Aloop circuit has a resistance of r1 and a current of 2.2 a. the current is reduced to 1.4 a when an additional 3.1 ω resistor is added in series with r1.
what is the value of r1? assume the inter- nal resistance of the source of emf is zero.
answer in units of ω.

Okay so here's the approach I took:

The potential difference in each of the circuits must be the same so if we derive equations for both the potential differences we can set them equal to each other and solve for R1:

In the first circuit

V = 2.2(R1)

In the second we have to find the equivalent resistor, since they are connected in series:

1/R1 + 1/R2 + 1/R3... = Rt

We have R2 so...

1/R1 + 1/3.1 = Rt

1/R1 + 0.323 = Rt

So...

V = 1.4(1/R1 + 0.323)

Set those equal:

2.2R1 = 1.4(1/R1 + 0.323)

2.2R1 = 1.4(1/R1) + 0.4522

Now multiply everything by R1 so we can combine like terms:

2.2R1^2 = 1.4 + 0.4522R1

Isolate to form a quadratic

2.2R1^2 - 0.4522R1 - 1.4 = 0

Solving this quadratic:

R1 = 0.90708 or R1 = -0.701

Since R cannot be negative

R1 = 0.907 ohms

when a device is used in a circuit in which the voltage is 81 v, the current flowing through the device is 3 a. what is the resistance of the device?

a. 27 ω

b. 40.5 a

c. 243 ω

d. 243 a

the correct answer is a. 27