**G5B03 from the General License Course Section 6.1, Power & Principles:
**

*Q. How many watts of electrical power are used if 400 VDC is supplied to an 800-ohm load?*A. 0.5 watts

B. 200 watts

C. 400 watts

D. 3200 watts

Let’s define a couple of terms within this question first, and then we’ll see how to solve it by deriving the proper mathematical relationship from our two old Technician Class friends, *Ohm’s Law* and *Power Law*.

Electrical *power* is the rate at which electrical energy is transferred in a circuit. It is a rate of doing work, measured in units of watts. As power increases, more work can be done in a given unit of time. In this question some work is being accomplished at an 800-ohm load. What’s a load in electrical terms?

An electrical *load* is the resistance (or impedance for AC) imposed by the components in a circuit. A load may be the resistance presented by a device that converts electrical energy into another form of energy. For instance, an antenna converts electrical energy into electromagnetic radiation (and some heat), and the antenna imposes a load on the transmitter circuit, impeding the flow of AC. A radio receiver’s speaker in an audio circuit imposes a load when converting electrical energy into mechanical sound waves. A common fan imposes a load and converts electrical energy into mechanical energy, creating a breeze along with the resistance or impedance presented to the circuit. In this question there is an unspecified load to which 400 volts of direct current is being supplied.

We need to calculate power in watts based upon the voltage (EMF or ‘E’) and the load resistance (R) in ohms. Remember from your Technician Class studies Ohm’s Law and Power Law. You may remember them arranged in a graphic like these below in which you cover the quantity that you seek to compute, and the remaining quantity graphical relationship tells you what to calculate:

**Ohm’s Law: E = I x R**

*Also*, I = E ÷ R and R = E ÷ I

**Power Law: P = E x I**

*Also*, I = P ÷ E and E = P ÷ I

By the Power Law we get an initial relationship for calculating power: P = E x I. From our question we know that E = 400 volts. However, we are not provided any information on the current (I). Rather, we have been provided the resistance of the load, 800 ohms.

Aha! Check out the Ohm’s Law relationship, I = E ÷ R. We can substitute E ÷ R for I in the Power Law relationship! So, it goes like this:

P = E x I, and

I = E ÷ R, so

P = E x (E ÷ R), otherwise denoted

P = E² ÷ R, also written

**P = E²/R**

Let’s plug in the values for E and for R and see what we calculate:

P = (400v)² / 800 ohms

P = 160,000 / 800

**P = 200 watts**

Notice also from Ohm’s Law that E = I x R. If we make the substitution for E within the Power Law we obtain:

P = E x I, and

E = I x R, so

P = I x R x I, otherwise denoted

**P = I² x R**

Given a known current (I) and load resistance (R), you can also compute the power transferred in a circuit, even without knowing the voltage (E). [See G5B05]

Keep the two bold type formulas above in your hip pocket for exam time calculation of power, or run through the derivation from Ohm’s Law and Power Law a few times so that you can do it again when needed.

The answer to General Class question G5B03, “**How many watts of electrical power are used if 400 VDC is supplied to an 800-ohm load?**” is “**B. 200 watts**.”

Related Questions: G5B04, G5B05