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The heat generated in a circuit is given by $Q = I^{2}Rt$, where I is current, R is resistance and t is time. If the percentage errors in measuring I, R and t are 2%, 1% and 1% respectively, then the maximum error in measuring heat will be
Explanation
$\Delta Q/Q \times 100 = (2\Delta I/I) \times 100 + (\Delta R/R) \times 100 + (\Delta t/t) \times 100$ = $2 \times 2\% + 1\% + 1\% = 6\%$.
