Question

Find the current flowing through the following electric circuit:

Answer 1

Since, resistors $1\Omega$ and $3\Omega$ are in series,

$\therefore R_{eq1}=1+3=4\Omega$

Now, resistors $4\Omega$ and $6\Omega$ are in parallel,

$\therefore R_{eq2}=4\Omega \parallel 6\Omega =\frac{4\times 6}{4+6}=\frac{24}{10}=2.4\Omega$

Now, $3.6\Omega ,2.4\Omega$ and $3\Omega$ are again series,

$\therefore R_{eq}=3.6+2.4+3=9\Omega$

Now, the current flowing the electric current $I$

$I=\frac{V}{R_{eq}}$

$\Rightarrow I=\frac{4.9}{9}=0.5Amp$

$\Rightarrow I=0.5Amp$