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If the error in measuring resistance, current and time are 1%, 2% and 1% respectively, then maximum error in measuring the electrical heat is

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18%

12%

answer is C.

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Detailed Solution

$H = i^{2} R t$

So % error in Heat is

$\frac{ΔH}{H} \times 100 = (2 \frac{Δi}{i} + \frac{ΔR}{R} + \frac{Δt}{t}) \times 100$

So,

$\frac{Δ H}{H} \times 100 = (2 \times 2 + 1 + 1) % = 6 %$

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