Question

The total current supplied to the circuit by the battery is

• A. $4\text{A}$
• B. $2\text{A}$
• C. $1\text{A}$
• D. $6\text{A}$

Answer 1

The correct option is A $4\text{A}$

From the given circuit we can see that the $2\Omega \&6\Omega$ resistors are connected between same points i.e both the ends of resistors have same potential difference, hence they are in parallel combination.

The circuit can be redrawn as shown below


For two resistors in parallel connection we can use,

$R_{eq}=\frac{R_1{R}_2}{R_1+R_2}$

For $6\Omega$ and $2\Omega$,

$R_{eq}=\frac{6\times 2}{6+2}=\frac{3}{2}\Omega$

Thus replacing the combination of resistors by $R_{eq}$ we get,


The net resistance of circuit is $\frac{3}{2}\Omega$, thus current in the circuit as supplied by battery will be,

$i=\frac{V}{R_{\text{net}}}=\frac{6}{\left(\frac{3}{2}\right)}$

$\Rightarrow i=\frac{12}{3}=4\text{A}$

Why this question? $\underset{–}{\text{Tip}:}$ Whenever a lot of resistances are given in a circuit, then try to observe the points of connection of resistors for segregating into parallel and series combination. $\underset{–}{\text{Trick}:}$For quick solving we can use $R_{\text{eq}}=\frac{R}{2}$ for two resistors of same resistance connected in parallel.