At the end of many Indian lanes a grey steel tank sits on a platform between two poles. Three thick wires reach its top on tall porcelain insulators. They carry 11 000 volts from the feeder. Thinner cables leave its side and run to the houses at about 220 volts. The tank hums softly all day, yet nothing inside it moves.
Two coils on one core
A transformer is two coils of insulated wire wound on one core of soft iron. The coil joined to the supply is the primary coil, with \(N_p\) turns. The coil joined to the load is the secondary coil, with \(N_s\) turns. No wire joins the two coils. Energy passes from one to the other only through the magnetic field in the core.
An alternating voltage across the primary drives an alternating current in it. That current makes an alternating magnetic flux in the core. The iron guides almost all of this flux round the core, through every turn of the secondary as well. The flux never stops changing, so it induces an emf in every turn it threads. This is mutual induction, met in the chapter Eddy Currents, Generators and Inductance.
The same emf in every turn
The same flux \(\Phi\) passes through a turn of the primary and a turn of the secondary. So the emf in one turn, \(d\Phi/dt\), is the same in both coils. The turns of a coil are in series, so their emfs add. In an ideal transformer the windings have no resistance, so each coil's voltage equals its total emf.
\[ V_p = N_p\,\dfrac{d\Phi}{dt} \qquad V_s = N_s\,\dfrac{d\Phi}{dt} \]
Divide the second by the first. The rate of change of flux cancels, and only the numbers of turns remain.
An alternating current in the primary coil makes an alternating flux in the core. The flux passes through the secondary coil and induces an emf in it. In an ideal transformer every turn of both coils carries the same emf, so the voltages are in the ratio of the numbers of turns.
The turns ratio \(N_s/N_p\) sets the output. With more turns on the secondary, \(N_s > N_p\), the output voltage is larger than the input. The transformer is a step-up transformer. With fewer turns on the secondary it is a step-down transformer. The pole transformer of the lane steps down from \(11\,000\,\text{V}\) to \(220\,\text{V}\). Its primary has \(11\,000/220 = 50\) times as many turns as its secondary.
Try the figure above. With \(100\) turns on the secondary the output is \(110\,\text{V}\), and with \(2000\) turns it is \(2200\,\text{V}\). The output voltage follows the secondary turns in simple proportion. The secondary current follows the output voltage, since the load is fixed. The primary current does something more surprising, which the next topic explains.
(a) A transformer has \(200\) turns on its primary and \(2000\) on its secondary, and the primary is on the \(220\,\text{V}\) mains. Find the secondary voltage, and say whether the transformer steps up or down. (b) A doorbell needs \(12\,\text{V}\) from a \(240\,\text{V}\) supply. Find the turns ratio. If the primary has \(1000\) turns, how many has the secondary?
(a) \(V_s = V_pN_s/N_p\) \(= 220\times2000/200 = 2200\,\text{V}\). The secondary has more turns, so it is a step-up transformer.
(b) \(N_p/N_s = V_p/V_s = 240/12 = 20\), a turns ratio of \(20 : 1\). With \(N_p = 1000\), the secondary has \(1000/20 = 50\) turns.
Values used: the mains \(220\,\text{V}\) at \(50\,\text{Hz}\); part (b) takes the sheet's \(240\,\text{V}\) supply.
Why a battery cannot drive it
A transformer needs a changing flux. Join the primary to a battery instead, and the current settles to a steady value. A steady current makes a steady flux, and a steady flux induces no emf. So the secondary gives nothing once the current has settled. Worse, the primary's resistance is small, so the steady current is large and heats the winding.
This is one reason the mains is alternating. A home inverter shows the same rule from the other side. Its battery gives \(12\,\text{V}\) of direct current. The inverter first switches that current to and fro, fifty times a second, and only then can a transformer step it up to \(220\,\text{V}\).
A \(12\,\text{V}\) car battery is joined to the primary of a transformer with a turns ratio \(N_s/N_p = 20\). What does the secondary give once the primary current has become steady?
Power in, power out
An ideal transformer has windings of no resistance and a core that wastes no energy. Then every joule the primary takes from the supply reaches the load through the secondary. So the power in equals the power out, \(V_pI_p = V_sI_s\). Put this beside the voltage ratio and the currents come out in the inverse ratio of the turns.
\[ V_pI_p = V_sI_s \]
Why. an ideal transformer loses no power, so the power in equals the power out.
\[ \dfrac{I_s}{I_p} = \dfrac{V_p}{V_s} \]
Why. divide both sides by \(V_sI_p\).
\[ \dfrac{I_s}{I_p} = \dfrac{N_p}{N_s} \]
Read it. the voltages are in the ratio of the turns, \(V_p/V_s = N_p/N_s\).
A step-up transformer raises the voltage and lowers the current by the same factor. It does not make power. In the example above, the \(2200\,\text{V}\) secondary supplies \(1\,\text{A}\) to its load, which is \(2200\,\text{W}\). The primary must then take \(2200\,\text{W}\) from the \(220\,\text{V}\) mains, so it draws \(I_p = 2200/220 = 10\,\text{A}\). The voltage went up ten times, and the current came down ten times.
The primary feels the load
Open the secondary, so that no current flows in it. The primary is then only a coil of large inductance across the mains. Its self-induced emf almost equals the supply voltage at every instant. So only a small magnetising current flows, just enough to keep the flux alternating. The chapter Alternating Current and the LCR Circuit shows that such a current in a coil takes almost no average power.
Now close the secondary through a load. The secondary current makes its own flux in the core, against the flux of the primary. The flux, and with it the back emf of the primary, starts to fall. More current then flows in from the supply, until the flux is back to what the supply voltage demands. So the primary current grows exactly as the load draws power.
A transformer also changes how large a resistance looks. A load \(R\) on the secondary draws from the primary as if a resistance \((N_p/N_s)^2R\) were joined there directly. So an \(8\,\Omega\) loudspeaker behind a \(10 : 1\) step-down transformer looks like \(800\,\Omega\) to the amplifier. This is called impedance matching, and it is why transformers sit inside radios and amplifiers.
An ideal pole transformer takes \(11\,000\,\text{V}\) and gives \(220\,\text{V}\). The houses on its secondary draw \(100\,\text{A}\) in all. What current flows in its primary, in amperes?
A student says, "A step-up transformer gives out more voltage, so it must give out more power than it takes in." What is wrong?
Where a real transformer loses energy
A real transformer gives out a little less power than it takes in. The rest turns into heat inside it. Its efficiency \(\eta\) is the fraction of the input power that reaches the load. Large transformers reach about \(99\,\%\), and even a small one in a charger passes on most of its input.
The core is of soft iron for two reasons. Its high permeability keeps almost all the flux inside the core, so little leaks. It is also magnetised and demagnetised easily, with a thin hysteresis loop, as the chapter Magnetism and Matter showed. The windings are insulated from the core, because iron conducts and would short the turns together.
Now the hum of the pole transformer makes sense. Each time its iron sheets are magnetised, they change their length very slightly. That happens twice in every cycle, so the tank hums at \(100\,\text{Hz}\). The heat of the losses goes into the oil that fills the tank. The oil carries it to the cooling fins outside.
A transformer of efficiency \(90\,\%\) takes \(2\,\text{A}\) from the \(220\,\text{V}\) mains. Its secondary gives \(11\,\text{V}\). Find the output power and the secondary current.
The input power is \(V_pI_p\). The output is \(90\,\%\) of it. The secondary current is the output power over the secondary voltage.
\(P_\text{in} = V_pI_p\), \(P_\text{out} = \eta P_\text{in}\), \(I_s = P_\text{out}/V_s\).
\(P_\text{in} = 220\times2 = 440\,\text{W}\). \(P_\text{out} = 0.9\times440 = 396\,\text{W}\). \(I_s = 396/11 = 36\,\text{A}\).
The transformer gives \(396\,\text{W}\) at \(11\,\text{V}\), a current of \(36\,\text{A}\).
A transformer takes \(1000\,\text{W}\) from the supply and delivers \(900\,\text{W}\) to its load. What is its efficiency, in percent?
Which of these reduce the energy lost in a transformer?
Sending power a long way
A power station sends a power \(P\) to a town along a transmission line of resistance \(R\). The current in the line is \(I = P/V\), where \(V\) is the voltage at which the power is sent. The line heats at the rate \(I^2R\). So the loss is \(P^2R/V^2\). For the same power, the loss falls as the square of the line voltage.
A station sends \(100\,\text{kW}\) to a town through a line of total resistance \(2\,\Omega\). Find the line current and the power lost when the line voltage is (a) \(1\,\text{kV}\) and (b) \(10\,\text{kV}\). What fraction of the power is lost each time?
The power sent is fixed at \(10^5\,\text{W}\). Each line voltage gives its own current, and the loss is that current squared times \(2\,\Omega\).
\(I = P/V\), and loss \(= I^2R\).
(a) \(I = 10^5/10^3 = 100\,\text{A}\), loss \(= 100^2\times2 = 20\,\text{kW}\). (b) \(I = 10^5/10^4 = 10\,\text{A}\), loss \(= 10^2\times2 = 200\,\text{W}\).
At \(1\,\text{kV}\) the line wastes \(20\,\%\) of the power; at \(10\,\text{kV}\) only \(0.2\,\%\).
Do not work out the line loss as \(V^2/R\) with \(V\) the line voltage. That would make the loss grow with the voltage, the opposite of the truth. The line voltage is shared between the line and the town. Only the small drop \(IR\) across the line itself heats it. Always find the current first, \(I = P/V\), and then the loss \(I^2R\).
High voltage is dangerous in a home, and no appliance is built for it. So the voltage is raised by a step-up transformer at the power station, and lowered in steps near the users. The electricity from the Tehri dam in Uttarakhand leaves on lines of \(400\,\text{kV}\) and above. Substations step it down to \(220\), \(132\) and \(33\,\text{kV}\), and then to the \(11\,\text{kV}\) of the feeders. The grey box on the pole takes the last step, from \(11\,\text{kV}\) to the \(220\,\text{V}\) of your house.
A line of \(0.5\,\Omega\) carries \(22\,\text{kW}\) to a village, once at \(220\,\text{V}\) and once at \(11\,\text{kV}\). How many times smaller is the power lost in the line at \(11\,\text{kV}\)?
Why is electrical power stepped up to a high voltage for transmission, and stepped down again near the user?
For a given power \(P = VI\), a higher voltage means a smaller current. The heat lost in the line is \(I^2R\), so it falls as the square of the current, and the high voltage saves most of the loss. High voltages are dangerous and useless for appliances, so transformers step the voltage down again near homes and factories.
Many Indian homes keep an inverter and a battery for power cuts. The battery gives \(12\,\text{V}\) of direct current, which no transformer can use. So the inverter's electronics switch the current to and fro fifty times a second. A transformer with about \(18\) times as many turns on its secondary then steps the \(12\,\text{V}\) up to \(220\,\text{V}\). When the mains returns, the same transformer works backwards and steps the mains down to charge the battery.
- Using a transformer on a battery.A transformer needs a changing flux; direct current gives none once it is steady.
- Thinking a step-up transformer gives more power.\(V_pI_p = V_sI_s\) at best; the current is stepped down by the same factor as the voltage is stepped up.
- Writing \(I_s/I_p = N_s/N_p\).The currents go the other way: \(I_s/I_p = N_p/N_s\). More turns means more voltage and less current.
- Finding the line loss as \(V^2/R\) with the line voltage.Find the current \(I = P/V\) first; the loss is \(I^2R\).