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

[cristina045]

Homework Statement

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.

Homework Equations

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

V= IR

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

 

[jbsteele]

across R2 without knowing the current through it. If someone could help me out with this I would really appreciate it. Thank you

 

[tomcue]

across R2.
Your calculations for the series circuit are correct. In this case, the voltage across R2 can be calculated as V = IR = (0.0438 A)(56 ohms) = 2.45 V. This is because in a series circuit, the potential difference (voltage) is divided among the resistors according to their individual resistances.

In the parallel circuit, your calculations for the total resistance and current are correct. To find the current through R1, you can use the equation I = V/R, where V is the total voltage (6.00 V) and R is the individual resistance of R1 (132 ohms). Therefore, the current through R1 is 6.00 V / 132 ohms = 0.0455 A.

To find the voltage across R2, you can use the formula V = IR, where I is the total current (26.67 A) and R is the individual resistance of R2 (56 ohms). Therefore, the voltage across R2 is (26.67 A)(56 ohms) = 1493.52 V. This may seem like a large number, but it is because the resistors are placed in parallel and the total current is very high.

Overall, your understanding of series and parallel circuits and internal resistance is correct. Keep in mind that in series circuits, the total current is the same throughout the circuit, while in parallel circuits, the total voltage is the same across each branch. Additionally, in parallel circuits, the total resistance is less than the individual resistances, resulting in a higher total current.