Volts = V = iR = 2.2 R1
so
2.2 R1 = 1.7 (R1+3)
2.2 R1 = 1.7 R1 + 5.1
.5 R1 = 5.1
R1 = 10.2
A loop circuit has a resistance of R1 and a current of 2.2 A. The current is reduced to 1.7 A when an additional 3 Ω resistor is added in series with R1. What is the value of R1? Assume the internal resistance of the source of emf is zero.
I got 12.75 however that isn't right
2 answers
V2 = I*R2 = 1.7 * 3 = 5.1 Volts.
R1 = Change in voltage/change in current = -5.1/(1.7-2.2) = 10.2 Ohms.
R1 = Change in voltage/change in current = -5.1/(1.7-2.2) = 10.2 Ohms.
4 answers