Question

In the circuit, the value of ${{I}_{2}}$ is ?

Answer 1

Hint: Use concept of current divider rule. Current divider refers to the splitting of current between the branches of the divider. Remember the formula to calculate equivalent resistance for different resistance connected in parallel.

Complete step by step answer:
Current is flow of charge per unit time. It is denoted by ‘I’.
$I=\dfrac{q}{t}$
SI unit of current is ampere$(A)$.
Resistance is the property of a conductor by virtue of which it opposes the flow of electric current through it. It is the ratio of voltage applied to the electric current flowing through it.
$R=\dfrac{V}{I}$
Materials which have very low resistance are called conductors while materials which have very high resistance are called insulators. This is because as resistance is high the flow of current becomes minimal. SI unit of resistance is ohm$(\Omega )$. $1$ ohm is defined as the resistance of a conductor with a potential difference of $1$volt applied to the ends through which $1$ ampere current flows. Resistance mainly depends on 3 factors: Length of conductor, Area of cross section and nature of material.

Resistance is directly proportional to length of wire, inversely proportional to area of cross section. Formula of resistance is: $R=\rho \dfrac{l}{A}$where $\rho $is known as resistivity.Resistivity is a measure of how much an electrical conductor opposes the flow of current through it. SI unit of resistivity is ohms metre$(\Omega -m)$.A current divider is a simple linear circuit that produces an output current that is a fraction of its input current. Current divider refers to the splitting of current between the branches of the divider.For using current divider rule first calculate equivalent the resistance:
${{R}_{1}}=10\Omega $
$\Rightarrow {{R}_{2}}=15\Omega $
$\Rightarrow {{R}_{3}}=30\Omega $

Let ${{R}_{eq}}$ be the equivalent resistance, as all the resistance are connected in parallel
$\Rightarrow \dfrac{1}{{{R}_{eq}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}+\dfrac{1}{{{R}_{3}}}$
$\dfrac{1}{{{R}_{eq}}}=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{30}$
$\Rightarrow \dfrac{1}{{{R}_{eq}}}=\dfrac{3+2+1}{30}=\dfrac{6}{30}=\dfrac{1}{5}$
$\Rightarrow \dfrac{1}{{{R}_{eq}}}=\dfrac{1}{5}$
$\Rightarrow {{R}_{eq}}=5\Omega $
Now using current divider rule:
${{I}_{1}}=\dfrac{{{\operatorname{R}}_{eq}}}{{{R}_{1}}}\times I=\dfrac{5}{10}\times 1.2=0.6A$
$\Rightarrow {{I}_{2}}=\dfrac{{{R}_{eq}}}{{{R}_{2}}}\times I=\dfrac{5}{15}\times 1.2=0.4A$
$\therefore {{I}_{3}}=\dfrac{{{R}_{eq}}}{{{R}_{3}}}\times I=\dfrac{5}{30}\times 1.2=0.2A$

Therefore, the current ${{I}_{2}}=0.4\,A$.

Note: Remember the current divider formula ${{I}_{R}}=\dfrac{{{R}_{eq}}}{{{R}_{R}}}\times I$
In some cases, we use the voltage divider rule according to the requirement of the question.The voltage divider is also known as potential divider. It is a simple circuit used to change a large voltage into a small voltage.