Vol.
DIY Electronics Projects

Chapter 2
Basic Projects and Test Equipment

As you will learn in this project, the power in a resistive circuit can be calculated given any two of the three primary parameters in a simple circuit: resistance, voltage, and current. Most of the power dissipated by a resistor is given off as heat, so in this project, you will also measure the temperature increase of a resistor under load, as illustrated in Figure 1.

You will also get to see what happens when a resistor overheats!

- 6 V battery
- Two 1/4 (0.25) W resistors: 10 Ω and 330 Ω.
- Small thermometer

Do not use any battery size greater than 6 V for this experiment.

The resistor values need not be exact but within five percent of the figures specified (+/- 0.5 Ω for the 10 Ω resistor; +/- 16.5 Ω for the 330 Ω resistor). The color codes for these 5% tolerance resistors are given by:

- 10 Ω resistor: Brown, Black, Black, Gold (10, +/- 5%)
- 330 Ω resistor: Orange, Orange, Brown, Gold (330, +/- 5%).

The thermometer should be as small as possible to facilitate rapid detection of heat produced by the resistor. I recommend a medical thermometer, the type used to take body temperature.

- Use of Joule’s Law
- Importance of component power ratings
- Significance of electrically common points

**Step 1: **Measure each resistor’s resistance with your ohmmeter, noting the exact values for later reference.

**Step 2: **Connect the 330 Ω resistors to the 6 V battery using a pair of jumper wires, as shown in the schematic diagram of Figure 2 and the illustration of Figure 1.

**CAUTION:** Connect the jumper wires to the resistor terminals before connecting the other ends to the battery. This will ensure your fingers are not touching the resistor when battery power is applied. You might wonder why I advise no bodily contact with the powered resistor. This is because it will become hot when powered by the battery.

**Step 3: **With the 330 Ω resistor connected to the battery, measure the voltage with a voltmeter. When measuring voltage, there is more than one way to obtain a proper reading, namely directly across the battery or directly across the resistor.

Battery voltage is the same as resistor voltage in this circuit since those two components share the same set of electrically common points. As illustrated in Figure 4, one side of the resistor is directly connected to one side of the battery, and the other side of the resistor is directly connected to the other side of the battery.

In Figure 4, all points of contact along the upper wire (colored red) are electrically common to each other. All points of contact along the lower wire (colored black) are likewise electrically common to each other. Voltage measured between any point on the upper wire and any point on the lower wire should be the same. Voltage measured between any two common points, however, should be zero.

**Step 4: **Using an ammeter, measure the current through the circuit. Again, there is no one correct way to measure current, so long as the ammeter is placed within the current flow path through the resistor and not across a voltage source.

To do this, make a break in the circuit, and place the ammeter within that break. Connect the two test probes to the two wires or terminal ends left open from the break. One viable option is shown in Figure 5.

**Step 5: **Now that you’ve measured and recorded resistor resistance, circuit voltage, and circuit current, you are ready to calculate power dissipation. Whereas voltage is the measure of electrical push motivating electrons to move through a circuit, and current is the measure of electron flow rate, power is the measure of work rate (how fast work is being done in the circuit). It takes a certain amount of work to push electrons through a resistance, and power describes how rapidly that work is taking place.

In mathematical equations, power is symbolized by the letter “P” and measured in the unit of the Watt (W). Power may be calculated by any one of three equations—collectively referred to as Joule’s law—given any two out of three quantities of voltage (V), current (I), and resistance (R):

$$P = I \cdot V$$

$$P = I^2R$$

$$P = \frac{V^2}{R}$$

Calculating power in this circuit, using the three measured values of voltage, current, and resistance, and the three equations. However you calculate it, the power dissipation figure should be roughly the same.

Assuming a battery with 6.000 volts and a resistor of exactly 330 Ω, the power dissipation will be 0.1090909 W, or 109.0909 mW, to use a metric prefix. Since the resistor has a power rating of 1/4 W (0.25 W, or 250 mW), it is more than capable of sustaining this level of power dissipation.

The power rating of any electrical component does not tell us how much power it will dissipate but simply how much power it *may* dissipate without sustaining damage. If the actual amount of dissipated power exceeds a component’s power rating, that component will increase the temperature to the point of damage.

**Step 6: **Touch the thermometer end to the middle of the resistor and see how warm it gets. Since the actual power level is almost half the rated power, the resistor should become noticeably warm, but it should not overheat.

**Step 7: **To illustrate what happens when a resistor exceeds its power ratings, disconnect the 330 Ω resistor and replace it with the 10 Ω resistor. Again, avoid touching the resistor once the circuit is complete, as it will heat up rapidly.

The safest way to do this is to disconnect one jumper wire from a battery terminal, then disconnect the 330 Ω resistor from the two alligator clips, then connect the 10 Ω resistor between the two clips, and finally reconnect the jumper wire back to the battery terminal.

**CAUTION: Keep the 10 Ω resistor away from any flammable materials when it is powered by the battery!**

You may not have enough time to take voltage and current measurements before the resistor begins to smoke. At the first sign of distress, disconnect one of the jumper wires from a battery terminal to interrupt the circuit's current and give the resistor a few moments to cool down.

**Step 8: **With power still disconnected, measure the resistor’s resistance with an ohmmeter and note any substantial deviation from its original value. If the resistor still measures within +/- 5% of its advertised value (between 9.5 and 10.5 Ω), re-connect the jumper wire and let it smoke a bit more.

What trend do you notice with the resistor’s value as it is damaged more and more by overpowering?

It is typical for resistors to fail with a greater-than-normal resistance when overheated. This is often a self-protective mode of failure, as an increased resistance results in less current and (generally) less power dissipation, cooling it down again. However, the resistor’s normal resistance value will not return if sufficiently damaged. Performing some Joule’s law calculations for resistor power again, we find that a 10 Ω resistor connected to a 6 V battery dissipates about 3.6 W of power, about 14.4 times its rated power dissipation. Little wonder it smokes so quickly after connection to the battery.

Learn more about the fundamentals behind this project in the resources below.

**Textbook:**

**Calculators:**

**Worksheets:**

Published under the terms and conditions of the Design Science License

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