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

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Solution

Heat produced $H = i^{2} R T$
$l o g H = 2 l o g i + l o g R + l o g T$
$\frac{Δ H}{H} = \frac{2 Δ i}{i} + \frac{Δ R}{R} + \frac{Δ T}{R}$
$\Rightarrow$ Error in $H = 2 \times 3 + 4 + 6 = 16$ %.

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According to Joule's law of heating, heat produced H=l2Rt, where I is current, R is resistance, and t is time. If the errors in the measurement of I,R and t are 3%,4% and 6%, respectively, find error in the measurement of H.

Heat produced H=i2RTlogH=2logi+logR+logTΔHH=2Δii+ΔRR+ΔTR⇒ Error in H=2×3+4+6=16%.

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

Heat produced $H = i^{2} R T$
$l o g H = 2 l o g i + l o g R + l o g T$
$\frac{Δ H}{H} = \frac{2 Δ i}{i} + \frac{Δ R}{R} + \frac{Δ T}{R}$
$\Rightarrow$ Error in $H = 2 \times 3 + 4 + 6 = 16$ %.