# Current through a branch of a circuit

1. Oct 17, 2009

### exitwound

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

2. Relevant equations

V=ir
E=V-ir

3. The attempt at a solution

First part is solved. The EMF of the battery is higher than the actual output voltage due to the internal resistance of the battery. E=11.25, V measured is 10.5.

The second part is boggling my mind.

When both switches are closed, the circuit acts as resistors in parallel. Voltages across the branches would be the same. Current through each branch would vary.

Using Kirkhoff's loop rules, starting the LOWER RIGHT corner of the diagram:

Loop 1:
$i_b(R_b)+E -i_s=0$

Loop 2:
$i_b(R_b)+e-i_L=0$

Equation 3:
$i_b = i_s + i_L$

The current through the lights $(i_L)$ is given as 14A. Therefore, we can use the idea that the voltage across the battery is the same as the voltage across the lights to solve for the resistance of the lights.

V=IR
$10.5 = 14(R_L)$
$R_L = .75$

At this point, I don't know how to proceed to get any closer to the solution.

2. Oct 17, 2009

### cepheid

Staff Emeritus
That's wrong because the output voltage of the battery is not going to be the same (i.e. it won't be 10.5 V) in the second case (when both the starting motor and the lights are on). The reason is because adding another load means that more total current is drawn from the battery, and therefore there will be a larger potential drop across the battery's internal resistance. You should use the data given in the *first* situation to deduce the resistance of the lights (which doesn't change):

10.5 V / (15 A) = R_L

Then, the sequence will go something like this:

1. What battery output voltage is required across R_L for it to draw only 14 A?
2. By how much must the *total* current drawn have increased in order for the battery output voltage to be reduced from 10.5 V to the value calculated in (1)?
3. Based on the *total* current calculated in (2), how much is being drawn by the starting motor?

3. Oct 17, 2009

### exitwound

Okay so if:

$V_L=i_L(R_L)$
$10.5=15(R_L)$
$R_L=.7$Ω

$V_L=i_L(R_L)$
$V_L=14(.7) = 9.8V$

So this says that there's 9.8V across the terminals of the battery and the lights, and the starting motor, right?

I don't quite know what you're asking in the 2nd question. If I now compute the voltage difference across the battery terminals I get (V_a is lower right corner of that circuit):

$V_a-i_b(R_b) + E = V_b$
$V_b-V_a = E-i_b(R_b)$
$9.8=11.25-i_b(.05)$
$i_b=29 A$

This would mean that there is now a 28A current flowing from the battery. Am I on the right track so far?

4. Oct 17, 2009

### cepheid

Staff Emeritus
Yeah, that's exactly what I was asking in (2). If the output voltage is 9.8 V, how much total current needs to be drawn from the battery to make that happen (given that internal resistance). That is exactly what you have calculated. Now you know the total current drawn from the battery.

5. Oct 17, 2009

### exitwound

So if the battery is pushing 28A, and the light branch is eating up 14A, then the starting motor branch must be using the other 14A, correct?

6. Oct 17, 2009

### cepheid

Staff Emeritus
Well, yes, except that the answer you calculated was 29 A, and you suddenly changed it to 28 A in mid-post by mistake.

7. Oct 17, 2009

### exitwound

Oops! Typo.

Thanks for the help!