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3.2: Ohm’s Law, Joules Law, and Series/Parallel Formulas

  • Page ID
    2335
  • Ohm’s Law

    V applied to it. The German physicist Georg Simon Ohm (1787–1854) was the first to demonstrate experimentally that the current in a metal wire is directly proportional to the voltage applied: I V.

    Ohm’s law. It can be viewed as a cause-and-effect relationship, with voltage the cause and current the effect. This is an empirical law like that for friction—an experimentally observed phenomenon. Such a linear relationship doesn’t always occur.

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    Making Connections: Real World Connections

    The figure illustrates the relationship between current and voltage for two different resistors. The slope of the graph represents the resistance value, which is 2Ω and 4Ω for the two lines shown.

    1012Ω or more. A dry person may have a hand-to-foot resistance of 105Ω, whereas the resistance of the human heart is about 103Ω. A meter-long piece of large-diameter copper wire may have a resistance of 105Ω, and superconductors have no resistance at all (they are non-ohmic). Resistance is related to the shape of an object and the material of which it is composed, as will be seen in Resistance and Resistivity.

    I=\frac{V}{R} for V, yielding V=IR.

    V can be interpreted as the voltage drop across a resistor produced by the current I. The phrase IRdrop is often used for this voltage. For instance, the headlight in Example has an IR drop of 12.0 V. If voltage is measured at various points in a circuit, it will be seen to increase at the voltage source and decrease at the resistor. Voltage is similar to fluid pressure. The voltage source is like a pump, creating a pressure difference, causing current—the flow of charge. The resistor is like a pipe that reduces pressure and limits flow because of its resistance. Conservation of energy has important consequences here. The voltage source supplies energy (causing an electric field and a current), and the resistor converts it to another form (such as thermal energy). In a simple circuit (one with a single simple resistor), the voltage supplied by the source equals the voltage drop across the resistor, since PE=qΔV, and the same q flows through each. Thus the energy supplied by the voltage source and the energy converted by the resistor are equal. (See Figure.)

    The voltage drop across a resistor in a simple circuit equals the voltage output of the battery.

    Making Connections: Conservation of Energy

    Resistors in Series and Parallel

    Figure. The total resistance of a combination of resistors depends on both their individual values and how they are connected.

    Image shows (a) A series connection of resistors. (b) A parallel connection of resistors.

    Joules Law

    Figure(a).) Since both operate on the same voltage, the 60-W bulb must draw more current to have a greater power rating. Thus the 60-W bulb’s resistance must be lower than that of a 25-W bulb. If we increase voltage, we also increase power. For example, when a 25-W bulb that is designed to operate on 120 V is connected to 240 V, it briefly glows very brightly and then burns out. Precisely how are voltage, current, and resistance related to electric power?

    qV, where q is the charge moved and V is the voltage (or more precisely, the potential difference the charge moves through). Power is the rate at which energy is moved, and so electric power is P = \frac{PE}{t} = \frac{qV}{t}.

    I = \frac{q}{t} (note that Δt=t here), the expression for power becomes P = IV.

    P) is simply the product of current times voltage. Power has familiar units of watts. Since the SI unit for potential energy (PE) is the joule, power has units of joules per second, or watts. Thus, 1 AV = 1 W. For example, cars often have one or more auxiliary power outlets with which you can charge a cell phone or other electronic devices. These outlets may be rated at 20 A, so that the circuit can deliver a maximum power P = IV = (20 A) (12 V) = 240 W. In some applications, electric power may be expressed as volt-amperes or even kilovolt-amperes 1 kA⋅V = 1 kW.

    I = V/R gives P = (\frac{V}{R})V = V2/R. Similarly, substituting V = IR gives P = I(IR) = I2R. Three expressions for electric power are listed together here for convenience:

    P = IV

    P = V2/R

    P = I2R.

    P can be the power dissipated by a single device and not the total power in the circuit.)

    P = V2/Rimplies that the lower the resistance connected to a given voltage source, the greater the power delivered. Furthermore, since voltage is squared inP = V2/R, the effect of applying a higher voltage is perhaps greater than expected. Thus, when the voltage is doubled to a 25-W bulb, its power nearly quadruples to about 100 W, burning it out. If the bulb’s resistance remained constant, its power would be exactly 100 W, but at the higher temperature its resistance is higher, too.