Three resistors R1 = 81.3 , R2 = 23.4 , R3 = 70.0 , and two batteries e m f1 = 4

Question

Three resistors R1 = 81.3 , R2 = 23.4 , R3 = 70.0 , and two batteries e m f1 = 40.0 V, and e m f2 = 369 V are connected as shown in the diagram below. The circuit has three branches that begin at the top middle of the diagram and re-connect at the bottom middle of the diagram. The left branch passes down through resistor R_1, then the positive pole of emf_1, then reaches the bottom middle. The middle branch passes through resistor R_2, then the negative pole of emf_2 then reaches the bottom middle. The right branch passes down through resistor R_3 then ends at the bottom middle. (a) What current flows through R1, R2, and R3?

Explanation / Answer

Given,

R1 = 81.3 , R2 = 23.4 , R3 = 70.0

e1 = 40 V ; e2 = 369 V

let left loop be loop1 and right be 2

-e1 + (81.3 + 23.4)I1 - 23.4 I2 - e2 = 0

104.7 I1 - 23.4 I2 = 409

For loop 2

e2 + (23.4 + 70)I2 - 23.4I1 = 0

93.4 I2 - 23.4 I1 = -369

I2 = (23.4 I1 - 369)/93.4

putting in 1

104.7 I1 - 23.4 (23.4 I1 - 369)/93.4 = 409

9778.98 I1 - 547.56I1 + 8634.6 = 38200.6

9231.42 I1 = 29566

I1 = 3.203 A

I2 = (23.4 x 3.203 - 369)/93.4 = -3.15

current through R2 = 3.20 + 3.15 = 6.35 A

Hence, Current through R1 ; I1 = 3.15 A

Current through R2 ; I2 = 6.35 A

Current through R3 ; I3 = 3.2 A

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