Watt’s Law

DC Experiment 4

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Background

In electric circuits, power is a measure of the electrical energy per unit of time. The power dissipation of a component, e.g. a resistor, can be represented in the following form:

Where,

If the element has a constant resistance R, using the knowledge learnt from Ohm’s Law, we can rewrite Watt’s Law in the following forms:

The unit of power is in Watts (W).

For an element in a power circuit, if the element delivers or generate energy, then it is called a source. For example, batteries or the power supply we used are sources. On the other hand, if the element consumes or dissipates energy, it is called a load. Obviously the light bulbs are load devices since they consume electrical energy and transform into heat and light.

The rated power of a load represents its maximum capability to consume or dissipate electrical power. In general, with same fabrication technology, a device with large power rating indicates for larger size since it requires bigger contact surface to dissipate heat into ambient environment. In Figure 4.1, resistors with larger power ratings are bigger in size.

In the experiment section, we will construct two circuits in Figure 4.2 and examine Watt’s Law by visualizing the effects when the light bulbs are operating at different power levels. We will also become more familiar with the basic concepts of power rating when dealing with simple power elements.

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Procedure

This experiment you are dealing with components that dissipate heat, so DO NOT touch the resistors and light bulbs during operation or you may get burned. Follow the instructions very carefully.

Part I: Single Light Bulb

1. Using the bulb and big 10Ω resistor to construct the circuit as shown in Figure 4.3. Note that you must NOT use any small size 10Ω resistors with less than 2W power rating.

1. Set the output voltage of MEGO to 4.5V then connect onto the board. You should see the bulb is glowing. DO NOT touch the resistor or the bulb as it may get hot.

2. Measure the voltages across the light bulb and the resistor. Record the measured voltages of V R1 and V bulb in column 2 of Table 4.1.

3. Set VEGO to small current mode [mA] to measure the current through light bulb, see in the setup in Figure 4.4. Record the value in column 3 of Table 4.1.5

1. Calculate the power dissipation of the resistor and the light bulb using Watt’s Law,

Also, enter the calculated values in column 4 of Table 4.1.

Part II: Two Light Bulbs

1. Construct the circuit shown in Figure 4.5. Use the same 10 Ω resistor but meanwhile add another bulb in parallel this time.

1. Set power supply to 4.5V and observe the light intensity of two bulbs, which should be expected to get dimmer.

2. Slowly increase power supply voltage until each bulb gets roughly the same brightness as in Part 1 experiment. This voltage do not be higher than 7V.

3. Measure the voltage across the light bulbs and the resistor. Record the measured voltage of V R1, V bulb1, and V bulb2 in Table 4.2.

4. Set VEGO in small current mode [mA] and measure the current through each bulb right and the resistor. Follow the setup in Figure 4.6. Record the values in Table 4.2.

Figure 4.6 (a)

Figure 4.6 (b)

Figure 4.6 (c)

1. Based on the measured voltages and currents, calculate the power dissipation of the resistor and light bulbs. Note the results in Table 4.2.

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Exercises

Part I: Single Light Bulb

1. Comparing the dissipated power of the resistor and the light to their rated power in percentage, is it safe to operate this circuit? (Assuming we want to keep these loads running less than 50% of their ratings)

2. Based on your measurement and calculation results, how much power is the power supply delivering to this circuit?

Part II: Two Light Bulbs

1. In Part II step 3, why was it necessary to increase the power supply voltage to maintain the brightness of both light bulbs?

2. Comparing the dissipated power of the resistor and the light bulbs to their rated power in percentage, is it safe to operate this circuit?