Thank you in advance for your help. I spent hours trying to figire Parts a-g out

Question

Thank you in advance for your help. I spent hours trying to figire Parts a-g out. thank you so much

(7%) Problem 8: Consider a circuit shown in the figure. Ignore the internal resistances of the batteries. Randomized variables 01 48 V 28 R3 92. Otheexpertta.com A 14% Part (a) Express the current Il going through resistor RI in terms of the currents I2 and I going through resistors R2 and R3. Use the direction of the currents as specified in the figure Grade Summary Deductions Potential 100% Submissions HOME Attempts remaining: 8 4 5 6 0% per attempt) detailed view 1 2 3 CLEAR Feedback Igive up! Submi Hint Hints: 0 deduction per hint. Hints remaining Feedback: 0% deduction per feedback. 14 Part (b) Write the equation of potential change in loop EBAF in terms of the circuit elements 14% Part (c) Write the equation of potential change in loop DCAF in terms of the circuit elements. A 14% Part (d Solve the three equations to get 13 A 14% Part (e) Calculate the numerical value of I3 in A A 14% Part (f Calculate the numerical value of I2 in A A 14% Part (g Calculate the numerical value of Il in A

Explanation / Answer

Given

   e1= 48V,e2= 28 V
   R1= 7 ohm,R2 = 9 ohm, R3 = 9 ohm

writing the kirchhoff voltage rule for upper loop

e1-i1R1 -i3R3 =0

for the lower loop

e2+e1-i1R1-i2R2 = 0

and the jucntion at middle near e1 is applying kirchhoff current rule

Part a)

   i2+i3 = i1

Part b) equaitio of potential change in the loop EBAF in terms of circuit elements is

   e1-i1R1 -i3R3 =0

Part c) equaitio of potential change in the loop CDAF in terms of circuit elements is

   e2+e1-i1R1-i2R2 = 0

e1- R1(i2+i3) -i3R3 = 0

e2+e1-R1(i2+i3) -i2R2 =0
- - + +       (subtraction)
----------------------------

-e2 +i2R2-i3R3 =0

substitute the values

-28 +i2*9 -i3*9 =0

9(i2-i3) = 28 ===> i2-i3 = 3.11

2e1+e2 -2R1(i2+i3)-i3R3-i2R2 = 0

   ===> 2*48+28-2*9(i2+i3)-9i3-9i2 = 0
   96+28 -18(i2+i3) - 9i3 -9i2 =0

   124 - 18(i1) -9(i2+i3) = 0

   124 -18i1 -9i1 = 0 ==> i1 = 4.5926 A

now

   e2+e1-i1R1-i2R2 = 0 implies

   48+28 -4.5926*7-i2*9 = 0 ====> i2 = 4.8724 A

and substitute the values of i1,i2 in the equation i1 = i2+i3

       i3 = i1-i2 = 4.5926 - 4.8724 = -0.2798 A

Part e) i3 = -0.2798 A

PArt f) i2 = 4.8724 A

part g) i1 = 4.5926 A

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