Problem?unitltemID 2757563&enrollmentiD-247708; A circuit is constructed with si

Question

Problem?unitltemID 2757563&enrollmentiD-247708; A circuit is constructed with six resistors and two batteries as shown. The battery voltages are V1-18 V and V2 - 12 V. The positive terminals are indicated with a +sign, The values for the resistors are: Ri . RS " 45 , R2-R6 . 101 R3-740, and R4-1280. The positive directions for the currents l1, 12 and l3 are indicated by the directions of the arrows. T R cated by the directions R3 b R, R4 1) What is V4, the magnitude of the voltage across the resistor R? 8.87 Submit Your submissions: 8.87 Computed value: 8.87 Submitted: Thursdsy, July 13 at 9:44 PM Feedback: Your answer has been judged correct; the exact answer is: 8.878612716763. 2) what is l,, the current that flows through the resistor Ry? A positive value for the current is defined to be in the direction of the arrow. Submit Your submissions:-158 x Computed value: 158 Submitted: Thursday, July 13 at 9:45 PM Feedback: a) What is tz, the curent that flows through the resistor Ry? A positihe value for the current is defined to be in the direction of the arroww. AL submit 49 What blj,the current that flows thrugh the reito RiA ptie value for the current is defined to be in the direction of the arrow. A Submt

Explanation / Answer

R1 = R5 = 45 Ohm ; R2 = R6 = 101 Ohm ; R3 = R4 = 128 Ohm

V1 = 18 V and V2 = 12 V

1)V4 = I4 R4

I4 = [V2/(R4 + R5)]

V4 = [V2/(R4 + R5)] R4

V4 = 12/(128 + 45)]x128 = 8.88 Volts

Hence, V4 = 8.88 Volts

2)from junction rule

I2 = I1 + I3

from loop rule

I1R1 - V1 - I3R3 = 0

I2 R2 + I3R3 + V1 + I2R6 - V2 = 0

solving the above three simultaneously, we get

I3 = R1/ [R1R3 + (R1 + R3)(R2 + R6)][V2 - V1[1 + (R2 + R6)/R1]]

I3 = 45/[45 x 74 + (45 + 74)(101 + 101)] [12 - 18[1 + (101 + 101)/45] = -0.139 A

Hence, I3 = -0.139 A

3)Solving the three eqn in above part for I2 we get

I2 = [ V1/R1 + I3(R1 + R3)/R1]

I2 = [18/45 + -0.139(45 + 74)/45] = 0.032 A

Hence, I2 = 0.032 A

4)I1 will be such that

I1 = I2 - I3

I1 = 0.032 - (-0.139) = 0.171 A

Hence, I1 = 0.171 A

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