Fabio Phys 208 HW A7 Begin Date: 3/5/2017 12:00:00 AM Due Date 3/14/2017 11:59:0

Question

Fabio Phys 208 HW A7 Begin Date: 3/5/2017 12:00:00 AM Due Date 3/14/2017 11:59:00 PM End Date 3/14/2017 11:59:00 PM (6%) Problem 18: Consider the circuit depicted in the diagram. 0.10 E2 480 5.0 200 0.500 400 0200 36.0V 0.050 h otheexpertta.com A 50% Part (a) Write the equation which results from applying the loop rule to loop akledcba. Grade Summary Potential 100% a 7 8 9 Submissions Attempts remaining: 4 5 6 per attempt) 2 3 detailed view CLEAR Submit Feedback: deduction per feedback. Hints: 0% deduction per hint. Hints remaining: 2.

Explanation / Answer

part a:

loop rule uses Kirchoff's Voltage Law which states sum of voltage drop across a closed loop is zero.

so here the equation will be:

E2-I2*r2-I2*R2+I1*R5+I1*r1-E1+I1*R1=0

==>48-0.5*I2-40*I2+20*I1+0.1*I1-24+5*I1=0

==>25.1*I1-40.5*I2+24=0

part b:

given I2=55 A,

from the equation obtained in part a,

25.1*I1-40.5*55+24=0

==>I1=(40.5*55-24)/25.1=87.788 A

hence current in top of the loop is 87.788 A

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