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

Hint: First calculate the equivalent resistance for the circuit by deducing series and parallel combinations and find the current flowing throughout the circuit. Use the branch current formula to find the current passing through 2k$\Omega$ resistor.

Complete step-by-step solution
In a group of resistances the equivalent resistance is of two types.
Series grouping: Req is greater than maximum value of resistance in the combination.

${R_{eq}} = {R_1} + {R_2} + {R_3} + {R_4}$
Parallel grouping: Req is smaller than the minimum value of resistance in the combination.

$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + \dfrac{1}{{{R_4}}}$
Here, 4$k\Omega $ resistor is parallel to 2$k\Omega$ resistor so
$
  {R_1} = 4 + 2 \\
  {R_1} = 6k\Omega \\
  $
Now, $R_1$ is parallel to 3$k\Omega$ resistor
\[
  {R_2} = \dfrac{1}{{{R_1}}} + \dfrac{1}{3} \\
  {R_2} = \dfrac{{6 \times 3}}{{6 + 3}} \\
  {R_2} = \dfrac{{18}}{9} \\
  {R_2} = 2k\Omega \\
  \]
Now $R_2$ is in series with 6 $k\Omega$ resistors, Hence the final equivalence resistance of the circuit is,
$
  {R_{eq}} = {R_2} + 6 \\
  {R_{eq}} = 2 + 6 \\
  {R_{eq}} = 8k\Omega \\
  $
From Ohm’s law, current (I) for the circuit is,
\[
  I = \dfrac{V}{{{R_{eq}}}} \\
  I = \dfrac{{72}}{8} \\
  I = 9mA \\
  \]
To find the current through 2$k\Omega $resistor, use the given branch formula

$
  {i_2} = I\left( {\dfrac{{{R_1}}}{{{R_1} + {R_2}}}} \right) \\
  {i_2} = 9\left( {\dfrac{3}{{6 + 3}}} \right) \\
  {i_2} = \dfrac{9}{3} \\
  {i_2} = 6mA \\
  $
Hence the current passing through the $k\Omega$ resistor is 6 mA and the correct option is B.

Note: For n identical resistance in series combination is
${R_{eq}} = nR$
For n identical resistance in parallel combination is
${R_{eq}} = \dfrac{R}{n}$