Current Electricity Charging Discharging Of Capacitors Question 478
Question: A 100 watt bulb working on 200 volt and a 200 watt bulb working on 100 volt have
Options:
A) Resistances in the ratio of 4 : 1
B) Maximum current ratings in the ratio of $ 1:4 $
C) Resistances in the ratio of 2 : 1
D) Maximum current ratings in the ratio of 1 : 2
Show Answer
Answer:
Correct Answer: B
Solution:
$ R=\frac{V^{2}}{P} $
Therefore $ R _{1}=\frac{200\times 200}{100}=400,\Omega $ and $ R _{2}=\frac{100\times 100}{200}=50,\Omega . $ Maximum current rating $ i=\frac{P}{V} $ So $ i _{1}=\frac{100}{200} $ and $ i _{2}=\frac{200}{100} $
Therefore $ \frac{i _{1}}{i _{2}}=\frac{1}{4} $ .
BETA
