Question

What is the total current supplied by the battery to the circuit shown in the following figure?

$\left( A \right).$ $2A$
$\left( B \right).$ $4A$
$\left( C \right).$ $6A$
$\left( D \right).$ $9A$

Answer 1

Hint : In this problem the current can be found out by finding the total resistance of the given circuit. The equivalent resistance can be found out by having basic ideas about series and parallel combinations. That is in series combination the current will remain the same whereas in parallel combination the current will be different. After finding the total resistance the current can be found out by using the formula $I = \dfrac{V}{R}$.

Formulas used:
$\Rightarrow {R_{S1}} = {R_1} + {R_2}$
$\Rightarrow \dfrac{1}{{{R_{P1}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$
$\Rightarrow V = IR$

Complete step-by-step solution:
The given circuit can be redrawn as

The resistors $4\Omega$ and $4\Omega$ are connected in series because the current remains same in these resistor, therefore equivalent resistance between these resistor is
${R_{S1}} = {R_1} + {R_2}$
Substituting the resistor values in above equation, we get
${R_{S1}} = 4 + 4$
Therefore, ${R_{S1}} = 8\Omega$
The resistors $8\Omega$ and $4\Omega$ are connected in parallel because the current different in these resistor, therefore equivalent resistance between these resistor is
$\dfrac{1}{{{R_{P1}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$
Substituting the resistor values in above equation, we get
$\dfrac{1}{{{R_{P1}}}} = \dfrac{1}{8} + \dfrac{1}{4}$
$\dfrac{1}{{{R_{P1}}}} = \dfrac{{4 + 8}}{{8 \times 4}}$
$\dfrac{1}{{{R_{P1}}}} = \dfrac{{12}}{{24}}$
$\dfrac{1}{{{R_{P1}}}} = \dfrac{1}{2}$
Therefore, ${R_{p1}} = 2\Omega$

The resistors $\dfrac{{16}}{3}\Omega$ and $2\Omega$ are connected in series because the current remains same in these resistor, therefore equivalent resistance between these resistor is
${R_{S2}} = {R_1} + {R_2}$
Substituting the resistor values in above equation, we get
${R_{S2}} = \dfrac{{16}}{3} + 2$
${R_{S2}} = \dfrac{{16 + 6}}{3}$
${R_{S2}} = \dfrac{{24}}{3}$
${R_{S2}} = 8\Omega$

The resistors $8\Omega$ and $8\Omega$ are connected in parallel because the current different in these resistor, therefore equivalent resistance between these resistor is
$\dfrac{1}{{{R_{P2}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$
Substituting the resistor values in above equation, we get
$\dfrac{1}{{{R_{P2}}}} = \dfrac{1}{8} + \dfrac{1}{8}$
${R_{P2}} = 4\Omega$
Therefore, Total resistance $R = 4\Omega$
Then, according to the ohm’s law
$V = IR$
Therefore,
$I = \dfrac{V}{R}$
$I = \dfrac{{24}}{4}$
$I = 6A$
Hence, the correct option is $C$ .That is $6A$ .

Note: It should be noted that the resistance of the resistor will be more when connected in series compared to that of resistor when connected in parallel. The current should be found only after finding the total resistance and the S.I unit electric current is ampere $\left( A \right)$ .