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Two resistors of resistances $R_{1}=100 \pm 3$ $ohm$ and $R_{2}=200 \pm 4$ $ohm$ are connected $(a)$ in series, $(b)$ in parallel. Find the equivalent resistance of the $(a)$ series combination, $(b)$ parallel combination. Use for $(a)$ the relation $R=R_{1}+R_{2}$ and for $(b)$ $\frac{1}{R^{\prime}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$ and $\frac{\Delta R^{\prime}}{R^{\prime 2}}=\frac{\Delta R_{1}}{R_{1}^{2}}+\frac{\Delta R_{2}}{R_{2}^{2}}$
Solution
$(a)$ The equivalent resistance of serles combination
$R=R_{1}+R_{2}=(100 \pm 3)$ $ohm$ $+(200 \pm 4)$ $ohm$
$=300 \pm 7 \text { ohm. }$
$(b)$ The equivalentl resistance of parallel combination
$R^{\prime}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{200}{3}=66.7$ $ohm$
Then, from $\frac{1}{R^{\prime}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$
we get,
$\frac{\Delta R^{\prime}}{R^{2}}=\frac{\Delta R_{1}}{R_{1}^{2}}+\frac{\Delta R_{2}}{R_{2}^{2}}$
$\Delta R^{\prime}=\left(R^{2}\right) \frac{\Delta R_{1}}{R_{1}^{2}}+\left(R^{2}\right) \frac{\Delta R_{2}}{R_{2}^{2}}$
$=\left(\frac{66.7}{100}\right)^{2} 3+\left(\frac{66.7}{200}\right)^{2} 4$
$=1.8$
Then, $R^{\prime}=66.7 \pm 1.8$ $ohm$
(Here, $\Delta R$ is expresed as $1.8$ instead of $2$ to keep in confirmity with the rules of stgnificant figures.
