# Voltage and Current in circuits( paral. and series) involv. internal resistance

1. Oct 19, 2009

### cristina045

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

Two resistors R1= 132 ohms and R2= 56 ohms are connected to a 6.00 V battery with an internal resistance of 5.00 ohms. Find the voltage across R2 and the current through R1
a) if the resistors are connected in series, and
b) if they are connected in parallel.

2. Relevant equations

epsilon = IR + Ir
r : internal resistance
epsilon : emf of the battery

V= IR

3. The attempt at a solution

a) So i'm really not sure but since were given the internal resistance of the battery , i think the 6.00 V represents the EMF ( thing is , it could also just represent the voltage and i don't know if i have to actually calculate the emf using the internal resistance) so what i did , i used the equation : epsilon = IR1 + Ir to find the current through R1.
Since it's the resistors would be connected in series, the current would remain the same throughout the circuit. So :

6.00= (137)I so I = 0.0438A -->corrected to 6.00 = I(132+5.00+56.0) I= 0.311A

i'm not sure this is right so if someone could help me and let me know if i'm on the right track.

For the voltage across R2 i tried using the formula : V=IR
so 0.0438(56 ohms) = 2.45v across R2 --> corrected to 0.0311(56) = V across R2

thing is, since the voltage across R2 was asked first, i'm wondering if I should be able to find it before actually finding my current , but don't know what formula I would use in that case.

b) using the formula Rtot = 1/R1 + 1/R1+1/Rr I found my total resistance to be 0.225 ohms if my resistors are placed in parallel. Using I = V/R i found my total current to be 26.67 A. Now though , i don't really know how to calculate my current across R1 and my voltage across R2.

I used :

Current through R1 = Itot ( R1/Rtot)
so IR1 = 26.67 (132/193)
I through R1 = 18.2 A

and again, i'm not sure how to get my voltage

Last edited: Oct 20, 2009