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Chapter 4: Q35E (page 843)

The power rating of a light bulb (such as a 100-W bulb) is the power it dissipates when connected across a 120-V potential difference. What is the resistance of (a) a 100-W bulb and (b) a 60-W bulb? (c) How much current does each bulb draw in normal use?

Short Answer

Expert verified
  1. The resistance of 100-W bulb is144惟.
  2. The resistance of 60-W bulb is 240惟.
  3. The current drawn by each bulb is 0.833 A and 0.500 A.

Step by step solution

01

Determination of the resistance of 100-W and 60-W bulbs.

The connection of the bulb is across a 120-V potential difference.

The expression of power is,

P=V2R

Solve for R and 100 W bulb,

role="math" localid="1655719758468" R=V2P=(120V)2(100W)=144惟

Repeat the calculation for 60 W bulb,

R=V2P=(120V)2(60W)=240惟Thus,theresistancesfor100-Wbulbisandfor60Wbulbis144惟andfor60Wbulbis240惟

02

(b) Determination of the current drawn by each bulb is

Use the Ohm鈥檚 Law

V=lRl=VRForthe100Wbulb:l=(120V)(144惟)=0.833AForthe60Wbulb:l=(120V)(240惟)=0.500AThus,thecurrentdrawnby100-Wbulbandthe60-Wbulbis0.833Aand0.500A

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