Question #23094

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?

Expert's answer

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%.

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%.

## Comments

## Leave a comment