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NEET 2003
Exam
NEET
Year
2003
Subject
Physics
Chapter
Alternating Current

NEET 2003 · Physics · Alternating Current

Two 220 volt, 100 watt bulbs are connected first in series and then...

Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time combination is connected to a 220 volt a.c. supply line. The power drawn by the combination in each case respectively will be:Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time combination is connected toTwo 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time combination is connected to

Options

  1. A50 watt, 10 watt
  2. B100 watt, 50 watt
  3. C200 watt, 150 watt
  4. D50 watt, 200 watt

Correct answer

D. 50 watt, 200 watt

Step-by-step solution

Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time combination is connected to R1=V2P1=(220)2100=484ΩR2=V2P2=(220)2100=484ΩI=220484×2=522P=I2(R1+R2)=2522×22×(484×2)=50 W\begin{aligned} & \mathrm{R}_1=\frac{\mathrm{V}^2}{\mathrm{P}_1}=\frac{(220)^2}{100}=484 \Omega \\ & \mathrm{R}_2=\frac{\mathrm{V}^2}{\mathrm{P}_2}=\frac{(220)^2}{100}=484 \Omega \\ & \mathrm{I}=\frac{220}{484 \times 2}=\frac{5}{22} \\ & \mathrm{P}=\mathrm{I}^2\left(\mathrm{R}_1+\mathrm{R}_2\right)=\frac{25}{22 \times 22} \times(484 \times 2) \\ & =50 \mathrm{~W}\end{aligned}) Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time combination is connected to P=V2R1+V2R2=2×(220)2484=200 W\begin{aligned} & P=\frac{V^2}{R_1}+\frac{V^2}{R_2} \\ & =\frac{2 \times(220)^2}{484}=200 \mathrm{~W}\end{aligned})
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