Question

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

• A. 3 mA
• B. 6 mA
• C. 12 mA
• D. 36 mA

Answer 1

Answer: (A)

In the given circuit $4k\Omega$ and $2k\Omega$ are in series.
This combination is in parallel with $3k\Omega$ resistance. Their effective resistance is
$=\frac{(4+2)\times 3}{(4+2)+3}=2k\Omega$
This $2k\Omega$ resistance is in series with $6k\Omega$.
Thus, the equivalent resistance of the circuit is
$R_{eq}=(2k\Omega )+(6k\Omega )=8k\Omega$
Current in the circuit is
$I=\frac{72V}{8\times 10^3\Omega }=9\times 10^{-3}A=9mA$
Current through $2k\Omega$ resistor is
$I_1=I\times \frac{1}{3}=\frac{9mA}{3}=3mA$