Physics - Current Electricity Question with Solution | TestHub

Given $R_1=6\Omega$ with $10\%$ tolerance and $R_2=4\Omega$ with $10\%$ tolerance.

Absolute error in $R_1$ is $\Delta R_1=\frac{10}{100}\times 6=0.6\Omega$

Absolute error in $R_2$ is $\Delta R_2=\frac{10}{100}\times 4=0.4\Omega$

For series combination, the equivalent resistance is $R_{eq}=R_1+R_2=6+4=10\Omega$

The absolute error in the series combination is $\Delta R_{eq}=\Delta R_1+\Delta R_2=0.6+0.4=1.0\Omega$

The tolerance of the combination is the percentage error, given by $\frac{\Delta R_{eq}}{R_{eq}}\times 100\%$

Tolerance $=\frac{1.0}{10}\times 100\%=10\%$

Answer: $10\%$