Topic 6 of 6
Transformer voltage and current
A transformer uses a changing magnetic flux to transfer energy between electrically separate windings. The turns ratio sets the ideal voltage ratio; conservation of energy then sets the inverse current ratio.
The primary winding connects to an alternating source. Its changing current establishes changing flux in a closed iron core, which links the secondary turns and induces e.m.f. A connected secondary load completes a separate circuit and draws current.
A changing core flux links two separate circuits
Each winding is insulated from the core and from the other winding. Use the supplied turn counts, not the number of drawn turns. In the ideal rms model, 230 V becomes 12.0 V, with input and output both 24.0 W. The separate real example has 30.0 W input, 24.0 W output and 6.0 W loss; it is not the ideal current-ratio model. The supplied 230 V calculation is not a mains investigation.
The core guides flux linking both windings. Charge does not travel through the iron from the primary to the secondary. A steady direct current provides no continuing flux change and therefore no sustained secondary e.m.f. after the switching transient; it is not an appropriate continuous input for this transformer.
State the ideal conditions before using the ratios
In the ideal model, every turn in each winding links the same changing core flux. Winding resistance, flux leakage and core losses are negligible. Faraday's law then makes each winding's e.m.f. proportional to its number of turns. With negligible winding voltage loss, the terminal voltage magnitudes satisfy:
For the sinusoidal source and resistive load used here, V and I are rms values. The ideal input and output powers are equal:
Ns/Np = Vs/Vp = Ip/Is
A step-up transformer increases voltage and reduces the corresponding current; it does not increase ideal power. A step-down transformer does the reverse. Keep peak and rms values distinct when using the mean-power relationship.
Worked ideal transformer
Find the output before finding input current
First estimate the scale. Roughly 1000 primary turns and 50 secondary turns, with a supplied primary voltage near 200 V, suggest an output near 10 V. A load of a few ohms then draws a few amperes and receives tens of watts, requiring primary current of order 0.1 A.
Now use Np = 1150, Ns = 60, Vp = 230 V rms and a secondary load R = 6.00 Ω:
Is = Vs/R = 12.0/6.00 = 2.00 A rms
Pout = VsIs = 24.0 W
Ip = Pout/Vp = 24.0/230
= 0.104 A rms
The smaller primary current is consistent with the larger primary voltage and equal ideal power. This is a supplied calculation model; the operating principle can be investigated with suitable low-voltage laboratory equipment or supplied observations.
Keep a real-loss account separate
In a real transformer, winding resistance produces heating, while changing core magnetisation and induced eddy currents dissipate energy in the core. Flux leakage reduces the flux common to both windings; leakage itself is not simply another name for heating.
Insulated laminations divide the iron core and restrict eddy-current paths. A suitable core material reduces magnetisation losses. These changes reduce losses but do not make the device create energy.
For a separate supplied nonideal case, input power is 30.0 W and useful output is 24.0 W:
Power dissipated = 30.0 - 24.0 = 6.0 W
Do not insert this loss into a calculation still using the ideal equal-power current ratio. Identify whether the question supplies an ideal transformer or a real input/output account before choosing an equation.