Vol. Direct Current (DC)
Chapter 15 Inductors

# Inductor Voltage and Current Relationship

### Inductor Voltage and Current Relationship

The instantaneous voltage drop across an inductor is directly proportional to the rate of change of the current passing through the inductor. The mathematical relationship is given by v = L (di/dt).

Inductors do not have a stable “resistance” as conductors do. However, there is a definite mathematical relationship between voltage and current for an inductor that you can think of as Ohm's law for an inductor:

$$v = L \frac{di}{dt}$$

Where:

• v = instantaneous voltage across the inductor
• L = inductance in henries (H)
• $$\frac{di}{dt}$$ = instantaneous rate of current change in amperes per second (A/s)

This equation is similar to that for capacitors. It relates one variable (in this case, inductor voltage drop) to a rate of change of another variable (in this case, inductor current). Both the voltage (v) and the rate of current change (di/dt) are instantaneous. That is, concerning a specific point in time, the use of lowercase letters v and i.

Furthermore, the current rate of change is expressed in units of amps per second, with a positive number representing an increase and a negative number representing a decrease.

### Voltage Drop Across an Inductor with a Constant Current

Like a capacitor, an inductor’s behavior is rooted in the variable of time. Aside from any resistance intrinsic to an inductor’s wire coil (which we will assume is zero for the sake of this section), the voltage dropped across the terminals of an inductor is purely related to how quickly its current changes over time.

Suppose we were to connect a perfect inductor (one having zero ohms of wire resistance) to a circuit where we could vary the amount of current through it with a potentiometer connected as a variable resistor, as illustrated in Figure 1.

##### Figure 1. Measuring the voltage drop across an inductor.

If the potentiometer mechanism remains in a single position (the wiper is stationary), the series-connected ammeter will register a constant (unchanging) current, and the voltmeter connected across the inductor will register 0 V. In this scenario, the instantaneous rate of current change is equal to zero because the current is stable.

The inductor voltage equation tells us that with 0 A/s change for di/dt, there must be zero instantaneous voltage across the inductor.

$$v = L \frac{di}{dt} = L \cdot 0 = 0 \text{ V}$$

The instantaneous inductor current and voltage over time are illustrated in Figure 2.

##### Figure 2. A constant inductor current results in an inductor voltage of zero.

From a physical perspective, with no current change, the inductor will generate a steady magnetic field. With no change in magnetic flux (dΦ/dt = 0 in Webers per second), there will be no voltage dropped across the coil length due to induction.

### Voltage Drop Across an Inductor with a Constantly Increasing Current

If we move the potentiometer wiper slowly in the “up” direction, its resistance from end to end will slowly decrease. This attribute has the effect of increasing current in the circuit, thus, the ammeter indicator should be increasing at a slow rate (Figure 3).

##### Figure 3. Decreasing the potentiometer resistance increases the inductor current.

Assuming that the potentiometer wiper is being moved such that the rate of current increase through the inductor is steady, the di/dt term of the formula will be a fixed value. This fixed value, multiplied by the inductor’s inductance in Henrys (also fixed and the unit of electrical inductance), results in a fixed voltage of some magnitude, as shown in Figure 4.

##### Figure 4. A constantly increasing inductor current results in a fixed, positive inductor voltage.

From a physical perspective, the gradual increase in current results in a magnetic field that is likewise increasing. In this scenario, the inductor will be acting as a load, with the negative side of the induced voltage on the end where electrons are entering and the positive side of the induced voltage on the end where electrons are exiting

This gradual increase in magnetic flux causes a voltage to be induced in the coil, which can be calculated using Michael Faraday’s induction equation e = N(dΦ/dt).

Due to a gradual change in current magnitude through the coil, this self-induced voltage across the coil happens to be of a polarity that attempts to oppose the change in current. In other words, the induced voltage polarity resulting from an increase in current will be oriented in such a way as to push against the direction of the current to try to keep the current at its former magnitude.

This phenomenon exhibits a more general principle of physics known as Lenz’s Law, which states that an induced effect will always be opposed to the cause producing it.

### Voltage Drop Across an Inductor With a Variable, Increasing Current

Changing the rate of current increase through the inductor by moving the potentiometer wiper “up” at different speeds results in different amounts of voltage being dropped across the inductor, all with the same polarity (opposing the increase in current), as shown in Figure 5.

##### Figure 5. A changing, increasing inductor current results in a variable, positive inductor voltage.

Here again, we see the derivative function of calculus exhibited in the behavior of an inductor. In calculus terms, we would say that the induced voltage across the inductor is the derivative of the current through the inductor: that is, proportional to the current’s rate of change concerning time.

### Voltage Drop Across an Inductor With a Decreasing Current

Reversing the direction of the wiper motion on the potentiometer (going “down” rather than “up”) will result in its end-to-end resistance increasing (Figure 6).

##### Figure 6. Increasing the potentiometer resistance decreases the inductor current.

This results in the circuit current decreasing and a negative value for di/dt. The inductor, always opposing any change in current, will produce a voltage drop opposite to the change's direction.

With that in mind, how much voltage the inductor will produce depends on how rapidly the current through it is decreased. As described by Lenz’s Law, the induced voltage will be opposed to the change in current. With a decreasing current, the voltage polarity will be oriented to try to keep the current at its former magnitude.

In this scenario, the inductor will act as a source, with the negative side of the induced voltage on the end where electrons exit and the positive side of the induced voltage on the end where electrons enter. The more rapidly current is decreased, the more voltage will be produced by the inductor in its release of stored energy to try to keep the current constant.

Again, the amount of voltage across a perfect inductor is directly proportional to the rate of current change through it. The only difference between the effects of a decreasing current and an increasing current is the polarity of the induced voltage.

For the same rate of current change over time, either increasing or decreasing, the voltage magnitude (volts) will be the same. For example, a di/dt of -2 A/s will produce the same amount of induced voltage drop across an inductor as a di/dt of +2 A/s, just in the opposite polarity.

### Voltage Drop Across an Inductor with Rapid Current Changes

High voltages will be produced if the current through an inductor is forced to change rapidly. Consider the following circuit of Figure 7.

##### Figure 7. Inductor circuit with a neon lamp and an open switch.

In this circuit, a lamp is connected across the terminals of an inductor. A switch is used to control the current in the circuit, and power is supplied by a 6 V battery. When the switch is closed, the inductor will briefly oppose the change in current from zero to some magnitude but will drop only a small amount of voltage. It takes about 70 V to ionize the neon gas inside a neon bulb like this; thus, the bulb cannot be lit on the 6 V produced by the battery or the low voltage momentarily dropped by the inductor when the switch is closed (Figure 8).

##### Figure 8. Closing the switch allows current to flow in the circuit.

When the switch is opened, however, it suddenly introduces an extremely high resistance into the circuit (the resistance of the air gap between the contacts). This sudden introduction of high resistance into the circuit causes the circuit current to decrease almost instantly. Mathematically, the di/dt term will be a large negative number.

Such a rapid change of current (from some magnitude to zero in very little time) will induce a high voltage across the inductor, oriented with negative on the left and positive on the right, to oppose this decrease in current (Figure 9).

##### Figure 9. Opening the circuit results in a rapid decrease in the current and a high voltage across the inductor that lights the neon lamp.

The voltage produced is usually more than enough to light the neon lamp, if only for a brief moment until the current decays to zero.

For maximum effect, the inductor should be sized as large as possible (at least 1 Henry of inductance).

### Review of the Inductor Voltage and Current Relationship

• The instantaneous voltage drop across an inductor is directly proportional to the rate of change of the current passing through the inductor.
• The inductor's self-induced voltage has a polarity that opposes the change in current (Lenz's Law).
• The mathematical relationship is v = L (di/dt)

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