Question

Anichrome wire used as a heater coil has
the resistance of 2 82/m. For a heater of
1 kW at 200 V, the length of wire re-
quired will be
(a) 80 m
(6) 60 m
(c) 40 m
(d) 20 m

Answer 1

Solution:

We know that:

$\boxed{\sf{Power = \frac{V^{2}}{R}}}$

Here:

$\implies$ 'V' is the potential difference

$\implies$ 'R' is the resistance of wire.

Given:

Resistance of wire per unit length = 2 Ω/m

Let length of wire be 'L'.

Then,

Resistance of wire:

$\implies$ Per unit length × Length of wire

$\implies$ 2L Ω

We know that:

$\boxed{\sf{Power = 1\:kW = 1000\:watt}}$

So:

Potential difference (V) = 200 V

Therefore:

$\implies$ 1000 = (200)² / 2L

$\implies$ 1000 = 40000 / 2L

$\implies$ L = 20 m

Hence,

Length of wire is 20 m.

Therefore:

Correct option: (d) 20 m

Answer 2

✿━━━━@Mg━━━━✿

$\boxed{Explained\:Answer}$

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✿━━━━@Mg━━━━✿

Anichrome wire used as a heater coil has

the resistance of 2 82/m. For a heater of

1 kW at 200 V, the length of wire re-

quired will be

(a) 80 m

(6) 60 m

(c) 40 m

(d) 20 m❤❤