when a battery is connected to a 100-Ω resistor, the current is 4.00 A. When the same battery is connected to a 449-Ω resistor, the current is 1.09 A. Find the emf supplied by the battery and the internal resistance of the battery.

Answer 1

E = Battery terminal voltage(e.m.f.).
Ri = Internal resistance.
R1 = 100 Ohms.
R2 = 449 Ohms.

I*R1 + I*Ri = E
4*100 + 4*Ri = E
Eq1: -E + 4Ri = -400

I*R2 + I*Ri = E
1.09*449 + 1.09*Ri = E
Eq2: -E + 1.09Ri = -489.41
Subtract Eq2 from Eq1:

Eq1: -E + 4Ri = -400
Eq2: -E + 1.09Ri = -489.41
Diff: 0 + 2.91Ri = 89.41
Ri = 30.73 Ohms.

In Eq1, replace Ri with 30.73 Ohms.
-E + 4*30.73 = -400
E = 400 + 122.9 = 522.9 Volts.

Answer 2

A shorter method:

R1 = 100 Ohms
R2 = 449 Ohms
Ri = Internal resistance.

V1 = I1*R1 = 4 * 100 = 400 Volts.

V2 = I2*R2 = 1.09 * 449 = 489.41 Volts.

Ri = (V2-V1)/(I1-I2) =
(489.41-400)/(4-1.09) = 30.73 Ohms.

E = I1*R1 + I1*Ri = 400 + 122.9 = 522.9
Volts.