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.

Answer 1

2.2*R1=1.7(3+R1)
2.2R1-1.7R1=5.1
.4R1=5.1
R1=...

Answer 2

Got 12.75 however that answer is incorrect.

Answer 3

well, it is right. Recheck to see you typed the problem right. I would be concerned with significant digits on this one, you gave the answer to four digits,I think I would have assumed the 3 ohm was 3.0, and take the answer to two digits.

Here is almost identical quiz problem: http://myspace.upike.edu/rarts/public_html/courses/physics/Physics/Physics2/hw/2015/HW_4%20-%20Current,%20Resistance,%20and%20Basic%20Circuits-preview3.pdf

on page 2, continued on 3.

Answer 4

The correct answer was actually 10.2

Answer 5

V2 = I*R2 = 1.7 * 3 = 5.1 Volts.

R1 = Change in voltage/Change in I = -5.1/(1.7-2.2) = 10.2 Ohms.

Answer 6

2.2R=1.7(3+R)
2.2R=5.1+1.7R
.5R=5.1
R=10.2