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)
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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Question
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9e543a2e-d396-4207-bcbf-7aa1f6de7a19%2F71853fe5-864a-4ca9-ad50-d143076b3eaa%2Fy1n1al_processed.png&w=3840&q=75)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9e543a2e-d396-4207-bcbf-7aa1f6de7a19%2F71853fe5-864a-4ca9-ad50-d143076b3eaa%2Fqm6ykvh_processed.png&w=3840&q=75)
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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