A battery with an emf of 12.0 V shows a terminal voltage of 11.1V when operating in a circuit with two lightbulbs, each rated at 4.0 W (at 12.0 V), which are connected in parallel. What is the batterys internal resistance?

Internal resistance of the battery is 1.45Ω

The bulbs are connected in parallel. This means that the same voltage will pass across them. And since they both have the same power rating, then the values of the resistance in the two bulbs are the same.

(1) Get the resistance of each of the bulbs, using the following relation between power(P), voltage (V) and resistance (R);

P =   -----------------(i)

Given;

P = 4W

V = 12V

Substitute for the values of P and V in equation (i)

4 =

Making R the subject of the formula and solving gives;

R = 144/4 = 36Ω

Therefore the resistance in each of the bulbs is 36Ω

(2)Calculate the effective resistance (Rₓ) in the parallel connection.

= +

Where;

R₁ and R₂ are values of resistance in the two bulbs = 36Ω each

Substitute the values of R₁ and R₂ into equation (ii)

= +

Solve for R

Rₓ = 18Ω

(3) Calculate the total current (I) flowing through the circuit using the following relation;

=> V = I x Rₓ

Where;

V is the terminal voltage = 11.1V

Rₓ is the effective resistance in the circuit = 18Ω

=> 11.1 = I x 18

=> I = 11.1 / 18

=> I = 0.62A

(4) Calculate the internal resistance of the battery

The emf (E) is related to the lost voltage and terminal voltage by the following relation;

E = Terminal Voltage + Lost Voltage

But,

Lost Voltage = I x r

Where;

I = current flowing through.

r = internal resistance.

=> E = V + Ir     ------------------(iii)

Where;

E = 12V

V = terminal voltage = 11.1V

I = 0.62A

Substitute these values into equation (iii)

=> 12 = 11.1 + 0.62r

=> 12 = 11.1 + 0.62r

=> 0.62r = 12 - 11.1

=> 0.62r = 0.9

=> r = 0.9 / 0.62

=> r = 1.45Ω

Therefore, the internal resistance is 1.45Ω