Finding ##R_1## & ##R_2## in Series Circuit

[kaspis245]

Homework Statement

Two resistances, $R_1$ and $R_2$, are connected in series across a 12 V battery. The current increases by 0.2 A when $R_2$ is removed, leaving $R_1$ connected across the battery. However, the current increases by just 0.1 A when $R_1$ is removed, leaving $R_2$ connected across the battery. Find $R_1$ and $R_2$.

Homework Equations

Ohm's law.

The Attempt at a Solution

$ΔI_1=0.2 A$
$ΔI_2=0.1 A$(1) $I=\frac{V}{R_1}-ΔI_1$

(2) $I=\frac{V}{R_2}-ΔI_2$

(3) $I=\frac{V}{R_1+R_2}$

From (1) and (2):
(4) $R_2=\frac{120R_1}{R_1+120}$

From (2) and (3):
(5) $\frac{V}{R_2}-ΔI_2=\frac{V}{R_1+R_2}$

Then I put (4) into (5) and get:
$R_1=0.503Ω$
$R_2=0.5009Ω$

The correct answer is $R_1=35Ω$, $R_2=50Ω$.

I fail to spot my mistake. Please help.

 

[gleem]

Recheck eq. 4

 

[kaspis245]

I see, it's suppose to be $R_2=-\frac{120R_1}{R_1-120}$, but then I get two $R_1$ values:
$R_1=35.15$
$R_1=204.85$
Both seem correct. How come the authors chose only one of them?

 

[gneill]

I see, it's suppose to be $R_2=-\frac{120R_1}{R_1-120}$, but then I get two $R_1$ values:
$R_1=35.15$
$R_1=204.85$
Both seem correct. How come the authors chose only one of them?

Calculate the value of $R_2$ associated with each choice of $R_1$.

 

[gleem]

In solving a quadratic eq you sometimes get extraneous roots. You must determine if it is a plausible soution. the 204.85 ohm solution does not satisfy eq 4 predicting a neg value for R2. So discard it.