω W V₁A VEQ R₁ 30 Ω ww V₁B R2 15 Ω W 2B + (b) 1B V3B R3 10 Ω www For the circuit in Fig Finally, we dete results. If you com ities. Therefore, REQ 60 Figure 7.6 also s Finally, V3A and R₂ 15 Ω W + V2B These are the c Practice Prob A circuit like VB 9 V VEQ REQ 7.5 Ω VB = 7V, R₁ 9V and V3 for the FIGURE 7.6 Solution: ow) diw bobal (c) i The first step is to split the circuit into two single source circuits. These circuits (along with their series equivalents) are shown in Figures 7.6b and 7.6c. For Figure 7.6b, REQ = R2|| R3 = 15 || 10 N = 6 N and burn The results Figure 7.7a sho ple. Note the greater voltage P ■ Since V ■ Since W Polar REQ 60 VEQ = VA = (12 V) =2 V ✓ VA 12 V REQ + R₁ 36 Ω Since this voltage is across the parallel combination of R2 and R3, and V2A = V3A = 2 V E=12 V-2 V10 V₁ = VA - VEQ = 12 V 2V = 10 V Circuit Analysis Techniques * ✓ FIGURE 7
Transcribed Image Text:Note OBJECTIVE 2 Although each circuit in this ection contains two voltage ources, the superposition corem can be applied to cuits containing any number voltage and/or current The following example demonstrates the superposition analysis of a multisource circun EXAMPLE 7.1 Determine the values of V1, V2, and V₁ for the circuit shown in Figure 7.6a. 12 V A₁ 30 Ω ww www R3 R2 15 Ω ww 10 Ω e At this point, we have cale Using the same appro R VEQ Since this voltage is ac and (a) R₁ VA 12 V R₁₁ 30 Ω W V1A V3A R3 R2 15 Ω W 10 Ω V2A VA 12 V 30 Ω W V₁A VEQ R₁ 30 Ω ww V₁B R2 15 Ω W 2B + (b) 1B V3B R3 10 Ω www For the circuit in Fig Finally, we dete results. If you com ities. Therefore, REQ 60 Figure 7.6 also s Finally, V3A and R₂ 15 Ω W + V2B These are the c Practice Prob A circuit like VB 9 V VEQ REQ 7.5 Ω VB = 7V, R₁ 9V and V3 for the FIGURE 7.6 Solution: ow) diw bobal (c) i The first step is to split the circuit into two single source circuits. These circuits (along with their series equivalents) are shown in Figures 7.6b and 7.6c. For Figure 7.6b, REQ = R2|| R3 = 15 || 10 N = 6 N and burn The results Figure 7.7a sho ple. Note the greater voltage P ■ Since V ■ Since W Polar REQ 60 VEQ = VA = (12 V) =2 V ✓ VA 12 V REQ + R₁ 36 Ω Since this voltage is across the parallel combination of R2 and R3, and V2A = V3A = 2 V E=12 V-2 V10 V₁ = VA - VEQ = 12 V 2V = 10 V Circuit Analysis Techniques * ✓ FIGURE 7
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