Question

Three DC currents, $I_1,I_{2}and{I}_3$ meet at a node with $I_1$ entering, $I_{2}and{I}_3$ leaving the node. $I_{1}and{I}_2$ are measured as 100 mA and 99 mA with $\pm 1$% accuracy. Then the value $I_3$ and the accuracy of $I_3$ are

• A. $10mA\pm 2\%$
• B. $199mA\pm 2\%$
• C. $1mA\pm 2\%$
• D. $1mA\pm 199\%$

Answer 1

The correct option is D $1mA\pm 199\%$

$\text{Sum of all the currents meeting of a node = 0}$

$\text{incoming currents}\to \text{+ve}$

$\text{outgoing currents}\to \text{-ve}$

$\therefore I_1-I_2-I_3=0$

$I_1=I_2+I_3$

$I_3=I_1-I_2$

$=100-99=1mA$

$Percentageaccuracy$

$\Rightarrow \frac{\partial I_3}{I_3}=\frac{\partial I_1}{I_1}\frac{I_1}{I_3}+\frac{\partial I_2}{I_2}\frac{I_2}{I_3}$

$=1\left(\frac{100}{1}\right)+1\left(\frac{99}{1}\right)=199\%$

$\therefore I_3=1mA\pm 199\%$