Current Electricity Charging Discharging Of Capacitors Question 561
Question: A heater coil connected to a supply of a 220 V is dissipating some power $ P _{1}. $ The coil is cut into half and the two halves are connected in parallel. The heater now dissipates a power $ P _{2}. $ The ratio of power $ P _{1}:P _{2} $ is
[AFMC 2004]
Options:
A) 2 : 1
B) 1 : 2
C) 1 : 4
D) 4 : 1
Show Answer
Answer:
Correct Answer: C
Solution:
$ P=\frac{V^{2}}{R} $ . If resistance of heater coil is R, then resistance of parallel combination of two halves will be $ \frac{R}{4} $ So $ \frac{P _{1}}{P _{2}}=\frac{R _{2}}{R _{1}}=\frac{R/4}{R}=\frac{1}{4} $
BETA
