Find the internal resistance of the battery if the efficiency of the circuit has doubled when the external resistance

Find the internal resistance of the battery if the efficiency of the circuit has doubled when the external resistance is increased from 3 ohms to 10.5 ohms.

In a closed circuit, some of the power will be dissipated on the external resistance R (net power), and some wasted on the internal resistance r.
Net power: Pp = R1 * I ^ 2 (resistance multiplied by the current squared).
Total power: Pc = (R1 + r) * I ^ 2.
Efficiency before replacement: k1 = Pp / Pc = (R1 * I ^ 2) / ((R1 + r) * I ^ 2) = R1 / (R1 + r).
Efficiency after replacement: k2 = (R2 * i ^ 2) / ((R2 + r) * i ^ 2) = R2 / (R2 + r).
By assumption, k2 / k1 = 2.
(R2 / (R2 + r)) / (R1 / (R1 + r)) = 2;
R2 / (R2 + r) = 2 * (R1 / (R1 + r)).
Substitute the values ​​- R1 = 3 Ohm, R2 = 10.5 Ohm:
10.5 Ohm / (10.5 Ohm + r) = 2 * 3 Ohm / (3 Ohm + r);
10.5 * (3 ohms + r) = 6 * (10.5 ohms + r);
31.5 ohm + 10.5r = 63 ohm + 6r;
4.5r = 31.5 ohm;
r = 7 ohms.
Answer: 7 ohms.