What current will flow through the 2 kOmega r | vedclass

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(A) First, find the equivalent resistance of the circuit. The $4 k\Omega$ and $2 k\Omega$ resistors are in series, so their equivalent resistance is $

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$3$

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(A) First, find the equivalent resistance of the circuit. The $4 k\Omega$ and $2 k\Omega$ resistors are in series, so their equivalent resistance is $R_s = 4 k\Omega + 2 k\Omega = 6 k\Omega$.This $6 k\Omega$ equivalent resistance is in parallel with the $3 k\Omega$ resistor. Their equivalent resistance $R_p$ is given by:$\frac{1}{R_p} = \frac{1}{6 k\Omega} + \frac{1}{3 k\Omega} = \frac{1+2}{6 k\Omega} = \frac{3}{6 k\Omega} = \frac{1}{2 k\Omega}$So, $R_p = 2 k\Omega$.Now, the total resistance of the circuit is $R_{total} = 6 k\Omega + R_p = 6 k\Omega + 2 k\Omega = 8 k\Omega$.The total current $I$ from the $72 V$ battery is:$I = \frac{V}{R_{total}} = \frac{72 V}{8 k\Omega} = 9 mA$.This current $I = 9 mA$ flows through the $6 k\Omega$ resistor and then splits at the junction into the $3 k\Omega$ branch and the branch containing the $4 k\Omega$ and $2 k\Omega$ resistors in series.Let $i$ be the current through the $2 k\Omega$ resistor. The voltage across the parallel branches is the same:$i 	imes (4 k\Omega + 2 k\Omega) = (I - i) 	imes 3 k\Omega$$i 	imes 6 k\Omega = (9 mA - i) 	imes 3 k\Omega$$2i = 9 mA - i$$3i = 9 mA$$i = 3 mA$.

What current will flow through the $2 k\Omega$ resistor in the circuit shown in the figure (in $mA$)?

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