1. A battery with an emf of 8.2 V and internal resistance of 1.11 Ω is connected across a load resistor R. If the current in the circuit is 1.04 A, what is the value of R?

2. What power is dissipated in the internal resistance of the battery? Answer in units of W.

1) R = 6.77 Ω

2) P = 1,2 W

Explanation:

In this case there is a cicuit with two series resistances

If EMF of battery is 8.2 (V) and the circuit current is 1.04 (A)

According to Ohm's Law

V = I * R then R = V/I R = 8.2 (V) / 1.04 (A)

R = 7.88 Ω That is the total resistance in the circuit which happens to be the sum of two series resistance one the internal resistance of the battey and the other the resistance of the electrical load

So R + 1.11 = 7.88 ⇒ R = 6.77 Ω

2. - The dissipated power in the battery is equal to = P = I²*R

Then P = 1,2 W