Question

How is many 135 ohm resistor is parallel are required to carry 5A on 22V Line

Answer 1

Answer:

Explanation:

The total current in the line should be 5A current

The circuit is in series.

we know

$\frac{1}{R_{p} } = \frac{1}{R_{1} }+ \frac{1}{R_{2} }+ \frac{1}{R_{3} }$

${R_{} } = 135$

we dont know the no.of resistors required

$\frac{1}{R_{p} } = \frac{1}{135} }+ \frac{1}{135} } +\frac{1}{135} }$

we consider the no.of resistors as x

so we can rewrite it as

$\frac{1}{R_{p} } = \frac{1}{135} } * x$

$\frac{1}{R_{p} } = \frac{x}{135} }$

${R_{p} } = \frac{135}{x}$

Using Ohms Law

V=IR

220= 5 * $\frac{135}{x}$           ( V, voltage must be 220 V, I think you were mistaken)

solving we get

  x= 135/44

I do think this is a wrong question