Question

8 39 o In Fig. 27-50, two batteries with an emf 12.0 V and an internal resistance r 0.300 ? 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?

Answer 1

(A) Here as we know that dissipation rate or maximum energy will transfer when the internal resistance of the battery is equalled to load resistance

As we can see that from the diagram that Internal resistance (r) of batteries are parallel to each other

So, the effective resistance is = (r*r) /(r+r) = r/2 = 0.3/2 = 0.15 ohms

And maximum energy transfer occurs when R = 0.15 ohms

(B) The maximum energy dissipation rate into R = [E^2] /(r/2)

Maximum energy dissipation = 2(E^2)/r = 2(12^2)/0.3 = 960W