Three resistors $R_{1} = 3 Ω, R_{2} = 4 Ω$ and $R_{3} = 5 Ω$ are connected as shown in figure. Find the equuivalent resistance of following circuit.

Open in App

Solution

Given, $R_{1} = 3 Ω$ $R_{2} = 4 Ω$ $R_{3} = 5 Ω$
On series, the equivalent resistance between and y is $R_{1} + R_{2} = 3 + 4$
$R_{e q} = 7 Ω$
In parallel the equivalent resistance is
$\frac{1}{R_{e q}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$
$\frac{1}{R_{e q}} = \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$
$\frac{1}{R_{e q}} = \frac{60}{47} Ω$
$\frac{1}{R_{e q}} = 1.276$
So the overall resistance of the circuit
$R_{e q} = R_{s e q} + R_{p e q}$
$R_{e q} = 7 + 1.276$
$R_{e q} = 8.276 Ω$

Suggest Corrections

Similar questions

R_{1}, R_{2}

R_{3}

R_{3}

1 Ω

R_{3}

R_{1}

R_{2}

R_{1}

R_{2}

r_{1}, r_{2}

r_{3}

\frac{i_{3}}{i_{1}}

Three resistors $2 Ω, 3 Ω$ and $4 Ω$ are connected in series in a circuit. The total resistance of the circuit is

Join BYJU'S Learning Program