# Internal Resistance in a Battery

1. Mar 1, 2016

### mattbeatlefreak

1. The problem statement, all variables and given/known data
Three resistors are connected in series to a battery. The resistances are R1 = 15 Ω ,R2 = 25 Ω , and R3 = 30 Ω and the current through the 15-Ω, resistor is 2.7 A .

What is the potential difference across the battery terminals if the battery is ideal?

What is the potential difference across the battery terminals if the battery has an internal resistance of 5.0 Ω?

2. Relevant equations
V=iR

Vbatt = ξ*[R/(Rbatt+R)] (this equation was derived in my textbook from Vbatt = ξ - iRbatt)

3. The attempt at a solution
This problem is very straightforward. The total resistance for the first part is just the sum of the resistors since they are in parallel. Solving for the potential difference across the battery I get 189 V; this answer was accepted as correct.

However, when I use the next equation, I get 176.4 V for the second part. Neither 176 or 177 volts (177 in the case that you round 189 to 190) was accepted.
I even checked my answer with this website simulation thing http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dcex6.html

I feel confident that this is the correct answer, but it is not being accepted. Any thoughts on why? Thanks in advance!

2. Mar 1, 2016

### Staff: Mentor

Sounds a bit like a trick question to me. Note that you are not given the EMF of the battery.

3. Mar 1, 2016

### Staff: Mentor

The problem statement is a bit vague: It doesn't specify whether the same current, 2.7 A, is measured through the 15 Ω resistor in both cases. It also doesn't claim that the emf of the battery is the same in both cases, or if the second case involves a different battery altogether not just the addition of another resistance.

What voltage would you measure across the battery terminals if the current was the same in both cases?

Edit: Ah. I see that Doc Al got in there before me.

4. Mar 1, 2016

### mattbeatlefreak

I assumed that they were talking about the same battery since they reference it as "the" battery in both questions. Thanks for the tip, got the right answer now.