Current Electricity Charging Discharging Of Capacitors Question 99
Question: A 100 W bulb B1, and two 60-W bulbs B2 and B3, are connected to a 250 V source, as shown in the figure. Now W1, W2 and W3 are the output powers of the bulbs B1, B2 and B3, respectively. Then
[IIT-JEE (Screening) 2002]
Options:
A) $ W _{1}>W _{2}=W _{3} $
B) $ W _{1}>W _{2}>W _{3} $
C) $ W _{1}<W _{2}=W _{3} $
D) $ W _{1}<W _{2}<W _{3} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ P=\frac{V^{2}}{R} $ so $ R=\frac{V^{2}}{P} $
Therefore $ R _{1}=\frac{V^{2}}{100} $ and $ R _{2}=R _{3}=\frac{V^{2}}{60} $ Now $ W _{1}=\frac{{{(250)}^{2}}}{{{(R _{1}+R _{2})}^{2}}}.R _{1} $ , $ W _{2}=\frac{{{(250)}^{2}}}{{{(R _{1}+R _{2})}^{2}}}.R _{2} $ and $ W _{3}=\frac{{{(250)}^{2}}}{R _{3}} $
$ W _{1}:W _{2}:W _{3}=15:25:64 $ or $ W _{1}<W _{2}<W _{3} $
BETA
