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Alternating Current question

2022 · 27 Jun · Shift 1 · Q61
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Alternating Current question

2022 · 27 Jun · Shift 1 · Q61

JEE MainPhysicsAlternating CurrentNumerical+4 / −1
A 220 V, 50 Hz AC source is connected to a 25 V, 5 W lamp and an additional resistance R in series (as shown in figure) to run the lamp at its peak brightness, then the value of R (in ohm) will be ‾\underline{\hspace{2cm}}​. JEE Main 2022 (Online) 27th June Morning Shift Physics - Alternating Current Question 87 English
Numerical answer
View written solutionFree

Correct answer: 975

  1. Rated values of the lamp

The lamp is rated at:

  • Voltage VL=25 VV_L = 25\,\text{V}VL​=25V
  • Power PL=5 WP_L = 5\,\text{W}PL​=5W

So the rated current of the lamp is

I=PLVL=525=0.2 AI = \frac{P_L}{V_L} = \frac{5}{25} = 0.2\,\text{A}I=VL​PL​​=255​=0.2A

Its resistance at rated operation is

RL=VL2PL=2525=125 ΩR_L = \frac{V_L^2}{P_L} = \frac{25^2}{5} = 125\,\OmegaRL​=PL​VL2​​=5252​=125Ω
  1. Condition for peak brightness

For the lamp to glow at its normal (peak) brightness, it must receive its rated voltage/current.

Since the lamp and resistor RRR are in series with a 220 V220\,\text{V}220V AC source, we use rms values.

Thus,

  • Supply rms voltage =220 V= 220\,\text{V}=220V
  • Lamp rms voltage required =25 V= 25\,\text{V}=25V
  • Series current required =0.2 A= 0.2\,\text{A}=0.2A
  1. Voltage across the series resistor

The remaining rms voltage must drop across RRR:

VR=220−25=195 VV_R = 220 - 25 = 195\,\text{V}VR​=220−25=195V
  1. Calculate the required resistance

Using Ohm's law,

R=VRI=1950.2=975 ΩR = \frac{V_R}{I} = \frac{195}{0.2} = 975\,\OmegaR=IVR​​=0.2195​=975Ω
  1. Final answer
975 Ω\boxed{975\,\Omega}975Ω​

The derived answer matches the stored correct answer.

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