DC Power

When we talk about power in a DC circuit, we are talking about the rate at which a device in the circuit delivers energy to the circuit, or removes energy from the circuit. A resistor converts some of the electrical energy flowing through it into heat and thus it consumes power. The rate at which a resistor converts electrical energy into heat is the power consumed by the resistor.  We can compute the power consumed by a resistor from any of the three following formulas:

P = VI

P = I2R

P = V2/R

The unit of measure of electrical power is the Watt (symbol W). We can use remember these formulas by using 3 pies that work the same way that our Ohm's Law pie did:

For those of you who like math, it is not necessary to remember all this. It is possible to derive any of the power equations by using P = VI and substituting either V or I with the appropriate form of Ohm's Law. For example:

P = VI and V = IR. Thus,by substitution for V:

P = (IR)I = I2R

Let's work some examples:

Example 1: The voltage across a resistor is measured as 55 V, and the current flowing through it is measured as 120 mA. What is the power consumed by the resistor?

Solution:

We know V and I and we would like to find P. If we look at our three power pie charts, only the first one contains V, I, and P. We use that pie chart and black out P, the value we wish to find. Here is what we are left with:

Therefore, P = VI = 55V*1.2A = 66 W.

The next example is a little trickier:

Example 2: A 1.8KW resistor has a current of 75 mA flowing through it. The maximum power that this resistor can dissipate is 1/2 Watt. Is the resistor big enough, or do we need to replace it with one that has a higher power dissipation?

Solution:

We want to know how much power is being dissipated by the resistor. Then we can compare that to the maximum allowable dissipation of 1/2W. If the actual power is greater, we need to replace the 1/2 W resistor with one that can dissipate more power.

We know R and I and we would like to find P. Looking at our power pie charts, we see that only the second chart contains R and I and P. We black out P and we are left with:

Thus, P = I2R. We can put in values for I and R to find P:

P = (75mA)2(1.8 KW) = (0.075A)2(1800W) = (0.005625)(1800) = 10.1W

The resistor is dissipating a little more than 10 W. This is a lot more than its rated maximum dissipation, and if we leave the 1/2 W resistor in this circuit it will quickly overheat and explode. We need to put in a resistor that can dissipate more than 10W. A 20W resistor would do nicely.

Now that we have tried to blow up a resistor, it is probably time to get down to the more serious business of analyzing DC circuits, which we will begin in the next section.

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