Electric Circuits Legacy Problem #16 Guided Solution
Problem*
A 3-way light bulb for a 110-V lamp has two different filaments and three different power ratings. Turning the switch of the lamp toggles the light from OFF to low (50 W) to medium (100 W) to high (150 W) brightness. These three brightness settings are achieved by channeling current through the high resistance filament (50 W), the low resistance filament (100 W) or through both filaments. Determine the resistance of the 50 W and the 100 W filaments.
Audio Guided Solution
It�s quite likely that in your home you have a three-way lamp. And if you plug a three-way light bulb into such a three-way lamp, then when you turn the switch from off to on the first time, you�ll notice the light bulb lights at a low setting. If you turn it again, you�ll get it to light a little bit brighter. And if you turn it again, you�ll get it to light with ultimate brightness. Finally turning it a fourth time turns it off. Now you need to have the right light bulb in the right lamp. And if you do, you�ll notice this. If you don�t have the right light bulb in such a lamp, what you�ll do is turn it once to turn it on, turn the switch once to turn it on, turn it twice for no change, turn it three times for no change, and turn it a fourth time to turn it off. Now these three-way light bulbs have two filaments inside of them. These two filaments allow for the three different possible brightnesses of the bulb. One of the filaments, when on, will light it at low setting. It�s a 50-watt filament. When the other one is on, the 100-watt filament, it will light it at medium brightness. When you flip the switch again, current will channel through both of the filaments such that you�ll get the high brightness. Here we�re asked to calculate the resistance of the 50-watt and the 100-watt light bulbs when they�re plugged into a 110-volt circuit. So we need an equation that relates the voltage drop across the light bulb and the power and the resistance. The equation of choice would be the one that goes power equal the delta V squared divided by R, where the delta V, or voltage impressed across the circuit, is 110 volts, and the P is 50 watts in one instance and 100 watts in the other. So using that equation and rearranging it in order to solve for R, I have R equal delta V squared over P. Now I can plug and chug my 110 volts in for delta V and square it, and I can divide by 50 watts in order to get about 242 ohms, and I can divide by 100 watts in order to get about 121 ohms. I round these two figures to two significant digits.
Solution
50-watt filament: R = 240 Ω (rounded from 242 Ω)
100-watt filament: R = 120 Ω (rounded from 121 Ω)
Habbits of an Effective Problem Solver
- Read the problem carefully and develop a mental picture of the physical situation. If necessary, sketch a simple diagram of the physical situation to help you visualize it.
- Identify the known and unknown quantities and record them in an organized manner. Equate given values to the symbols used to represent the corresponding quantity - e.g., \(\descriptive{\text{δV}}{δV,change in voltage} = 9.0\unit{\volt}\); \(\descriptive{R}{R,resistance} = 0.025\unit{\ohm}\); \(\descriptive{I}{I,current} = \colorbox{gray}{Unknown}\).
- Use physics formulas and conceptual reasoning to plot a strategy for solving for the unknown quantity.
- Identify the appropriate formula(s) to use.
- Perform substitutions and algebraic manipulations in order to solve for the unknown quantity.
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