Electro Magnetic Induction And Alternating Currents Question 482
A resistance less ring has 2 bulbs A and B rated at 2V, 19 W and 2V, 29 W respectively. The ring encloses an ideal solenoid whose magnetic field is as shown. The radius of solen,oid is 1 m and the number of turns/length =1000/m. The current changes at rate of 9 A/sec Find the value of P if power dissipated in bulb B is $ 180 $ watt.
Options:
4
6
8
11
Show Answer
Answer:
Correct Answer: A
Solution:
- Resistance of bulb $ R=\frac{v^{2}}{P}=\frac{4}{10}=0.4 $
Resistance of bulb $ B=\frac{v^{2}}{P}=0.2 , \Omega $
$ emf=\frac{d\phi }{dt}=\frac{d}{dt}({\mu_{0}}nI A) $
$ ={\mu_{0}}n\times A\times \frac{dI}{dt}={{10}^{-7}}\times 4\pi \times 1000\times \pi {{(1)}^{2}}\times 9 $
$ v=36\times {{10}^{-3}} $
$ I=\frac{v}{{{R}_{eq}}}=\frac{36\times {{10}^{-3}}}{0.6}=6\times {{10}^{-2}},A $ Power dissipated through bulb $ B=I^{2}R $ $ =\left(6\times {{10}^{-2}}\right)^{2}\times 0.2=7.2\times {{10}^{-4}},watt $
BETA
