Help with multiloop circuit problem

[lcybulsk]

Homework Statement

In Fig. 27-50, two batteries of emf ε = 12.0 V and internal resistance r = 0.300 Ω are connected in parallel across a resistance R. (a) For what value of R is the dissipation rate in the resistor a maximum? (b) What is that maximum?

The figure can be seen here...problem 27.39 http://www.cabrillo.edu/~cfigueroa/4B/4Bproblem_sets/Chap27_problems.pdf

Homework Equations

V=IR
Loop/Junction Rules
P=I^2R (Internal resistance dissipation)

The Attempt at a Solution

I used the loop rule to find an equation for both emf devices and resistor (same emf, so same equation i presume)
E - IR - Ir = 0

However when i get to the junction rule and finding an equation for current, i cannot figure out if the current in the emf and the current in the resistor are the same

If the currents are the same i did E- (R-r)I=0, then solved for I

Once i found I as I= E/(R-r) i substituted it in the equation for power P=I^2R in order to get a function where i can differentiate it

P= (E/(R-r)^2R

Is this correct? I am confused on if i can assume the currents are the same in the resistor R and the batteries

 

[gneill]

You might be better off finding the Thevenin equivalent of the two batteries first. That will simplify things considerably since you'll be dealing with a single source and a single internal resistance.

If the above paragraph leaves you mystified because you've never heard of Thevenin or his equivalent, then I suggest that you write out the Kirchoff loop equations. Solve for the current through the load resistor.

 

[SammyS]

When you write: E - IR - Ir = 0, you are assuming that the current through each battery is the same as the current through R. That violates the junction rule.