Question

how many 330 ohm resistor in parallel are required to carry 20 ampere on 220 volt line​

Answer 1

To find:

Number of 330 ohm resistors to be attached in parallel in order to carry 20 ampere current on a 220 volt line.

Calculation:

Let the number of resistors required be "n".So, net resistance of Circuit will be :

$\dfrac{1}{R} = \dfrac{1}{330} + \dfrac{1}{330} + .... \: n \: times$

$= > \dfrac{1}{R} = \dfrac{n}{330}$

$= > R = \dfrac{330}{n}$

Applying Ohm's Law:

$\therefore \: V = I \times R$

$= > \: 220 = 20 \times \dfrac{330}{n}$

$= > \: \dfrac{330}{n} = 11$

$= > \:n = \dfrac{330}{11}$

$= > \: n = 30 \: resistors$

So, final answer is

$\boxed{ \sf{ \: n = 30 \: resistors}}$

Answer 2

Question:-

How many 330 ohm resistor in parallel are required to carry 20 ampere on 220 volt line?

Answer:-

We know, when n resistors are connected in parallel, their Equivalent Resistance is given by

$\boxed{\sf{\frac{1}{{R}_{eq}}\:=\:\frac{1}{{R}_{1}}\:+\:\frac{1}{{R}_{2}}\:+\:...+\:\frac{1}{{R}_{n}}}}$

Here each resistors are of 330 ohm and there are n such resistors connected in parallel. So their equivalent resistance:

$\large{\sf{\frac{1}{{R}_{eq}}\:=\:\frac{1}{330}\:+\:\frac{1}{330}\:+\:...\:+\:(n\:times)}}$

$\implies\large{\sf{\frac{1}{{R}_{eq}}\:=\:\frac{n}{330}}}$

$\implies\large{\sf{\frac{{R}_{eq}}{1}\:=\:\frac{330}{n}}}$

$\implies\large{\sf{{R}_{eq}\:=\:\frac{330}{n}\ohm}}$

And by Ohm's law

$\sf{\boxed{V\:=\:I\:*\:{R}_{eq}}}$

Here, given V = 220V, and I = 20A

$\implies\large{\sf{220\:=\:20\:*\:\frac{330}{n}}}$

$\implies\large{\sf{\frac{11}{330}\:=\:\frac{1}{n}}}$

$\implies\large{\sf{n\:=\:30\: resistors}}$

Ans. n = 30 resistors