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  1. Problems
  2. /A bulb is marked 220 V, 40 W. Find the current it takes at 220 V…
MediumPhysicsHigh School (9–10)Electricity & Magnetism › Current ElectricityNumericalBoard ExamAEEE (Amrita)AMUEEEAP EAPCETBCECECOMEDK UGET
+15 more exams
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A bulb is marked 220 V, 40 W. Find the current it takes at 220 V…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A bulb is marked 220 V220\ \text{V}220 V, 40 W40\ \text{W}40 W.

  1. Find the current it takes at 220 V220\ \text{V}220 V.
  2. Find the resistance of the filament.
  3. What power does it give on 165 V165\ \text{V}165 V? (Take the resistance as fixed.)

Check your answer

Work it out on paper, then type your number. The answer itself stays hidden.

A
Ω
W
Need a reminder? The key ideaShowHide

Joule's law: H=I2Rt=VIt=V2R tH = I^2 R t = VIt = \dfrac{V^2}{R}\,tH=I2Rt=VIt=RV2​t. Power P=VI=I2R=V2RP = VI = I^2 R = \dfrac{V^2}{R}P=VI=I2R=RV2​, in watts. A fuse, in the live wire, melts when the current is too large.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. (a) I=40 W220 V=0.182 AI = \dfrac{40\ \text{W}}{220\ \text{V}} = 0.182\ \text{A}I=220 V40 W​=0.182 A.

Step 2. (b) R=(220 V)240 W=1210 ΩR = \dfrac{(220\ \text{V})^2}{40\ \text{W}} = 1210\ \OmegaR=40 W(220 V)2​=1210 Ω.

Step 3. (c) P=(165 V)21210 Ω=22.5 WP = \dfrac{(165\ \text{V})^2}{1210\ \Omega} = 22.5\ \text{W}P=1210 Ω(165 V)2​=22.5 W.

Answer: (a) 0.182 A0.182\ \text{A}0.182 A. (b) 1210 Ω1210\ \Omega1210 Ω. (c) 22.5 W22.5\ \text{W}22.5 W.

What do we see? Three quarters of the voltage gives (34)2=916\left(\dfrac{3}{4}\right)^2 = \dfrac{9}{16}(43​)2=169​ of the power.

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