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A resistor of $6 \mathrm{k} \Omega$ with tolerance $10 \%$ and another resistance of $4 \mathrm{k} \Omega$ with tolerance $10 \%$ are connected in series. The tolerance of the combination is about
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The correct answer is:
$10 \%$
In series combination equivalent resistance,
$\begin{array}{l}
\mathrm{R}=\mathrm{R}_{1}+\mathrm{R}_{2} \\
=6+4=10 \mathrm{k} \Omega
\end{array}$
Error in combination,
$\begin{array}{l}
\Delta \mathrm{R}=\Delta \mathrm{R}_{1}+\Delta \mathrm{R}_{2} \\
=\frac{10}{100} \times 6+\frac{10}{100} \times 4 \\
=0.6+0.4=1 \\
\frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{1}{10}=10 \%
\end{array}$
$\begin{array}{l}
\mathrm{R}=\mathrm{R}_{1}+\mathrm{R}_{2} \\
=6+4=10 \mathrm{k} \Omega
\end{array}$
Error in combination,
$\begin{array}{l}
\Delta \mathrm{R}=\Delta \mathrm{R}_{1}+\Delta \mathrm{R}_{2} \\
=\frac{10}{100} \times 6+\frac{10}{100} \times 4 \\
=0.6+0.4=1 \\
\frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{1}{10}=10 \%
\end{array}$
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