Question

The resistance of a bulb filament is 100 ohm at a temperature of 100 degrees centigrade . If it's temperature coefficient of resistance be 0.005 per degree centigrade , it's resistance will become 200 ohm at a temperature of

Answer 1

Rt= Ro(1+alpha t)

Hence, 100ohm= Ro(1+0.005 x 100)

therefore, Ro= 100/1.5

Now, 200= 100/1.5(1+0.005 x t2)

0.005 x t2= 2

t2= 400 degree centigrade

But by this formula, alpha=(R2-R1)/R1(t2-t1) we have

0.005=200-100/ 100(t2-100)

t2=300 degree centigrade

Answer 2

Answer:

a.) The resistance of the bulb becomes 200 ohm at a temperature of 300 degrees celsius.

Explanation:

R 100=100Ω ,

T1=100oC ,

R′=200Ω ,

α=0.005 per oC

Step 1: Resistance-temperature relationship

R=R0 (1+α(T2−T1))

where is the coefficient of resistance to temperature.

So, R′=R 100 (1+α(T2−T1)) .... (1)

Step 2: The final temperature

Putting R values in 100,

T1 ,

R′, we obtain in equation (1)

200=100(1+0.005(T2−100))

Or, T2−100=200

← T2=300 degrees celsius

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Answer 3

The resistance will become 200 ohms at a temperature of $300^{o} C$.

Step-by-step explanation:

Given:

The resistance of a bulb filament $=100$ ohm.

Temperature $T=100^{o}C$

$R^{1} =200ohm$.

Temperature coefficient of resistance ∝$=0.005$ per$^{o}C$.

To find:

Find the resistance will become 200 ohms at a temperature of.

Formula used:

$R^{1} =R_{100} (1+$∝$T_{2}-T_{1}))$

Solution:

By using $R^{1} =R_{100} (1+$∝$T_{2}-T_{1}))$

$200=100(1+0.005(T_{2}-100))$

$T_{2}=300^{o}C$

Hence, the temperature is $300^{o}C$.

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