Ohm’s Law and Ohm’s Law Calculations

Because Ohm’s law is a relationship between volts, amps, and ohms, and because they must always balance out, if we know any two of the values, then we can calculate the third. If resistance stays the same but voltage rises, then the greater force pushes more current through the circuit. If resistance stays the same but voltage decreases, then less current will flow through the circuit. That means that the total current flow of a circuit in amps always equals the voltage divided by the resistance.

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FIGURE 36-24
Ohm’s law circle (see note to left).
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FIGURE 36-25
Calculating current flow in a circuit.

In calculating Ohm’s law, R stands for resistance, V for voltage, and A for current (A is for amps).* Depending on which value you wish to solve for, you will apply one of the following three formulas:

  • A = V/R
  • V = A × R
  • R = V ÷ A

Voltage can vary across different points in a circuit, but determining the current at any point can be found without an ammeter by using Ohm’s law.

Using the Ohm’s law circle will help you remember which math operation to use Figure 36-24. All you have to do is place your finger over the value you are looking for. If you place your finger on the top value (volts), then you would multiply amps by resistance. If you place your finger on one of the side values, then you would divide volts by the other value. This means two things. First, the values always have to balance. And second, as long as you know any two values, the third can be calculated. This is especially helpful when determining current flow, because instead of breaking into a circuit to measure current with a meter, if the voltage and resistance are known, Ohm’s law may be used to calculate amps.

For example, battery voltage can be measured. Let’s use 12 volts. The value of the resistor, 4 ohms, is on its casing. Current then equals voltage (12 volts) divided by resistance (4 ohms). We can quickly calculate that there are 3 amps of current flowing through every point in the circuit Figure 36-25.

Solving Ohm’s Law

Using the rules of Ohm’s law gives an accurate method of determining values in an electrical circuit. To assist with these problems, use the Ohm’s law circle (Figure 36-26). If the value of V and R are known, then to find A, V is divided by R. Place your thumb over A and the circle tells you this formula. Similarly, if V and A are known, then R can be found by dividing V by A. If A and R, are known, then V is found by multiplying A by R.

In the circuit diagram in Figure 36-27, the value of the amperage is not known. The applied voltage is 12 volts (V) and the resistance is 4 ohms (Ω). To find the value of A, simply divide V by R, or 12 by 4. The value of A is 3 amps (Α).

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FIGURE 36-26
To find the value of V, multiply A by R, or 6 amps by 4 ohms.
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FIGURE 36-27
To find the value of A, divide V by R, or in this case, 12 volts divided by 3 ohms.

In the circuit diagram in Figure 36-26, the value of the applied voltage is not known but the amperage and resistance are. To find the value of V, multiply A by R, or 6 amps by 4 ohms. The answer is 24 volts.

In the circuit diagram in Figure 36-27, the value of the current flow is unknown. To find the value of A, divide V by R, or in this case, 12 volts divided by 3 ohms. The value of A is 4 amps.

* Note that some sources use E for voltage and I for current. Also, we intentially reversed the R and A in the Ohm’s law circle to highlight the fact that Amps is ALWAYS the result (product) of Volts and Resistance. If the amps are not correct, it is because either the Volts or Resistance is wrong.

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AS-70: Ohm’s Law: The technician can demonstrate an understanding of and explain the use of Ohm’s law in verifying circuit parameters (resistance, voltage, amperage).

Ohm’s law quantifies the relationship between amperage, voltage, and resistance in any given circuit. It provides a means to calculate any one of these parameters from the values of the other two. Ohm’s law is easiest applied through the use of a circle diagram like the one shown like the one shown in the "magic circle" illustration.

These diagrams are used by placing your finger over the parameter you are trying to calculate, then reading the remainder of the diagram to identify the calculation you need to perform. For example, if you need to work out the amperage flowing in the circuit, block off the A, and your calculation is V ÷ R, or voltage divided by resistance. For a 12-volt automotive circuit with 20 ohms of resistance, your calculation is 12 ÷ 20, to arrive at 0.6 amps.

AS-71: Resistance: The technician can demonstrate an understanding of the relationship of resistance to heat, voltage drop, and circuit parameters.

The electrical resistance of a component is the component’s opposition to the flow of electrical current. Any resistance in a circuit creates a voltage drop across the resistive component, be it a resistor, a bulb, an electric motor, or merely a wire. The amount of voltage drop is proportional to the amount of resistance. Resistors work by dissipating energy in the form of heat; therefore, for a given amount of current flowing in a circuit, the higher the resistance of a component, the more heat it will produce.

Let’s look at an example using a rear window defogger. The defogger elements form a resistor in the circuit. In this case, heat is produced by the resistor grid to defog the window. The resistance of each grid line creates a steady voltage drop across the length of the grid line. Current flowing through the grid line then creates heat. Broken grid lines cannot conduct current flow, so there is no voltage drop, nor heat generated in that grid line. Battery voltage (12 volts) is applied across the element when you switch on the defogger. If you were to measure the voltage at the center of one of the grid lines in the element, you should expect to see a voltage of approximately 6 volts. If the grid is burned open on the ground side, then the voltage would read about 12 volts, since there cannot be a voltage drop if there is no current flow. If the grid was burned open on the power side, the voltage would be near 0 volts because the current flow cannot get past the open circuit to the middle of the grid line. Moving the voltmeter toward the side with the break can pinpoint the exact location of the broken grid line so that repairs can be made.