Question

a cell of e.m.f. 3.0 volts and internal resistance 0.10 ohms is connected through an ammeter of 0.05 ohm resistance to a 5.0 ohms rheostat by means of wires having a total resistance of 0.85 ohms. In the calculation of the current, what percentage error would be made by neglecting all resistance except that of the rheostat?

Answer 1

Actual net resistance of thecircuit R = 0.1 ohms + 0.05 ohms + 5.0 ohms + 0.85 ohms = 6.0 ohms.
The actual current is thus I = V / R = 3 V / 6 ohms = 0.5 A.

If taking into account only the resistance of the rheostat, calculated current
would be Ir = V / Rr = 3 V / 5 ohms = 0.6 A.

The percentage error with respect to the true value can be estimated as
PE = 100% * abs(I - Ir) / I = 100% * 0.1 A / 0.5 A = 0.2 * 100% = 20%.