Current Electricity Charging Discharging Of Capacitors Question 568
If in the circuit, power dissipation is 150 W, then R is 150 W
[AIEEE 2002]
Options:
2 W
6 W
5 W
4 W
Show Answer
Answer:
Correct Answer: B
Solution:
$ P=\frac{V^{2}}{R _{eq}}\Rightarrow 150=\frac{{{(15)}^{2}}}{[2R/(R+2)]}=\frac{225\times (R+2)}{2R} $
Therefore $ R=\frac{450}{75}=6\Omega $ .
BETA
