Answer:
Here, V = 4.4V, I = 2.2A
\[\Delta
\text{V}=0.\text{1V},\,\,\Delta \text{I}=0.\text{2A}\]
\[\text{R
}=\frac{V}{I}=\frac{4.4}{2.2}=2\,ohm\]
Maximum permissible error
\[\frac{\Delta
R}{R}=\frac{\Delta V}{V}+\frac{\Delta I}{I}=\frac{0.1}{4.4}+\frac{0.2}{2.2}\]
\[=0.0\text{23}+0.0\text{91}=0.\text{114}\]
\[\Delta
\text{R}=0.\text{114}\times \text{R}=0.\text{114}\times \text{2}=0.\text{228}\]
\[\Delta
\text{R}=0.\text{2W}\]
(rounding off to one decimal place)
Resistance of wire with max. permissible error R = (2.0+ 0.2) ohm
Max. percentage error =
\[\frac{\Delta R}{R}\times
\text{1}00=0.\text{114}\times \text{1}00=\text{11}.\text{4}%\]
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