Current Electricity Charging Discharging Of Capacitors Question 511
Question: A (100 W, 200 V) bulb is connected to a 160 V power supply. The power consumption would be
[CBSE PMT 1997; JIPMER 2000]
Options:
A) 64 W
B) 80 W
C) 100 W
D) 125 W
Show Answer
Answer:
Correct Answer: A
Solution:
$ P _{consumed}={{( \frac{V _{A}}{V _{R}} )}^{2}}\times P _{R}=\frac{{{(160)}^{2}}}{{{(200)}^{2}}}\times 100 $
$ =64W $
BETA
