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can you explain the steps of the Va-Vc/R2+R3... and the next step where it says "Substitute in known components values. We know that Vb is 20V less than Va:"

can you explain the steps of the Va-Vc/R2+R3... and the next step where it says "Substitute in known components values. We know that Vb is 20V less than Va:"

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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can you explain the steps of the Va-Vc/R2+R3...

and the next step where it says "Substitute in known components values. We know that Vb is 20V less than Va:"

R1 1k R2 5ΚΩ b www 14 E2 20 V 13 www www C R3 3k n a R5 R4 2ΚΩ I1 1kΩ www 12
Transcribed Image Text:R1 1k R2 5ΚΩ b www 14 E2 20 V 13 www www C R3 3k n a R5 R4 2ΚΩ I1 1kΩ www 12
1₁ + 12 + 12-14 = 0 Use Ohm's law to replace currents with voltages and resistances Note how R2 and R3 are connected in series. We can combine them to form one equivalent resistor which will simplify our calculations: Gather terms: Va-Vc Va-Vc, Va - Vc + R₁ R4 R₂ + R₂ Substitute in known component values. We know that point Vb is 20 V less than Va: + N 7=1 Simplify: (multiply both sides by 8 kn) GE152 Now we can directly solve for Va: In = 0 Va-OV V - OV Va-OV OV - (Va - 20V) + 1 kn 2 ΚΩ 1 kn + = 0 8 kn 1₁ = Va +8V + 4V - 8(-Va+20V) = 0 V 21% = 160 V Eq 1 1₁ Vc - V₂ R₁ V₂ = 7.619 V With the voltage of Va known all currents can be solved for. V - OV 8 kn 7.619 V - OV 8 k = 0 = 0.952 mA Page 12 of 18 Fall Term, 2023
Transcribed Image Text:1₁ + 12 + 12-14 = 0 Use Ohm's law to replace currents with voltages and resistances Note how R2 and R3 are connected in series. We can combine them to form one equivalent resistor which will simplify our calculations: Gather terms: Va-Vc Va-Vc, Va - Vc + R₁ R4 R₂ + R₂ Substitute in known component values. We know that point Vb is 20 V less than Va: + N 7=1 Simplify: (multiply both sides by 8 kn) GE152 Now we can directly solve for Va: In = 0 Va-OV V - OV Va-OV OV - (Va - 20V) + 1 kn 2 ΚΩ 1 kn + = 0 8 kn 1₁ = Va +8V + 4V - 8(-Va+20V) = 0 V 21% = 160 V Eq 1 1₁ Vc - V₂ R₁ V₂ = 7.619 V With the voltage of Va known all currents can be solved for. V - OV 8 kn 7.619 V - OV 8 k = 0 = 0.952 mA Page 12 of 18 Fall Term, 2023
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