Question

Consider the circuit shown in figure. Calculate the current through the $3\Omega$ resistor.

• A. $1.33\text{A}$
• B. $0.67\text{A}$
• C. $2\text{A}$
• D. None of these

Answer 1

The correct option is A $1.33\text{A}$


The $3\Omega$ resistor and the $6\Omega$ resistor are joined in parallel. Their equivalent resistance is
$R=\frac{\left(3\Omega \right)\times \left(6\Omega \right)}{\left(3\Omega \right)+\left(6\Omega \right)}=2\Omega$
Thus, the two resistor may be replaced by a single resistor of resistance $2\Omega$. The circuit can be redrawn as shown in Figure b. The two resistor in the figure are joined in series. The equivalent resistance is $4\Omega$+ $2\Omega$= $6\Omega$
The current through the battery is $i=\frac{12V}{6\Omega }=2\text{A}$
Now, look at Figure $\left(a\right)$. The current through the battery and the $4\Omega$ resistor is $2\text{A}$.
This current is divided in the two resistors ( $3\Omega$ and $6\Omega$) which are joined in parallel.

Using $i_1=\frac{R_{2}i}{R_1+R_2}$ ,the current though the $3\Omega$ resistor is
$i_1=\frac{\left(6\Omega \right)\times \left(2A\right)}{\left(3\Omega \right)+\left(6\Omega \right)}=\frac{12\Omega A}{9\Omega }=1.33\text{A}$