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What current will flow through the $2 \mathrm{k} \Omega$ resistor in the circuit shown in the figure? 
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2595 Upvotes
Verified Answer
The correct answer is:
$3 \mathrm{mA}$
From KCL we have
$i=\frac{72}{8 \times 10^{3}}=9 \times 10^{-3} \mathrm{A}$
$$
\begin{aligned}
i \times 6 &=(9-i) \times 3 \\
i &=3 \mathrm{mA}
\end{aligned}
$$
$i=\frac{72}{8 \times 10^{3}}=9 \times 10^{-3} \mathrm{A}$
$$
\begin{aligned}
i \times 6 &=(9-i) \times 3 \\
i &=3 \mathrm{mA}
\end{aligned}
$$
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