The current in a loop that has a resistance of R1 has acurrent of 2.00 A. The cu

Question

The current in a loop that has a resistance of R1 has acurrent of 2.00 A. The current is reduced to 1.6 A when anadditional resistor R2 = 3.00 ohm is added to the series withR1. What is the value of R1? Any help? I think I solved it and got the values V = 24, andR1 = 12. Just want to make sure its correct. The current in a loop that has a resistance of R1 has acurrent of 2.00 A. The current is reduced to 1.6 A when anadditional resistor R2 = 3.00 ohm is added to the series withR1. What is the value of R1? Any help? I think I solved it and got the values V = 24, andR1 = 12. Just want to make sure its correct.

Explanation / Answer

V=IR in the first case V=I1R1 In the second case V= I2(R1 +R2) subtructing first equation from the second 0 = I1R1 -I2(R1 +R2) solving for R1 we hav e (I2   -I1)R1   = I2R2 R1  =I2R2 /(I2   - I1) R1  = 1.60 x 3.00 /(2.00  - 1.60) R1  = 12.0 R1  =I2R2 /(I2   - I1) R1  = 1.60 x 3.00 /(2.00  - 1.60) R1  = 12.0 R1  = 1.60 x 3.00 /(2.00  - 1.60) R1  = 12.0

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