The voltage across a lamp

unitsandmeasurement
cbse

#1

The voltage across a lamp is V = (6.0 ±0.1) Volt and the current passing through it I=- (4 ± 0.2) ampere. Find the power consumed by the electric lamp. Give that power, P = VI


#2

As, V = (6.0±0.1)'V, I = (4.0±0.2)A
Power, P = VI =6.0 x 4.0 = 24 W = 24 W
and maximum error in power measurement
∆P/P = ∆V/V + ∆I/I = 0.1/6.0 + 0.2 / 4.0
= 0.017 + 0.050 = 0.067
∆P = 0.067 x P =0.067 x 24 = 1.6 W
Power consumed by the electric lamp within error limit is (24±1.6)W.