Physics · Ch 4 — Electromagnetic Induction and Alternating Current

Construction and Working of Transformer

4.6.1

Construction and Working of Transformer

A transformer's principle is the mutual induction between two coils: when the electric current passing through one coil changes with time, an emf is induced in a magnetically-linked neighbouring coil. Its construction places two coils of high mutual inductance -- the primary coil P, across which the alternating voltage to be transformed is applied, and the secondary coil S, from which the transformed output power is drawn -- wound over the SAME transformer core, generally made of a laminated, good magnetic material such as silicon steel (laminated specifically to minimise eddy-current loss). The two coils are electrically insulated from each other but magnetically linked through the shared core, and the whole assembled core-and-coils unit is typically housed in a container filled with a suitable medium for improved insulation and cooling.

Working. With the primary connected to an alternating voltage source, an alternating magnetic flux is set up in the core; assuming no flux leakage, this SAME flux is linked with both the primary and secondary windings, changing at the same rate for each. The back emf induced in the primary is $\varepsilon_P=-N_P\,d\Phi_B/dt$ , and since the applied primary voltage equals this back emf, $v_P=-N_P\,d\Phi_B/dt \qquad (4.24)$ . Because the same $d\Phi_B/dt$ threads the secondary too, the induced secondary emf is $\varepsilon_S=-N_S\,d\Phi_B/dt$ , and (for an open secondary circuit) this equals the secondary terminal voltage, $v_S=-N_S\,d\Phi_B/dt \qquad (4.25)$ . Dividing (4.25) by (4.24) eliminates the shared flux term entirely, giving

\dfrac{v_S}{v_P} = \dfrac{N_S}{N_P} = K \qquad (4.26)

where K is the voltage transformation ratio. For an IDEAL transformer, input power equals output power, $v_Pi_P=v_Si_S$ , which combined with (4.26) gives

$\dfrac{v_S}{v_P} = \dfrac{N_S}{N_P} = \dfrac{i_P}{i_S} \qquad (4.27)$ …

Figure 4.32Transformer construction and a roadside transformer

What this figure shows. Figure 4.32(a) shows the schematic construction of a transformer: a rectangular laminated 'Transformer core' carries two separate windings -- 'Primary winding of $N_P$ turns', across which the primary voltage $V_P$ is applied and through which primary current $I_P$ flows, and 'Secondary winding of $N_S$ turns', across which the secondary voltage $V_S$ appears and through which secondary current $I_S$ is drawn by the external load -- with both windings electrically insulated from each other but magnetically linked through the shared iron core. Figure 4.32(b) is a photograph of a real 'Roadside transformer', the pole-mounted or ground-mounted grey cylindrical unit familiar from residential streets, which performs exactly this step-down function to bring high-voltage distribut …

Misc Example 4.16Voltage per turn and power delivered by an ideal transformer

Worked out. An ideal transformer with 460 primary turns and 40,000 secondary turns is connected to 230 V AC mains, and its secondary feeds a load of resistance $10^4\ \Omega$ ; the voltage developed per turn of the secondary, and the power delivered to the load, are required. The secondary voltage is $V_S = V_P(N_S/N_P) = 230\times(40000/460) = 20{,}000$ V, so the voltage per turn is $V_S/N_S = 20000/40000 = 0.5$ V (the SAME value that also holds for the primary, since flux linkage per turn is identical on both sides of an ideal transformer). The power delivered to the load is $P=V_S^2/R_S = (20000)^2/10^4 = 40{,}000$ W $=40$ kW, obtained directly from the secondary voltage and load resistance us …

Misc Example 4.17Inverter step-up transformer -- secondary turns and primary current

Worked out. A household inverter's built-in step-up transformer converts 12 V AC to 240 V AC, with a 100-turn primary and a secondary that delivers 50 mA to the external circuit; the number of secondary turns and the primary current are required. The transformation ratio is $K=V_S/V_P=240/12=20$ , so the secondary turns are $N_S=K\times N_P = 20\times100=2000$ turns. Using the ideal-transformer current relation $I_P/I_S = N_S/N_P = K$ , the primary current is $I_P = K\times I_S = 20\times50\ \text{mA} = 1$ A. The example is a practical, everyday application of the same turns-ratio relations used throughout the transformer section, applied this time to a step-up (rather than the mor …

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