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

Question

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 (A) 2 % (B) 3 % (C) 4 % (D) 6 %

Tardigrade Admin · Accepted Answer

(Δ Q/Q) × 100=(2 Δ I/I) × 100+(Δ R/R) × 100+(Δ t/t) × 100 =2 × 2 %+1 %+1 %=6 %

Q. 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

2 %

3 %

4 %

6 %

$\frac{Δ Q}{Q} \times 100 = \frac{2Δ I}{I} \times 100 + \frac{Δ R}{R} \times 100 + \frac{Δ t}{t} \times 100$
$= 2 \times 2% + 1% + 1% = 6%$

Q. The heat generated in a circuit is given by Q=I2Rt, 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

2 %

3 %

4 %

6 %

QΔQ​×100=I2ΔI​×100+RΔR​×100+tΔt​×100 =2×2%+1%+1%=6%

Q. 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

$\frac{Δ Q}{Q} \times 100 = \frac{2Δ I}{I} \times 100 + \frac{Δ R}{R} \times 100 + \frac{Δ t}{t} \times 100$
$= 2 \times 2% + 1% + 1% = 6%$