In a circuit consisting of a battery and a 10 ohm resistance, a voltmeter is first connected in series

In a circuit consisting of a battery and a 10 ohm resistance, a voltmeter is first connected in series, then in parallel with the resistance. The indications in both cases are the same. The resistance of the voltmeter is 1000 ohms. What is the internal resistance of the battery?

Given:
R = 1000 Ohm,
r = 10 Ohm
U1 = U2 – voltmeter readings.
Find x – battery resistance.
Let us denote by E the EMF of the battery.
From Ohm’s law for the common circuit, we find the current in both cases. In the first case, the voltmeter (connected to the circuit in series) measures the voltage across (r + x), in the second case, the voltage across r.
I1 = E / (R + r + x),
I2 = E / (r + x)
Since in the second case the voltmeter and resistance are connected in parallel, their total resistance can be considered equal to r, because R >> r.
U1 = I1 * (r + x) = E * (r + x) / (R + r + x),
U2 = I2 * r = E * r / (r + x).
E * (r + x) / (R + r + x) = E * r / (r + x);
(r + x) ^ 2 = r * (R + r + x)
x ^ 2 + 20 * x + 100 = 10000 + 100 + 10 * x;
x ^ 2 + 10 * x – 10000 = 0
Let’s solve the quadratic equation. Only one root is positive:
x = 5 + (10025) ^ 0.5 ≈ 5 + 100 = 105 ohms.