Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Physics
An electric bulb marked as 50 W-200 V is connected across a 100 V supply. The power consumed by the bulb at present is
Question Error Report
Question is incomplete/wrong
Question not belongs to this Chapter
Answer is wrong
Solution is wrong
Answer & Solution is not matching
Spelling mistake
Image missing
Website not working properly
Other (not listed above)
Error description
Thank you for reporting, we will resolve it shortly
Back to Question
Thank you for reporting, we will resolve it shortly
Q. An electric bulb marked as $50 W-200\, V$ is connected across a $100\, V$ supply. The power consumed by the bulb at present is
JIPMER
JIPMER 2013
Current Electricity
A
37.5 W
11%
B
25 W
33%
C
12.5 W
43%
D
10 W
13%
Solution:
We know that $R=\frac{V^{2}}{P}$
Given $V=200\, V ,\, P=40\, W$
and $V'=100\, V$
Hence, $R=\frac{V^{2}}{P}=\frac{200 \times 200}{50}$
$P'=\frac{V^{2}}{R}=\frac{100 \times 100 \times 50}{200 \times 200}$
$P'=12.5\, W$