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Three resistors $1 \Omega, 2 \Omega$ and $3 \Omega$ are connected to form a triangle. Across $3 \Omega$ resistor a $3 \mathrm{~V}$ battery is connected. The current through $3 \Omega$ resistor is
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Verified Answer
The correct answer is:
$1 \mathrm{~A}$
The arrangement is shown in figure.
Here, two reisistance of $1 \Omega$ and $2 \Omega$ are in series, which form $3 \Omega$ which is in parallel with $3 \Omega$ resistance. Therefore, the effective resistance $\frac{(1+2) \times 3}{(1+2)+3}=\frac{3}{2} \Omega$
$\therefore$ Current in the circuit,
$$
\begin{array}{l}
\mathrm{I}=\frac{3}{(3 / 2)}=2 \mathrm{~A} \\
\therefore \text { Current in } 3 \Omega \text { resistor }=\frac{\mathrm{I}}{2}=\mathrm{IA}
\end{array}
$$
Here, two reisistance of $1 \Omega$ and $2 \Omega$ are in series, which form $3 \Omega$ which is in parallel with $3 \Omega$ resistance. Therefore, the effective resistance $\frac{(1+2) \times 3}{(1+2)+3}=\frac{3}{2} \Omega$
$\therefore$ Current in the circuit,
$$
\begin{array}{l}
\mathrm{I}=\frac{3}{(3 / 2)}=2 \mathrm{~A} \\
\therefore \text { Current in } 3 \Omega \text { resistor }=\frac{\mathrm{I}}{2}=\mathrm{IA}
\end{array}
$$
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