Basic Electrical Theory

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1. According to Ohm’s Law, increasing resistance in a circuit

Question 1 of 10

2. A 5KW heatstrip rated at 230V would draw how many amps at 23V?

Question 2 of 10

3. Electrons contain a _________ Charge

Question 3 of 10

4. Atoms with _______ electrons in the outer layer (Valance) generally make better Insulators

Question 4 of 10

5. Volts X Amps = ________

Question 5 of 10

6. A common example of an inductive load is

Question 6 of 10

7. 1 Volt is

Question 7 of 10

8. A _________ is an example of a switch

Question 8 of 10

9. 1 HP = _______ BTUs per Hr

Question 9 of 10

10. When you install a dimmer on a incandescent light bulb and add resistance to the circuit the light will dim and the amperage of the circuit will decrease.

Question 10 of 10



  1. Avatar Jonathan Hoeppner says:

    I though P=EI and E=IR. To #5.

    1. Avatar Bryan Orr says:

      Watts(P) = Amps(I) X Volts (E)

  2. Avatar Martin M says:

    Question number 2: I = P / U gives 5000W / 23V = 217 A


    1. Avatar mputerio says:

      @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.

  3. Avatar Jamie Kitchen says:

    746 watts per hp. Should have read the question more closely and realized it was asking for Btu’s.

  4. Avatar Taylor Kristiansen says:

    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!

    1. Avatar Michael Liddle says:

      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.

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