l Help Chapter 21 & 22 Begin Date: 8/28/2017 12:00:00 AM Due Date: 9/28/2017 11:
ID: 1658547 • Letter: L
Question
l Help Chapter 21 & 22 Begin Date: 8/28/2017 12:00:00 AM Due Date: 9/28/2017 11:59:00 PM End Date: 12/8/2017 12:00:00 AM (8%) Problem, 5: Consider a circuit shown in the figure. Ignore the intemal resistances of the batteries. Randomized Variables 5-44 V R3 = 6S2. Otheexpertta.com 14% Part (a) Express the current 1, going through resistor R, in terms of the currents 12 and 13 going through resistors R2 and R3 Use the direction of the currents as specified in the figure Grade Summary 100% Attempts remaining 0 per attemp) detailed view Hint Feedback:% deduction per feedback Hints:-% doluction per hint. Hints rernaining 14% Part (b) Write the equation of potential change in loop EBAF in terms of the circuit elements. F4 14% Part (d) Write the equation of potential change in loop DCAF in terms of the circuit elements. Solve the three equations to get 13. 14% Part (c) 14% Part (e) Calculate the numerical value of 13 in A. 14% Part (f) Calculate the numerical value of, in A. 14% Part (g) Calculate the numerical value of 1, in A.Explanation / Answer
(a)
Apply KCL for point E
I1 - I2 - I3 = 0 ......(1)
(b)
Apply KVL loop EBAF,
E1 - I1*R1 - I3*R3 = 0 ......(2)
(c)
Apply KVL for loop DCAF,
-E2 + I2*R2 - I3*R3 = 0 .......(3)
(d)
From above equations,
I1 - I2 - I3 = 0
(l1 = l2 + l3)
44 - I1*9 - I3*6 = 0
put the value of l1,
44 - (12 + 13)*9 - l3*6 = 0
44 - 9*l2 - 12*l3 = 0 ......(4)
-24 + l2*9 - l3*6 = 0 .......(5)
From equation (4) and (5)
20 - 18*l3 = 0
(e)
by solving,
I3 = 1.11 A
(f)
put the vaue of l3 in eq(5)
-24 + l2*9 - 1.11*6 = 0
l2 = 3.40 A
(g)
l1 - 3.40 - 1.11 = 0
l1 = 4.51 A
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.