In the circuit in the Figure below, the high-resistance voltmeter reads 1.55 V when the switch is open and 1.49 V when the switch is closed.
(1) Explain why:
(a) the e.m.f. of the cell can be considered to be 1.55 V
(b) the voltmeter reading drops when the switch is closed.
(2) Calculate the internal resistance of the cell.
[ Solution ]
(1a) As the voltmeter has a very high resistance, it takes virtually no current. Therefore, with the switch open, there is negligible current in the cell, so the reading of 1.55 V can be taken as its e.m.f.
(1b) When the switch is closed, the 10 Ω resistor is brought into the circuit. This causes a current, I, in the circuit so that the potential difference, V, across the cell drops to V = ε − Ir. (With the switch
open, I = 0, so that V = ε.)
(2) From V = ε − Ir
=>Ir = ε − V
= 1.55 V − 1.49 V
= 0.06 V
For the 10 Ω resistor:
I = V/R = 1.49 V/10 Ω = 0.149 A
r = (ε − V)/I = 0.06 V/0.149 A
=>r= 0.40 Ω