Question

An electric bulb is rated 200V, 100W. What is the heat energy developed in the bulb when used for 1 minute in 250V supply. *
a)6000 J
b)1562.5 J
c)9375 J
d)3000 J​

Answer 1

Answer:

Correct option is

C

30 KJ

⇒R=PV2=100200×200=400

Q=RV2×t

=400100×100×20×60

=30000J

=30KJ

Answer 2

We know that,

• The resistance of a conductor remains constant, so let us first find the resistance;

Power = (Volt)²/Resistance

100 w = (200)²/R

100 = 40000/R

100 R = 40000

R = 400 $\Omega$

Now, let the us find the Heat energy;

• Heat = Power × Time

H = P × T

H = (V²/R) × T

H = 250²/400 × 60 - - - - - - { 1 min = 60 sec }

H = 62500/400 × 60

H = 625 × 15

H = 9375 Joule

Therefore, option c) 9375 j is correct.

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Thank youu!! :))