The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are 1 %,2 % and 3 % respectively. The maximum percentage error in the detection of the dissipated heat will be:

Question

The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are 1 %,2 % and 3 % respectively. The maximum percentage error in the detection of the dissipated heat will be: (A) 2 (B) 4 (C) 6 (D) 8

Tardigrade Admin · Accepted Answer

E H = I 2 R × t (Δ E / E ) × 100=(2 Δ I / I ) × 100+(Δ R / R ) × 100+(Δ T / T ) × 100 =2 × 2+1+3=8

Q. The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are 1\%,2\% and $3%$ respectively. The maximum percentage error in the detection of the dissipated heat will be:

2

4

6

8

$E_{H} = I^{2} R \times t$
$\frac{Δ E}{E} \times 100 = \frac{2Δ I}{I} \times 100 + \frac{Δ R}{R} \times 100 + \frac{Δ T}{T} \times 100$
$= 2 \times 2 + 1 + 3 = 8$

Q. The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are 1\%,2\% and 3% respectively. The maximum percentage error in the detection of the dissipated heat will be:

2

4

6

8

EH​=I2R×t EΔE​×100=I2ΔI​×100+RΔR​×100+TΔT​×100 =2×2+1+3=8

Q. The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are 1\%,2\% and $3%$ respectively. The maximum percentage error in the detection of the dissipated heat will be:

$E_{H} = I^{2} R \times t$
$\frac{Δ E}{E} \times 100 = \frac{2Δ I}{I} \times 100 + \frac{Δ R}{R} \times 100 + \frac{Δ T}{T} \times 100$
$= 2 \times 2 + 1 + 3 = 8$