Class Management Help Chapter 21 Begin Date: 9/30/2017 12:00:00 AM -- Due Date:

Question

Class Management Help Chapter 21 Begin Date: 9/30/2017 12:00:00 AM -- Due Date: 10/8/2017 11:59:00 PM End Date: 12/19/2017 12:00:00 AM (7%) Problem 10: Consider the circuit in the diagram, with sources of emf listed below. Randomized Variables 629 V 0.10 5.0 20 = 46 V ' I 0.50 4012 4=39 V 0.20 iO.05 h ©theexpertta.com G> 33% Part (a) Find 11 in amps Grade Summary Deductions Potential 0% 100% Submissions sin0 tan() | | ( cos Attempts remaining: 8 (0% per attempt) detailed view asin0 acos0 atan acotan0 sinhO cosh0 tanh0 cotanhO Degrees O Radians 0 Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining Feedback: 0% deduction per feedback. - 33% Part (b) Find 12 in amps 33% Part (c) Find 13 in amps All content © 2017 Expert TA, uc

Explanation / Answer

From the top loop, bdea,
E1 - I1(R1 + r1 + R5) - (I1 - I2)(r2 + R2) - E2 = 0
29 - I1 (5 + 0.1 + 20) - (I1 - I2)(0.5 + 40) - 46 = 0
- 17 - 25.1 I1 - (I1 - I2) 40.5 = 0        ...(1)

From the bottom loop aegj,
E2 - E4 + E3 - (I2 - I1)(r2 + R2) - I3(r4 + r3 + R3) = 0
46 - 39 + 15 - (I2 - I1)(0.5 + 40) - I3(0.2 + 0.05 + 78) = 0
22 - (I2 - I1) 40.5 - 78.25 I3 = 0     ...(2)

(1) + (2) gives
5 - 25.1 I1 - 78.25 I3 = 0 ...(3)
Consider the point a,
I3 = I1 + I2    ...(4)
Substituting (4) in (3),
5 - 25.1 - 78.25 (I1 + I2) = 0
5 - 103.35 I1 - 78.25 I2 = 0
I2 = 1/7825 [5 - 103.35 I1]       ...(5)

Substituting (5) in (1),
- 17 - 25.1 I1 - (I1 - 1/78.25 [5 - 103.35 I1]) 40.5 = 0
-17 + 2.59 - 119.09 I1 = 0
I1 = 0.121 A           ...(6)

Substituting (6) in (5),
I2 = 1/78.25 [5 - 103.35 x 0.121]
= - 0.096 A

I3 = I1 + I2 = 0.025 A

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