In the current carrying branch shown in the figure, the voltmeter has a resistance of 100Ω.If power dissipated in 2Ω,resistance is 18 watt, the reading of the voltmeter is
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
100 Volt
b
300 Volt
c
400 Volt
d
200 Volt
answer is D.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
i2×2=18 ⇒i=3ACurrent through the voltmeter, iv=3×200100+200=2A∴voltmeter reading =2×100 volt =200 volt.