b. E2 R3 Figure 1: 1. In figure 1 Kirchhoff's rules read: -12 R2 - €1 + IR1 - €2 = 0 (a) : €2 - I1 R1 - I3 R3 = 0 (True, False) I3 = I1 + 12 R3+e2 R3+e2 R2 R1 R2+R2R3+R3R1 Rj+jR3+e2 R3 RIR2+R2R3+R3R1 €2 R2- R I = (b) : I2 = (True, False) 13 = (c) : Va - Va = -13 R3 = €1 + 12 R2 = I R1 - €2 (True, False) (d) The direction of the current I2 is opposite of what is assumed in the figure (True, False) (e) If e1 = 12V ; €2 = 10V ; RỊ = 0.012; R2 = 1N ; and R3 = .06N , then |I1| = 171.7A ; |I2| = .3A ; and |13| = 171.4A (True, False) (f) For the same numerical values given in part (e) the energy released by the battery, 1, for a time interval of 1.5 second is : Energye = (I1)t = 3090.5J (True, False)
b. E2 R3 Figure 1: 1. In figure 1 Kirchhoff's rules read: -12 R2 - €1 + IR1 - €2 = 0 (a) : €2 - I1 R1 - I3 R3 = 0 (True, False) I3 = I1 + 12 R3+e2 R3+e2 R2 R1 R2+R2R3+R3R1 Rj+jR3+e2 R3 RIR2+R2R3+R3R1 €2 R2- R I = (b) : I2 = (True, False) 13 = (c) : Va - Va = -13 R3 = €1 + 12 R2 = I R1 - €2 (True, False) (d) The direction of the current I2 is opposite of what is assumed in the figure (True, False) (e) If e1 = 12V ; €2 = 10V ; RỊ = 0.012; R2 = 1N ; and R3 = .06N , then |I1| = 171.7A ; |I2| = .3A ; and |13| = 171.4A (True, False) (f) For the same numerical values given in part (e) the energy released by the battery, 1, for a time interval of 1.5 second is : Energye = (I1)t = 3090.5J (True, False)
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
Please do 1 D, E, and F
![R2
E2
R3
Figure 1:
1. In figure 1 Kirchhoff's rules read:
-12 R2 - €1 + I1 R1 - €2 = 0
(a) :
€2 - I1 R1 - I3 R3 = 0
(True, False)
I3 = I1 + 12
I = R3+€2 R3+e2 R2
R1 R2+R2 R3+R3R1
R+e R3+€2 R3
R R2+R2 R3+R3 R1
(b) :
I2 =
(True, False)
I3 =
(c) : Va - Va = -13 R3 = €1 + I2 R2 = I R1 - €2 (True, False)
(d) The direction of the current I2 is opposite of what is assumed in the figure
(True, False)
(e) If e1 = 12V ; €2 = 10V ; R1 = 0.012 ; R2 = 12 ; and R3 = .06n, then |I1| = 171.7A ; |I2| = .3A ; and |13| = 171.4A (True, False)
(f) For the same numerical values given in part (e) the energy released by the battery, 1, for a time interval of 1.5 second is : Energye =
(I€1)t = 3090.5J (True, False)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F90ba196a-a4da-4c5b-97fd-f3d58be17337%2F90d825c9-e072-43c2-a731-8f121d5ad576%2Fnlsyxa2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:R2
E2
R3
Figure 1:
1. In figure 1 Kirchhoff's rules read:
-12 R2 - €1 + I1 R1 - €2 = 0
(a) :
€2 - I1 R1 - I3 R3 = 0
(True, False)
I3 = I1 + 12
I = R3+€2 R3+e2 R2
R1 R2+R2 R3+R3R1
R+e R3+€2 R3
R R2+R2 R3+R3 R1
(b) :
I2 =
(True, False)
I3 =
(c) : Va - Va = -13 R3 = €1 + I2 R2 = I R1 - €2 (True, False)
(d) The direction of the current I2 is opposite of what is assumed in the figure
(True, False)
(e) If e1 = 12V ; €2 = 10V ; R1 = 0.012 ; R2 = 12 ; and R3 = .06n, then |I1| = 171.7A ; |I2| = .3A ; and |13| = 171.4A (True, False)
(f) For the same numerical values given in part (e) the energy released by the battery, 1, for a time interval of 1.5 second is : Energye =
(I€1)t = 3090.5J (True, False)
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