Physics, asked by karanrar2001, 10 months ago

Example 105. A nichrome heating element across
230 V supply consumes 1.5 kW of power and heats up to a
temperature of 750°C. A tungsten bulb across the same supply
operates at a much higher temperature of 1600°C in order to
be able to emit light. Does it mean that the tungsten bulb
necessarily consumes greater power ?-(b) Which of the two
has greater resistance : a1 kW heater or a 100 W tungsten
bulb, both marked for 230 V ?
[NCERT​

Answers

Answered by vk5544323
0

Answer:

Answer :

A::D

Solution :

E=220V

P=60W

,R = (V^2)/(P) = 220xx(220)/(60) = 806.67 Omega (a) E' = 180V P' = (V^2)/(R ) = 40.16 =40 W (b) E'' =240VP''= (V^2)/(R )=71.4 W= 71W`.

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Answered by arpitzarkart56
1

Answer:

I hope it's help you

Explanation:

at room temperature 27°C , the resistance of the heating element is R_{27^{\circ}}= \frac{230}{3.2}=71.875\OmegaR

27

=

3.2

230

=71.875Ω

[ according to Ohm's law, V = IR , so , R = V/I ]

at the steady temperature t°C , the resistance ,

R_tR

t

=230/2.8 = 82.143\Omega82.143Ω

now, use the relation, R=R_0[1+\alpha(T-T_0)]R=R

0

[1+α(T−T

0

)]

here, R=R_t = 82.143\OmegaΩ

R_0=R_{27^{\circ}}=71.875\OmegaR

0

=R

27

=71.875Ω

\alpha=1.7\times10^{-4}/^{\circ}Cα=1.7×10

−4

/

C

now, 82.143 = 71.875 [1 + 1.7 × 10^-4(t - 27)]

=> 82.143 - 71.875 = 71.875 × 1.7 × 10^-4(t - 27)

=> 0.084 × 10^4 = t - 27

=> 840 + 27 = t

=> t = 867°C

hence, temperature = 867°C

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