The battery in the circuit below has EMF 5.4 V and drives a current of 0.30 A through a lamp. The voltmeter reading is 4.8 V.
Explain why the voltmeter reading is less than the EMF of the cell.
Calculate values for (a) the internal resistance of the battery, and (b) the energy transformed per second in the lamp.
State two assumptions you made in order to complete these calculations.
Answers: (a) 2.0 Ω, (b) 1.44 W
2. The attempt at a solution
The voltmeter reading is less than the EMF of the cell because the EMF is a non-ideal power source that has internal resistance and therefore the output 4.8 V is lower than the input 5.4 V.
(a) E = I r + V → 5.4 V = 0.3 A * r + 4.8 V → r = 2 Ω.
(b) W = V I t = 4.8 V * 0.3 A * 1 s = 1.44 W.
I would assume that the ammeter has no resistance and therefore does not disturb the circuit. I would also assume that the voltmeter has infinite resistance and also does not disturb the circuit.
I am mostly unsure on the theory part. Is it correct? What's wrong?