# Series Wiring

1. Jul 10, 2015

### kaspis245

1. The problem statement, all variables and given/known data
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 $R_2$.

2. Relevant equations
Ohm's law.

3. 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 $R_1=35Ω$, $R_2=50Ω$.

2. Jul 10, 2015

### gleem

Recheck eq. 4

3. Jul 11, 2015

### 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?

4. Jul 11, 2015

### Staff: Mentor

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

5. Jul 11, 2015

### 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.