Basic Electrical Theory
Basic Electrical Theory
Related Tech Tips
I started working as a tech when I was 17 years old, fresh out of trade school. My first winter out on my own, I went to a service call in an older part of Orlando, a part of town I had never worked in before. It was an especially cold winter that year, and […]
Read more
When reading ANY article about electrical theory or application, keep in mind that it will only scratch the topic's surface. You can dedicate years of your life to understanding electrical theory and design the way many engineers do and still know just enough to be dangerous. In HVAC, we rarely need to have a DEEP […]
Read more
Back in the “good old days,” controls were all analog and mechanical; that simply means they acted in a directly connected and variable manner based on a change in force. Both pneumatic (air pressure) and hydraulic (fluid pressure) systems are examples of mechanical or analog controls. When the pressure increases or decreases on a particular […]
Read more
Comments
I though P=EI and E=IR. To #5.
I though P=EI and E=IR. To #5.
Watts(P) = Amps(I) X Volts (E)
Watts(P) = Amps(I) X Volts (E)
Question number 2: I = P / U gives 5000W / 23V = 217 A
No?
Question number 2: I = P / U gives 5000W / 23V = 217 A
No?
@Martin M — the heat strip uses 5kW at 230 volts. A heat strip is a resistive load, so we need to find the resistance, because the resistance will be the same across any voltage. Understanding that we’re talking about a resistor is key to this question.
By reducing the voltage across a resistor, you reduce the power consumed as well as the current flow (amps), so the 5kW only applies to 230v. The key is to find the resistance of the heat strip:
a) First, find amps @ 230v: Power = Amps (I) * Volts (V) –> 5000 = I * 230 –> I = 5000/230 –> I = 21.7 amps @ 230v
b) Now find the resistance of the heat strip, using Ohm’s law: Volts (V) = Amps (I) * Resistance (R) —> 230 = 21.7 * R —> R = 230/21.7 = 10.6 ohms is the heat strip resistance, which is a property of the heat strip that does not change for any voltage applied. The key to this problem is to understand how a resistor works in this manner.
c) Now that we know the resistance, we simply use Ohm’s law again, this time for 23 volts, to find the current at 23 volts: Volts (V) = Amps (I) * Resistance (R) —-> 23 = I * 10.6 —> I = 23/10.6 —> I = 2.17 amps @ 23 volts
Or, now that the theory behind the problem is explained, you can notice that 23 is 1/10 of 230. So the current at 23 volts will be 1/10 the current at 230 volts, i.e., just divide the current by 10: 5000 = 230 * amps —> I = 5000/230 = 21.7 amps at 230 volts
We’re using 10x less voltage, so the current will also drop 10x, since a resistive load is linear: A resistor that uses 21.7 amps at 230 volts, will use 2.17 amps at 23 volts.
The first method is the general way you’d solve this problem if the numbers don’t work out nicely. The second way is a little faster if you notice the way the numbers line up.
@Martin M — the heat strip uses 5kW at 230 volts. A heat strip is a resistive load, so we need to find the resistance, because the resistance will be the same across any voltage. Understanding that we’re talking about a resistor is key to this question.
By reducing the voltage across a resistor, you reduce the power consumed as well as the current flow (amps), so the 5kW only applies to 230v. The key is to find the resistance of the heat strip:
a) First, find amps @ 230v: Power = Amps (I) * Volts (V) –> 5000 = I * 230 –> I = 5000/230 –> I = 21.7 amps @ 230v
b) Now find the resistance of the heat strip, using Ohm’s law: Volts (V) = Amps (I) * Resistance (R) —> 230 = 21.7 * R —> R = 230/21.7 = 10.6 ohms is the heat strip resistance, which is a property of the heat strip that does not change for any voltage applied. The key to this problem is to understand how a resistor works in this manner.
c) Now that we know the resistance, we simply use Ohm’s law again, this time for 23 volts, to find the current at 23 volts: Volts (V) = Amps (I) * Resistance (R) —-> 23 = I * 10.6 —> I = 23/10.6 —> I = 2.17 amps @ 23 volts
Or, now that the theory behind the problem is explained, you can notice that 23 is 1/10 of 230. So the current at 23 volts will be 1/10 the current at 230 volts, i.e., just divide the current by 10: 5000 = 230 * amps —> I = 5000/230 = 21.7 amps at 230 volts
We’re using 10x less voltage, so the current will also drop 10x, since a resistive load is linear: A resistor that uses 21.7 amps at 230 volts, will use 2.17 amps at 23 volts.
The first method is the general way you’d solve this problem if the numbers don’t work out nicely. The second way is a little faster if you notice the way the numbers line up.
746 watts per hp. Should have read the question more closely and realized it was asking for Btu’s.
746 watts per hp. Should have read the question more closely and realized it was asking for Btu’s.
For Question #2 can you please explain how you got to that answer!? 1 Kw = 1000W, so using P= E x I, I just did 5000W (5Kw) / 23 = 217 .. I’m super confused!
For Question #2 can you please explain how you got to that answer!? 1 Kw = 1000W, so using P= E x I, I just did 5000W (5Kw) / 23 = 217 .. I’m super confused!
Instead of doing P=V x A. We convert 5 kW to 5000 watts. Divide 5000W / 230V = nominally 22 amps (21.73). We can now use ohms law to find the actual resistance of the 5kW strip with this information.
R= V/A: R= 230V/21.73A. R=10.58
now if we do 23V/10.58 we get 2.17 amps.
So the answer is really found in finding the resistance in the 5kW strip.
Instead of doing P=V x A. We convert 5 kW to 5000 watts. Divide 5000W / 230V = nominally 22 amps (21.73). We can now use ohms law to find the actual resistance of the 5kW strip with this information.
R= V/A: R= 230V/21.73A. R=10.58
now if we do 23V/10.58 we get 2.17 amps.
So the answer is really found in finding the resistance in the 5kW strip.
To leave a comment, you need to log in.
Log In