The battery, being closed to a resistance of 4 ohms, gives a current of 0.5 A. If it is closed to a resistance

The battery, being closed to a resistance of 4 ohms, gives a current of 0.5 A. If it is closed to a resistance of 8.95 ohms, then it gives a current of 0.24 A. Determine the EMF, internal resistance and “short” current.

Given:
R1 = 4 ohms,
R2 = 8.95 Ohm,
I1 = 0.5A,
I2 = 0.24A;
Find: EMF E, internal resistance r and short-circuit current Ik;
We use Ohm’s law for the general circuit:
E / (R + r) = I;
In the first case:
E / (R1 + r) = I1,
and in the second case:
E / (R2 + r) = I2;
We got a system of equations:
E / (4 + r) = 0.5,
E / (8.95 + r) = 0.24;
Let’s take their attitude:
(4 + r) / (8.95 + r) = 0.48;
r = 0.57 ohm.
E = I * ((R + r) = I1 * ((R1 + r) = 0.5A * 4.57 Ohm = 2.285V
Ik – current when R = 0;
Ik = E / r = 2.285V / (0.57 Ohm) = 4A.