Question

Part A : The resistance of an electric heater is 90 Ω when connected to 120 V. How much energy does it use during 15 min of operation? (answer in J)

Part B : The lightbulb used in a computer projector has a resistance of 71 Ω. What is the current through the bulb when it is operating on 120 V? ( answer in A )

Answer 1

Solution:

A) From the information,

P = V²/R = 120²/90

= 160 Watts,

The energy during 15 min of operation is,

E = Pt = 160*15*60

= 1.44*10⁵ J.

B)From the information,

I = V/R

= 120/71

= 1.70 A.

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