Bulbs Not Quite As Bright

Electricity and Magnetism Level 3

Three light bulbs are rated to dissipate 60 W 60 W per bulb at 120 V 120 V (AC RMS). Suppose the three bulbs are connected in parallel, and the parallel combination is supplied by a 120 V 120 V AC source with 5 Ω 5 \Omega of internal resistance.

How much voltage is across each bulb (to the nearest volt)?


The answer is 113.

Steven Chase by Steven Chase , 37, USA

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1 solution

Grant Bulaong
Oct 31, 2017

The resistance of each bulb is R = V 2 P = 240 Ω R=\dfrac{V^2}{P}=240\Omega . With three bulbs connected in parallel, their equivalent resistance is 240 3 = 80 Ω \tfrac{240}{3} = 80\Omega . Thus the current passing through the circuit is 120 80 + 5 = 24 17 A \tfrac{120}{80+5}=\tfrac{24}{17}A . This passes through the three bulbs, which are collectively in series with the internal resistance. The voltage across these bulbs is given by V = I R = ( 24 17 ) ( 80 ) = 113 V V=IR=(\tfrac{24}{17})(80)=113V .

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