Find initial current and resistance in a circuit

[cnh1995]

i) 12 - 10 = 0.5 I → I = 4 A.

(ii) 12 = 4 (0.5 + R) → R = 2.5 Ω

Right.

or 12 - 10.5 = 0.5 I → I = 3 A
12 = 3 (0.5 + R) → R = 3.5

Right.
Do you conceptually understand what is happening here? You got the temperature part right. If you can explain the drop in voltage from 12V to 10V, you have understood the concept of internal resistance of a voltage source.

 

[moenste]

Do you conceptually understand what is happening here? You got the temperature part right. If you can explain the drop in voltage from 12V to 10V, you have understood the concept of internal resistance of a voltage source.

The battery always has 12 V. Only the voltmeter changes from 10 V to 10.5 V.

And then we just use E = Ir + V and E = I (r + R), where E = 12 V, r = 0.5 Ohm and V is 10 V and 10.5 V.

 

[cnh1995]

The battery always has 12 V. Only the voltmeter changes from 10 V to 10.5 V.

Right. Internal resistance comes in series with the circuit and some voltage drop takes place across it.