Question

In the circuit shown in the figure, the current through

A. the $3\Omega$ resistor is 0.50 A
B. the $3\Omega$ resistor is 0.25 A
C. the $4\Omega$ resistor is 0.50 A
D. the $4\Omega$ resistor is 0.25 A

Answer 1

We will keep on simplifying the circuit by using equivalent resistances in place of a combination of resistors. Then on the final simplest circuit, we will find the current and then divide the current into branches and find the current through each branch.

Formula used:
Ohm’s Law
$V=IR$
Equivalent resistance of resistors arranged in series
$R={{R}_{1}}+{{R}_{2}}$
Equivalent resistance of resistors arranged in parallel
$\dfrac{1}{R}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}$

Complete complete:
First, we will find the equivalent resistance of resistors in the circle.

These resistances are in series. So, according to the formula, their equivalent resistance is equal to $2\Omega +4\Omega +2\Omega =8\Omega$. This equivalent resistor and the resistor in the box give an equivalent resistor of ${{\left( \dfrac{1}{8\Omega }+\dfrac{1}{8\Omega } \right)}^{-1}}=4\Omega$. This is shown in the next figure.

Repeating the same procedure again for the resistors in circle and box we again get the equivalent resistance of resistors in the circle as $2\Omega +4\Omega +2\Omega =8\Omega$ and the equivalent resistance of this and the resistor in the box gives an equivalent resistor of ${{\left( \dfrac{1}{8\Omega }+\dfrac{1}{8\Omega } \right)}^{-1}}=4\Omega$. This is shown in the next figure.

Here we can see that the total resistance in the circuit will be $2\Omega +4\Omega +3\Omega =9\Omega$. Now, according to ohm’s law, the current through this circuit will be
$\begin{aligned} & V=IR \\\ & I=\dfrac{V}{R}=\dfrac{9V}{9\Omega }=1A \\\ \end{aligned}$
Now when we go back to the more complex circuits, we see that the resistance along both the marked paths is the same. So, the current will divide equally.

Going to the original circuit again the resistance is equal on both paths so the current will again divide equally.

This way we get the current through the $3\Omega$resistor as 1 ampere and through the $4\Omega$ resistor as 0.25 ampere. Hence, the correct option is D, i.e. the $4\Omega$ resistor is 0.25 A.

Note:
Take care that the current does not always divide equally between two branches of the circuit. It is dependent on the resistance between them. Here, because the resistance on the two branches of the circuit was equal, that is why the current was divided equally.