An electric circuit is given below. V1 = 1 V and Resistance R = 1000 Ω. The current through the resistance R is very close to 1 mA and the voltage across point A and B, VAB = 1 V. Now the circuit is changed to: where value of V2 = 5 V. The internal resistances of both the batteries are 0·1 Ω. The current through the resistance R is about:

The problem describes a circuit where two batteries are connected in parallel across a load resistance R = 1000 Ω. To find the current through R, we calculate the equivalent electromotive force (EMF) of the parallel combination of the two batteries: V1 = 1 V and V2 = 5 V, with internal resistances r1 = r2 = 0.1 Ω.

The equivalent voltage (Veq) for batteries in parallel is given by the formula:

Veq = [(V1/r1) + (V2/r2)] / [(1/r1) + (1/r2)]

Substituting the values:
Veq = [(1/0.1) + (5/0.1)] / [(1/0.1) + (1/0.1)] = (10 + 50) / (10 + 10) = 60 / 20 = 3 V.

The equivalent internal resistance is negligible (0.05 Ω) compared to R. Therefore, the current through the resistance R is:
I = Veq / R = 3 V / 1000 Ω = 0.003 A = 3.0 mA.

Thus, Option C is the correct answer.