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JEE Main Physics Class 12 Current Electricity Part 4 Questions
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© examsnet.com
Question : 3
Total: 31
An electric bulb rated
50
W
−
200
V
is connected across a
100
V
supply. The power dissipation of the bulb is :
[4 Apr 2024 Shift 2]
12.5
W
25
W
50
W
100
W
Validate
Solution:
👈: Video Solution
Rated power & voltage gives resistance
R
=
V
2
P
=
(
200
)
2
50
=
40000
50
R
=
800
P
=
(
V
applied
)
2
R
=
(
100
)
2
800
P
=
12.5
watt
Hence option 1 is correct.
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