The heat generated in circuit $q = I^{2}$ Rt where I is current, R is resistance and t is time. If the percentage error in measuring I, R and t are $2 %, 1$ respectively, then the maximum error in measuring heat will be

Q. The resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and voltage are each

3

% respectively, the maximum percentage error in measurement of resistance is:
(

R = \frac{V}{I}

)

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According to Joule's law of heating, heat produced H=I2 Rt, where I is current, R is resistance and t is time. If the errors in the measurements of I, R and t are 3%, 4% and 6% respectively, then error in the measurement of H is

The correct option is B ± 16%H=I2Rt ∴Δ HH×100=(2Δ II+Δ RR+Δ tt)×100 =(2×3+4+6)%=16%

According to Joule's law of heating, heat produced $H = I^{2} R t,$ where $I$ is current, $R$ is resistance and $t$ is time. If the errors in the measurements of $I$ , $R$ and $t$ are 3%, 4% and 6% respectively, then error in the measurement of H is

The correct option is B $\pm 16 %$
$H = I^{2} R t$
$∴ \frac{Δ H}{H} \times 100 = (\frac{2 Δ I}{I} + \frac{Δ R}{R} + \frac{Δ t}{t}) \times 100$
$= (2 \times 3 + 4 + 6) % = 16 %$