Class Management I Help Electric Circuits Begin Date: 9/21/2018 12:01:00 AM -- D

Question

Class Management I Help Electric Circuits Begin Date: 9/21/2018 12:01:00 AM -- Due Date: 10/1/2018 11:59:00 PM End Date: 10/5/2018 11:59:00 PM (5 %) Problem 21: Consider the circuit diagram in the figure E1 18 V 0.5 2.5 12 R 6.0 1.5 0.52 45 V Otheexpertta.com 50% Part (a) What is the equation which results when applying the loop rule to loop acdcba, in terms of the variables given in the figure? Grade Summary Deductions Potential 5% 95% Submissions Attempts remaining: 15 % per attempt) detailed view 1% 1% 1% 1% END DEL CLEAR Submit Hint I give up! Hints: 1 % deduction per hint. Hints remaining: 4 Feedback: 2 for a 0% deduction No meaningful feedback available for the current submission. E 50% Part (b) What is the current through the top loop 12, in amps, if the current through the middle of the circuit is 1,-1.42 A? 12-3, 16 v' Correct!

Explanation / Answer

here,

a)

using KVL in the loop aedcba

- R1 * I1 - r1 * I1 - R2 * I2 = - E1

R1 * I1 + r1 * I1 + R2 * I2 = E1

b)

R2 * I2 = ( E1 - (R1 + r1) * I1)

I2 = ( E1 - (R1 + r1) * I1) /R2

I2 = ( 18 - ( 0.5 + 6) * 1.42) /2.5

I2 = 3.51 A

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