Question

Find the current flowing through the following electric circuit

Answer 1

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Answer 2

Explanation :

From the given figure :

$1\ \Omega$ and $3\ \Omega$ are in series. So, their equivalent is $4\ \Omega$.

Now the $4\ \Omega$ resistor and $6\ \Omega$ resistor are in parallel. So, their equivalent will be :

$\dfrac{1}{R}=\dfrac{1}{6}+\dfrac{1}{4}$

$R=2.4\ \Omega$

The resistors $3.6\ \Omega$, $2.4\ \Omega$ and $3\ \Omega$ are in series. So, equivalent becomes :

$R_{eq}=9\ \Omega$

Using Ohm's law :

$V=IR$

So, $I=\dfrac{V}{R}$

$I=\dfrac{4.5\ V}{9\ \Omega}$

$I=0.5\ A$

Hence, this is the required solution.