(33%) Problem 1: Consider the following circuit of three resistors (R1, R2, and

Question

(33%) Problem 1: Consider the following circuit of three resistors (R1, R2, and R), with batteries that have emfs = 18 V and 6-49 V, and internal resistances r1 and r2. 0.5 R2 2.5 Randomized Variables 1 18 V 4 =49 V 12R 6.0 1.5 0.5 ©theexpertta.com 33% Part (a) Find the current /i, in amps. Grade Summary Deductions 0% Potential 100% ' Submissions Attempts remaining: 20 (4% per attempt) detailed view sinO cosO cotan asin)acos) atan) acotansinh) coshO tanhOcotanhO o Degrees o Radians tan() 15 . 4 5 6 1 2 3 Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: Feedback: 0% deduction per feedback - A 33% Part (b) Find the current 12, in amps. 33% Part (c) Find the current 13 in amps.

Explanation / Answer

I1 = I2 + I3

For the upper loop KVL is -

18 = 0.5*I2 + 6*(I2 + I3) + 2.5*I2 = 9*I2 + 6*I3

=> I3 + 1.5*I2 = 3

=> I3 = 3 - 1.5*I2-------------------------------------------(a)

For the lower loop -

49 = 1.5*I3 + 6*(I2 + I3) + 0.5*I3 = 6*I2 + 8*I3

put the value of I3 -

6*I2 + 8*(3 - 1.5*I2) = 49

=> 6*I2 - 12*I2 = 25

=> I2 = - 25 / 6 = -4.17 A

and I3 = 3 - 1.5*(-4.17) = 9.3 A

I1 = I2 + I3 = -4.17 + 9.3 = 5.13 A.

So,

Part (A) - I1 = 5.13 A

Part (B) - I2 = -4.17 A

Part (C) - I3 = 9.3 A

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