According to Joule's law of heating,heat prod | vedclass

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(B) Given the formula for heat produced: $H = I^2Rt$.The relative error in $H$ is given by the propagation of errors formula:$\frac{\Delta H}{H} = 2\f

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$ \pm 16\%$

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(B) Given the formula for heat produced: $H = I^2Rt$.The relative error in $H$ is given by the propagation of errors formula:$\frac{\Delta H}{H} = 2\frac{\Delta I}{I} + \frac{\Delta R}{R} + \frac{\Delta t}{t}$.To find the percentage error,we multiply by $100$:$\frac{\Delta H}{H} 	imes 100 = \left( 2 	imes \frac{\Delta I}{I} 	imes 100 + \frac{\Delta R}{R} 	imes 100 + \frac{\Delta t}{t} 	imes 100 ight)$.Substituting the given percentage errors $(3\%, 4\%, 6\%)$:$\frac{\Delta H}{H} 	imes 100 = (2 	imes 3\% + 4\% + 6\%)$.Calculating the result:$\frac{\Delta H}{H} 	imes 100 = (6\% + 4\% + 6\%) = 16\%$.Thus,the error in the measurement of $H$ is $\pm 16\%$.

According to Joule's law of heating,heat produced $H = I^2Rt$,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,then the error in the measurement of $H$ is:

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