How to find the wattage of the 20 Ω load?
- Thread starter adamaero
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ron
Senior Member
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- New York, 40.7514,-73.9925
You got it
gar
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adamaero:
You obviously know the answer. But why draw the circuit as shown? You show no voltage source. So I will assume it is inductively coupled into your 1 turn loop. Now, if this is intended to be a trick question, then I don't know what the object is because you only supply 4 answers.
It is a simple question to answer. So what is the real purpose of the question as presented.
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adamaero:
You obviously know the answer. But why draw the circuit as shown? You show no voltage source. So I will assume it is inductively coupled into your 1 turn loop. Now, if this is intended to be a trick question, then I don't know what the object is because you only supply 4 answers.
It is a simple question to answer. So what is the real purpose of the question as presented.
.
.
ron
Senior Member
- Location
- New York, 40.7514,-73.9925
LarryFine
Master Electrician Electric Contractor Richmond VA
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- Henrico County, VA
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- Electrical Contractor
The simple-math approach:
The 20-ohm resistor is five times the 4, so its voltage drop will be (5 x 24) 120v.
E = I x R, so you need the current. 24 + 120 = 144, the circuit voltage, 4 + 20 = 24, the resistance.
I = E/R = 144 / 24 = 6a. E = I x R = 6 x 20 = 120v. P = E x I = 120 x 6 = 720w.
The 20-ohm resistor is five times the 4, so its voltage drop will be (5 x 24) 120v.
E = I x R, so you need the current. 24 + 120 = 144, the circuit voltage, 4 + 20 = 24, the resistance.
I = E/R = 144 / 24 = 6a. E = I x R = 6 x 20 = 120v. P = E x I = 120 x 6 = 720w.
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