Ohm's Law Help (page 2)

By — McGraw-Hill Professional
Updated on Sep 5, 2011

Resistance Calculations

Ohm’s law can be used to find a resistance between two points in a dc circuit when the voltage and the current are known. For the following problems, imagine that both the voltmeter and ammeter scales in Fig. 12-8 are visible but that the potentiometer is uncalibrated.

Resistance Practice Problem


If the voltmeter reads 24 V and the ammeter shows 3.0 A, what is the value of the potentiometer?


Use the formula R = E/I , and plug in the values directly because they are expressed in volts and amperes: R = 24/3.0 = 8.0 Ω.

Power Calculations

You can calculate the power P (in watts, symbolized W) in a dc circuit such as that shown in Fig. 12-8 using the following formula:

P = EI

where E is the voltage in volts and I is the current in amperes. You may not be given the voltage directly, but you can calculate it if you know the current and the resistance.

Remember the Ohm’s law formula for obtaining voltage: E = IR . If you know I and R but don’t know E , you can get the power P by means of this formula:

P = (IR) I = I 2 R

That is, take the current in amperes, multiply this figure by itself, and then multiply the result by the resistance in ohms.

You also can get the power if you aren’t given the current directly. Suppose that you’re given only the voltage and the resistance. Remember the Ohm’s law formula for obtaining current: I = E/R . Therefore, you can calculate power using this formula:

P = E (E/R) = E 2 / R

That is, take the voltage, multiply it by itself, and divide by the resistance.

Stated all together, these power formulas are

P = El = I 2 R = E 2 /R

Now we are all ready to do power calculations. Refer once again to Fig. 12-8.

Direct Current Voltage/Current/Resistance Circuits Current Calculations

Fig. 12-8 . Circuit for working Ohm’s law problems.

Power Practice Problem


Suppose that the voltmeter reads 12 V and the ammeter shows 50 mA. What is the power dissipated by the potentiometer?


Use the formula P = EI . First, convert the current to amperes, getting I = 0.050 A. Then P = EI = 12 × 0.050 = 0.60 W.

Practice problems of these concepts can be found at: Direct Current Practice Test

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