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

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

Class 11 Physics Units, Dimensions and Measurement Errors of Measurement

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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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A
$(24.0 \pm 1.8)$
B
$(24.0 \pm 0.75)$
C
$(22.0 \pm 0.4)$
D
$(18.0 \pm 0.4)$

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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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If the maximum percentage error in the measurement of distance $(h)$ and time $(t)$ are '$e_1$' and '$e_2$' respectively,the percentage error in the measurement of acceleration of gravity $(g)$ is [use $h=\frac{1}{2} gt^2$].

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In Ohm's experiment,the values of an unknown resistance were found to be $4.12 \; \Omega, 4.08 \; \Omega, 4.22 \; \Omega$,and $4.14 \; \Omega$. Calculate the mean absolute error and relative error in these measurements.

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