Question

Three resistors $R_1=3\Omega ,R_2=4\Omega$ and $R_3=5\Omega$ are connected as shown in figure. Find the equuivalent resistance of following circuit.

Answer 1

Given, $R_1=3\Omega$ $R_2=4\Omega$ $R_3=5\Omega$

On series, the equivalent resistance between and y is $R_1+R_2=3+4$

$R_{eq}=7\Omega$

In parallel the equivalent resistance is

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$\frac{1}{R_{eq}}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}$

$\frac{1}{R_{eq}}=\frac{60}{47}\Omega$

$\frac{1}{R_{eq}}=1.276$

So the overall resistance of the circuit

$R_{eq}=R_{seq}+R_{peq}$

$R_{eq}=7+1.276$

$R_{eq}=8.276\Omega$