The voltage across a lamp is (6.0 pm 0.3) tex | vedclass

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(A) Given: Voltage $V = (6.0 \pm 0.3) 	ext{ V}$ and Current $I = (4.0 \pm 0.1) 	ext{ A}$.Power $P = V 	imes I = 6.0 	imes 4.0 = 24.0 	ext{ W}$.Th

Vedclass · Accepted Answer

$(24.0 \pm 1.8)$

Vedclass · Answer

(A) Given: Voltage $V = (6.0 \pm 0.3) 	ext{ V}$ and Current $I = (4.0 \pm 0.1) 	ext{ A}$.Power $P = V 	imes I = 6.0 	imes 4.0 = 24.0 	ext{ W}$.The relative error in power is given by $\frac{\Delta P}{P} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$.Substituting the values: $\frac{\Delta P}{24.0} = \frac{0.3}{6.0} + \frac{0.1}{4.0}$.$\frac{\Delta P}{24.0} = 0.05 + 0.025 = 0.075$.$\Delta P = 24.0 	imes 0.075 = 1.8 	ext{ W}$.Thus, the power consumed is $(24.0 \pm 1.8) 	ext{ W}$.

The voltage across a lamp is $(6.0 \pm 0.3) \text{ V}$ and the current passing through it is $(4.0 \pm 0.1) \text{ A}$. The power consumed in watt will be (using percentage error):

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