Current Electricity Charging Discharging Of Capacitors Question 533
Question: Two electric bulbs rated $ P _{1} $ watt V volts and $ P _{2} $ watt V volts are connected in parallel and V volts are applied to it. The total power will be
[MP PMT 2001; MP PET 2002]
Options:
A) $ P _{1}+P _{2}watt $
B) $ \sqrt{P _{1}P _{2}} $ watt
C) $ \frac{P _{1}P _{2}}{P _{1}+P _{2}}watt $
D) $ \frac{P _{1}+P _{2}}{P _{1}P _{2}}watt $
Show Answer
Answer:
Correct Answer: A
Solution:
If resistances of bulbs are $ R _{1} $ and $ R _{2} $ respectively then in parallel $ \frac{1}{R _{P}}=\frac{1}{R _{1}}+\frac{1}{R _{2}} $
Therefore $ \frac{1}{( \frac{V^{2}}{P _{p}} )}=\frac{1}{( \frac{V^{2}}{P _{1}} )}+\frac{1}{( \frac{V^{2}}{P _{2}} )} $
Therefore $ P _{P}=P _{1}+P _{2} $
BETA
