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Two equal resistances, $400 \Omega$ each, are connected in series with a $8 \mathrm{V}$ battery. If the resistance of first one increases by $0.5 \%$, the change required in the resistance of the second one in order to keep the potential difference across it unaltered is to
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The correct answer is:
increase it by $2 \Omega$
In series combination,
$$
R=R_{1}+R_{2}
$$
Same current will pass in $A_{1}$ and $R_{2}$. According to question, $R_{1}$ is increased by $0.5 \%$
$\therefore$ Increment in the resistance
$$
\begin{array}{l}
=\frac{0.5}{100} \times 400 \\
=2.0 \Omega
\end{array}
$$
So to keep the potential unchanged in second resistance the change required will be $2.0 ~ \Omega$ increment.
$$
R=R_{1}+R_{2}
$$
Same current will pass in $A_{1}$ and $R_{2}$. According to question, $R_{1}$ is increased by $0.5 \%$
$\therefore$ Increment in the resistance
$$
\begin{array}{l}
=\frac{0.5}{100} \times 400 \\
=2.0 \Omega
\end{array}
$$
So to keep the potential unchanged in second resistance the change required will be $2.0 ~ \Omega$ increment.
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