A Practical Power Supply Problem
Electricity and Magnetism
Level
3
A DC power source reads 12 volts with a Voltmeter. The maximum power output with an optimum load is 40 Watts. To leave a safety headroom, we need to define a load that consumes 30 Watts. What is the resistance of the load in Ohms?
The answer is 2.7.
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1 solution
It is known that the maximum power output is obtained when the resistance of the load, rx, is equal to the internal resistance of the source, r. So let us find the internal resistance of the source knowing Pmax=40 W, the maximum power:
Pmax=Imax^2 X r= (Vs/2 r)^2 Xr= Vs^2/(4 r)
Inserting values:
40=12^2/(4 r), which yields r=0.9 Ohms
Now, under the working condition we want to design, the working current, Iw=12/(0.9+rx), where rx is the load resistance. So, with the load power of Pw=30 W, we have:
Pw=Iw^2 rx
30=144 rx /(0.9+rx)^2
Solving this quadratic equation for rx gives two answers 0.3 Ohm and 2.7 Ohm. We should only accept 2.7 Ohm as the answer, because if we chose 0.3 Ohms, we have almost shorted out the supply. So the answer is 2.7 Ohms.