Solve Non-Ideal Battery Homework: Find Power Through R2

[kjlchem]

Homework Statement

A circuit is constructed with five resistors and one real battery as shown above right. We model. The real battery as an ideal emf V = 12 V in series with an internal resistance r as shown above left. The values for the resistors are: R1 = R3 = 30 Ω, R4 = R5 = 78 Ω and R2 = 138 Ω. The measured voltage across the terminals of the batery is Vbattery = 11.56 V.

What is the power through R2?

Homework Equations

V=IR

P = IV
Kirchoff's laws

The Attempt at a Solution

I have already found the currents I1 and I3. Also, I think that I2 = I1 - I3. I tried plugging that into P = IV, and did not get the correct answer. It could be my units or my previous equation. I have current in amps and voltage in volts. I do not know what I am doing wrong.

 

[PeterO]

Homework Statement

A circuit is constructed with five resistors and one real battery as shown above right. We model. The real battery as an ideal emf V = 12 V in series with an internal resistance r as shown above left. The values for the resistors are: R1 = R3 = 30 Ω, R4 = R5 = 78 Ω and R2 = 138 Ω. The measured voltage across the terminals of the batery is Vbattery = 11.56 V.

What is the power through R2?

Homework Equations

V=IR

P = IV
Kirchoff's laws

The Attempt at a Solution

I have already found the currents I1 and I3. Also, I think that I2 = I1 - I3. I tried plugging that into P = IV, and did not get the correct answer. It could be my units or my previous equation. I have current in amps and voltage in volts. I do not know what I am doing wrong.

You said you had calculated the currents, what did you calculate the Voltage drop across R2 to be? (in order to use VI]
If you didn't know V you could have used the I²R expression instead.

 

[fmd826]

how did you find I1 and I3? I am doing the same problem and need help.

 

[fmd826]

nvm I got them. and using p=I^2R did get the answer for power through R2.

 

[CWatters]

This is how i'd do it..

Replace R3-5 and R2 with an equivalent Rx.

Then you have a battery supplying R1 + Rx. You know the measured battery voltage so you can work out the current I1.

The potential divider rule will give you the voltage across Rx (which is equal to the voltage across R2).

Knowing the voltage across R2 you can work out the current through R2 or calculate the power directly using P=I^2 * R

 

[ayajek]

I am working on the same problem, but can't figure out how to find r (resistance in the battery) . I found out I1 by ignoring the resistance in the battery, and finding total resistance, then dividing the battery's voltage by it, and yet got a correct answer.

 

[gneill]

I am working on the same problem, but can't figure out how to find r (resistance in the battery) . I found out I1 by ignoring the resistance in the battery, and finding total resistance, then dividing the battery's voltage by it, and yet got a correct answer.

Since you are given the voltage across the battery terminals as a measured value, presumably on the operating circuit, the internal resistance r is irrelevant to further calculations. Whatever the actual values of V and r, the measured value at the terminals under operating conditions is all you need to know provided the circuit is not altered. In other words you have a known, fixed potential change that you can plug into any equation that needs to "know" the potential across the battery component.