Question

5. An electric iron is rated 1kW-220V. Calculate the following-<br />(a) The resistance of its heating element.<br />(b) The amount of current that will flow through the element.<br />(c) The amount of heat that will be produced in 2 min.<br />(d) The power consumed if the line voltage falls to 200V.​

Answer 1

Explanation:

p=1000w

v=220v

p=v^2 /r

1000=(220)^2/r

r=48400/1000

r=48.4

p=vi

1000=220i

i=1000/220

i=100/22

H=i^2 rt

H=(100×100)/(22×22)×48.4×2×60

H=1000×120

H=120000j

p=v^2/R

P=(200)(200)/48.4

P =826.4

Answer 2

Answer:

(i) 48.4 Ω

(ii) 4.5454 A

(iii) 120 kJ

(iv) 826.44 W

Explanation:

Part (i) -

we know that,

$P = \frac{V^2}{R}$

given,

P = 1kW = 1000 watt

V = 220 V

then in the equation,

$P = \frac{V^2}{R} \\\\1000 = \frac{(220)^2}{R} \\\\R = \frac{48400}{1000} \\\\R = 48.4 \Omega$

We got resistance  = 48.4 Ω

Part (ii) -

we know that,

$P = VI$

given,

P = 1kW = 1000 watt

V = 220 V

then in the equation,

$P = VI\\\\1000 = 220\times I\\\\I = \frac{1000}{220}\\ \\I = 4.5454 amperes$

We got a current that will flow through the element = 4.5454 A

Part (iii) -

we know that,

$H = I^2RT$

putting known values,

$H = (\frac{100}{22} )^2\times 48.4\times 120\\\\H = 120000 joules$

The amount of heat that will be produced in 2 min(120 sec) = 120 kJ

Part (iv) -

Power consumed =

$P = \frac{V^2}{R} \\\\\\P = \frac{(200)^2}{48.4} \\\\P = 826.44 W$

So, the power consumed will be = 826.44 W

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