Question

Four resistances are connected in a circuit as shown in the figure. The electric current flowing through $4\Omega$ and $6\Omega$ resistance respectively will be

• A. $21\text{A}$ and $2\text{A}$
• B. $2\text{A}$ and $2\text{A}$
• C. $1\text{A}$ and $2\text{A}$
• D. $2\text{A}$ and $4\text{A}$

Answer 1

The correct option is B $2\text{A}$ and $2\text{A}$

The given circuit can be reduced as follows, by replacing the combination of resistance by an equivalent resistance.

$R_1=\frac{4}{2}=2\Omega$

$R_2=\frac{6}{2}=3\Omega$

($\because$ The resistance of equal value are connected in parallel )

Now these two resistance are in series, hence net resistance of the circuit is,

$\Rightarrow R_{\text{eq}}=2+3=5\Omega$


The current supplied by battery in circuit is,

$i=\frac{V}{R_{\text{eq}}}=\frac{20}{5}=4\text{A}$

The current through $2\Omega$ and $3\Omega$ will be same i.e $4\text{A}$.

Now, the current will distribute equally in the $4\Omega$ and $6\Omega$ resistors due to the symmetry of arrangement of resistors.


$\Rightarrow i_{4\Omega }=i_{6\Omega }=\frac{4}{2}=2\text{A}$

$$\begin{array}{c}\text{Why this Q?}\\ \text{Current entering parallel combination will distribute}\text{as per the value of resistance in each path}\\ \text{and current will be shared among the resistances in the inverse ratio of the value of resistances.}\\ \text{For equal resistances connected in parallel, current will divide equally in each resistance.}\\ i_1:i_2=\frac{1}{R_1}:\frac{1}{R_2}\end{array}$$