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Resistances $1 \Omega, 2 \Omega$ and $3 \Omega$ are connected to form a triangle. If a $1.5 \mathrm{~V}$ cell of negligible internal resistance is connected across the $3 \Omega$ resistor, the current flowing through this resistor will be
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The correct answer is:
$0.5 \mathrm{~A}$
Equivalent resistance between A and B= series combination of $1 \Omega$ and $2 \Omega$ in parallel with $3 \Omega$ resistor.
$\frac{1}{\mathrm{R}}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3} \quad$ or $\mathrm{R}=1.5 \Omega$
$\therefore$ Current in the circuit is $\mathrm{I}=\mathrm{V} / \mathrm{R}=1.5 / 1.5=$ $1 \mathrm{~A}$.
Since the resistance in $\operatorname{arm} \mathrm{ACB}=$ resistance in $\operatorname{arm} \mathrm{AB}=3 \Omega$, the current divides equally in the two arms. Hence the current through the $3 \Omega$ resistor $=\mathrm{I} / 2=0.5 \mathrm{~A}$.
$\frac{1}{\mathrm{R}}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3} \quad$ or $\mathrm{R}=1.5 \Omega$
$\therefore$ Current in the circuit is $\mathrm{I}=\mathrm{V} / \mathrm{R}=1.5 / 1.5=$ $1 \mathrm{~A}$.
Since the resistance in $\operatorname{arm} \mathrm{ACB}=$ resistance in $\operatorname{arm} \mathrm{AB}=3 \Omega$, the current divides equally in the two arms. Hence the current through the $3 \Omega$ resistor $=\mathrm{I} / 2=0.5 \mathrm{~A}$.
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