Question

Two scales on a voltmeter measure voltages up to 20.2 and 30.2 V, respectively. The resistance connected in series with the galvanometer is 1530Ω for the 20.2-V scale and 3090Ω for the 30.2-V scale. Determine (a) the coil resistance and (b) the full-scale current of the galvanometer that is used in the voltmeter.

Answer 1

a) From the problem statement, either way galvanometer undergoes maximum deflection. In the first case, with 20.2 V applied across it, and 30.2 in the second. This means that full-scale current flows in galvanometer, whatever the case. Let this current be "i". Then,

i = 20.2/(1530 + r) = 30.2/(3090 + r).

From the above,

20.2(3090 + r) = 30.2(1530 + r).

Solving for r,

r = 1621.2 ?.

b) Just derive i from either of the expressions equated above. For instance, if the first is chosen,

i = 20.2/(1530 + 1621.2) = 6.41mA is the answer