PHY294H - Lecture 32

Transformers - Using Mutual inductance

Transformers are the first example of a circuit in which we apply an alternating current (AC) source. When we consider the mutual inductance of two concentric solenoids, we found,

$${\cal{E}}_1 = - M {dI_2 \over dt}$$ (1)

The mutual inductance we found was,

$$M = {\mu_0 N_1N_2 A\over l_1}$$ (2)

In a transformer, we couple two coils using a high permeability magnetic core (e.g. Fe or Permalloy). If this coupling is perfect, we get the same mutual inductance as that of concentric solenoids. The self-inductance of a solenoid is,

$$L_1 = {\mu_0A N_1^2\over l_1}$$ (3)

Now we consider applying an oscillatory current (AC) source to one of the coils of a transformer. The applied current source applies an oscillatory voltage source,

$${\cal{E}} = V_0 sin(\omega t)$$ (4)

Let us assume that this produces a voltage across the first solenoid which is given by,

$${\cal{E}}_1 = V_1 sin(\omega t)$$ (5)

The current through the first solenoid is then found from,

$$V_1 sin(\omega t) = -L {dI_1 \over dt}$$ (6)

which implies,

$$I_1(t) = {V_1\over \omega L} cos(\omega t)$$ (7)

The voltage across the second solenoid is then,

$${\cal{E}}_2 = - M {dI_1 \over dt} = {M \over L} {\cal{E}}_1$$ (8)

Using Eq. () and (), we have,

$${{\cal{E}}_2 \over {\cal{E}}_1} = {V_2\over V_1} = {N_2 \over N_1}$$ (9)

If N₂>N₁ this is called a step up transformer, while if N₂<N₁ it is called a step down transformer.

In an ideal transformer no power is lost, so we must have,

V₁ I₁ = V₂ I₂ (10)

which implies

$${I_2 \over I_1} = {V_1 \over V_2} = {N_1 \over N_2}$$ (11)

A step up transformer has a step up in voltage, but a step down in current. A step down transformer steps down the voltage but steps up the current. Now consider the power lost in transmitting power down a transmission line which has resistance R. The loss is given by,

P_lost = I² R (12)

If we seek to deliver a power P_del. We do not know the load resistance, but we do know that the power delivered using the transmission line is P_del = VI. Using this to eliminate I in the above, we find

$$P_{lost} = {P_{del}^2 R\over V^2}$$ (13)

This means that the power lost is less if R is smaller and if V is large. That is, high voltage transmission lines are most efficient.

AC circuits

We already have treated some of the behaviors to be found in circuits containing inductors capacitors and resistors. We are going to analyse the ciruits again, but now with an AC voltage source. It is still true that ideal inductors and capacitors only store energy while resistors dissipate energy. The time constants for $\tau_L$ and $\tau_C$ are still important, however they only apply to transients in the circuit response. We shall consider the steady state response of circuits with AC sources. The natural frequency $\omega = 1/(LC)^{1/2}$, as we shall see, is the same as the resonant frequency of an LRC circuit with an AC source. To understand the behavior of AC circuits, we will introduce the concept of the reactance of a circuit element and also the idea of the phase of the current through the circuit and the phase relation between the current though a circuit element and the voltage across that circuit element.

A resistor in series with an AC source

This is the simplest case as usual, but it is useful to define the notation using this simple case. We have an AC voltage source given by,

$${\cal{E}} = V_0 sin(\omega t)$$ (14)

As usual Kirchhoff's loop law yields,

$${\cal{E}} - IR = V_0sin(\omega t) - I R = 0$$ (15)

Solving for the current we find,

$$I = {V_0\over R}sin(\omega t)$$ (16)

It is evident that this is the same as for a DC circuit. However for later reference, note that the phase of the current is the same as the phase as the voltage across the resistor.

A capacitor in series with an AC source

Kirchhoff's loop law now gives, ${\cal{E}} - {Q/ C} = 0$. Solving for the charge on the capacitor as a function of time, we thus find, $Q = C\ V_0 sin(\omega t)$. To find the current, we use,

$$I(t) = {dQ\over dt} = \omega C V_0 cos(\omega t) = \omega C\... ...\pi \over 2}) = {V_0\over X_C} sin(\omega t + {\pi \over 2})$$ (17)

From this expression we see that the current ``leads'' the voltage in a capacitor. This is natural as the capacitor has to charge up before it has a voltage so the current flow is ahead of the voltage. In the last expression we have defined the capacitive reactance, ${X_C} = {{1\over \omega C}}$. This acts like an effective resistance due to a capacitor. Note that if the frequency goes to zero, a capacitor has infinite reactance, while as frequency goes to infinite it has very small reactance.

An inductor in series with an AC source

Kirchhoff's loop law now gives,

$${\cal{E}} - L{dI\over dt} = 0$$ (18)

Solving for the charge on the capacitor as a function of time, we thus find,

$${dI\over dt} = {V_0\over L} sin(\omega t)$$ (19)

To find the current, we integrate to find,

$$I(t) = -{V_0\over \omega L} cos(\omega t) = {V_0\over X_L} sin(\omega t - {\pi \over 2})$$ (20)

From this expression we see that the current ``lags'' the voltage in an inductor. This is natural as the inductor resists a change in current (which would lead to a change in magnetic field in the coils of the inductor). We have defined the inductive reactance $X_L = \omega L$. The inductive reactance is larger at high frequency as there the rate of change of current is high and the inductor resists a change in current. At low frequencies the inductive reactance goes to zero as there the rate of change of current is slow and hence there is little induced emf, from Faraday's law.