Finding R1 when series resistor reduces loop current from 3A to 1.6A

[pattiecake]

ok, so the problem states: the current in a loop circuit that has a resistance of R1 is 3A. The current is reduced to 1.6A when an additional resistor R2=2ohms is added in series with R1. What is the value of R1?

I drew two diagrams. My logic was that the voltage would be the same, b/c the voltage is supplied by the battery which does not change. Thus you have V=I1R1 for the first scenario; and you have V=I2(R1+R2). Since V is the same, you can set the two equations equal to each other, and solve for R1.

Is the logic right so far? If so, maybe I just have an algebratic error...

(3A)R1=(1.6A)(R1+2ohms)
(3/1.6)=(R+2)/R
1.875=1+R/2
.875=R/2
R=1.75 ohms?

 

[Chen]

That is almost correct, you made a mistake here:
$$\frac{3}{1.6} = \frac{R+2}{R}$$
Needs to become:
$$1.875 = 1 + \frac{2}{R}$$

 

[ShawnD]

I get R1 = 2.285 from breaking that down 2 different ways.

$$(3)(R_1) = (1.6)(R_1 + 2)$$

$$3R_1 = 1.6R_1 + 3.2$$

$$1.4R_1 = 3.2$$

$$R_1 = 2.285$$



$$(3)(R_1) = (1.6)(R_1 + 2)$$

$$\frac{3}{1.6} = \frac{R_1 + 2}{R_1}$$

$$1.875 = 1 + \frac{2}{R_1}$$

$$0.875 = \frac{2}{R_1}$$

$$R_1 = \frac{2}{0.875}$$

$$R_1 = 2.285$$

 

[pattiecake]

thank you! thank you! thank you!