In the circuit shown below current ${{I}_{0}}$ flowing through $6\Omega $ resistance and voltage drop ${{V}_{0}}$ across $3\Omega $ resistance respectively are-
(A). $\dfrac{2}{3}A,\,4V$
(B). $\dfrac{2}{3}A,\,2V$
(C). $\dfrac{1}{3}A,\,2V$
(D). $\dfrac{1}{3}A,\,4V$

Hint: The figure shown above contains resistors connected in series and parallel. By resolving the circuit, we can calculate the total resistance in the circuit. Using the ohm’s law, we can calculate the current and potential drop for individual resistances and hence find the result.
Formulas used:
$R=\dfrac{V}{I}$

Complete answer:

In the given circuit, $6\Omega $ and $3\Omega $ are in parallel so their equivalent resistance will be-
$\begin{align}
  & R=\dfrac{6\times 3}{6+3} \\
 & \Rightarrow R=2\Omega \\
\end{align}$
The equivalent resistance is $2\Omega $. The figure can be simplified as-

In the above figure $2\Omega $ and $2\Omega $ are in series, therefore, the equivalent resistance will be-
$R'=2+2=4\Omega $
Therefore, the equivalent resistance is $4\Omega $. The above figure can be simplified as

In the above figure, $4\Omega $ and $4\Omega $ are in parallel. Therefore, their equivalent resistance will be-
$\begin{align}
  & R''=\dfrac{4\times 4}{4+4} \\
 & \Rightarrow R''=2\Omega \\
\end{align}$
The equivalent resistance is $2\Omega $ and this equivalent resistance is in series with $3\Omega $ aso the equivalent resistance is-
$R'''=2+3=5\Omega $
The total resistance in the circuit is $5\Omega $.
According to Ohm’s law,
$R=\dfrac{V}{I}$ - (1)
Here, $V$ is the potential difference
$R$ is the resistance
$I$ is the current
Given, $R=5\Omega $, $V=10V$. We substitute given values in the above equation to get,
$\begin{align}
  & 10=\dfrac{5}{I} \\
 & \Rightarrow I=\dfrac{10}{5} \\
 & \therefore I=2A \\
\end{align}$
Therefore, the total current in the circuit is $2A$.

The circuit given above can be simplified as-

Applying loop in DACD, we get,
$\begin{align}
  & 10-4(2-I)-3(2)=0 \\
 & \Rightarrow 10-8+4I-6=0 \\
 & \Rightarrow I=1A \\
\end{align}$
Therefore, $I=1A$
Applying both loops of ABCDA,
$\begin{align}
  & 10-2(1)-3I'-3(2)=0 \\
 & \Rightarrow I'=\dfrac{2}{3}A \\
\end{align}$
Therefore, $I'=\dfrac{2}{3}A$, so $\dfrac{I'}{2}=\dfrac{1}{2}\times \dfrac{2}{3}=\dfrac{1}{3}A$
Therefore, the value of ${{I}_{0}}=\dfrac{1}{3}A$.
The potential difference across $3\Omega $ between BC is ${{V}_{0}}$. From eq (1), the value of ${{V}_{0}}$ is-
$\begin{align}
  & {{V}_{0}}=\dfrac{2}{3}\times 3 \\
 & \Rightarrow {{V}_{0}}=2V \\
\end{align}$
Therefore, the value of ${{V}_{0}}$ and ${{I}_{0}}$ respectively are $2V,\dfrac{1}{3}A$.

Hence, the correct option is (C).

Note:
The Kirchhoff’s voltage law states that the sum of all potential differences in a loop is zero. It follows the law of conservation of energy. The potential in a wire is the same everywhere unless an element is introduced in it. The series current through the resistors is the same while in parallel, the potential drop on resistors is the same.