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A loop circuit has a resistance of R1 and a current of 1.9 A. The current is reduced to 1.5 A when an additional 3.1 Ω resistor is added in series with R1. What is the value of R1? Answer in units of Ω.

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MechEngineer

Answer:

11.625 Ohm

Explanation:

Let V be the Voltage charge of the loop, as this is constant we know that before the resistor addition the current I is:

V/R1 = 1.9 or V = 1.9R1

After the resistor addition to series R = R1 + 3.1

I = V/R = V/(R1 + 3.1) = 1.5

We can substitute V = 1.9R1

1.9R1 = 1.5R1 + 1.5*3.1

0.4R1 = 4.65

R1 = 4.65/0.4 = 11.625 Ohm

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