Question

Five resistors of 10 ohm each are connected in series.what will be their effective resistance?

Answer 1

$\large{\bf{\gray{\underline{\underline{\orange{Given:}}}}}}$

Five resistors of 10Ω are connected in series.

$\large{\bf{\gray{\underline{\underline{\green{To\:Find:}}}}}}$

Equivalent resistance of the series connection.

$\large{\bf{\gray{\underline{\underline{\pink{Solution:}}}}}}$

➠ Equivalent resistance of series connection is given by

$\dag\:\boxed{\bf{\red{R_s=R_1+R_2+...+R_{\infty}}}}$

Let's calculate :D

$:\implies\tt\:R_s=R_1+R_2+R_3+R_4+R_5$

$:\implies\tt\:R_s=R+R+R+R+R$

$:\implies\tt\:R_s=5R$

$:\implies\tt\:R_s=5(10)$

$:\implies\boxed{\bf{\purple{R_s=50\Omega}}}$

Answer 2

Answer:

$\boxed{\sf Effective \ resistance \ (R_{eff} = 50 \Omega}$

Given:

Five resistors of $\sf 10 \Omega$ each are connected in series.

To Find:

Effective resistance of the combination

Explanation:

When resistance are connected in series effective/equivalent resistance:

$\boxed{ \bold{R_{eff} = R_1 + R_2 + ... + R_n}}$

When the resistance connected in series combination are equal then effective/equivalent resistance:

$\boxed{ \bold{R_{eff} = nR}}$

Here;

n = Number of resistors connected in series

R = Value of resistance

In the given question,

n = 5

R = $\sf 10 \Omega$

$\therefore$

$\sf \implies R_{eff} = 5 \times 10$

$\sf \implies R_{eff} =50 \Omega$

Additional information:

When resistance are connected in parallel effective/equivalent resistance:

$\sf R_{eff} = (\frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n})^{-1}$

When the resistance connected in parallel combination are equal then effective/equivalent resistance:

$\sf R_{eff} = \frac{R}{n}$

Here;

n = Number of resistors connected in parallel

R = Value of resistance