Question

The primary coil having N_P turns of an ideal transformer is supplied with an alternating voltage V_P. Obtain an expression for the voltage V_S induced in its secondary coil having N_S turns.

Answer 1

For the simple transformer shown in the figure, the output voltage V_S depends almost entirely on the input voltage V_P and the ratio of the number of loops in the primary and secondary coils. Faraday's law of induction for the secondary coil gives its induced output voltage V_S to be

`V_S = -N_S((Deltaphi)/(Deltat))` .....(i)

Where N_S is the number of loops in the secondary coil and `(Deltaphi)/(Deltat)` is the rate of change of magnetic flux.

The cross-section area of the coils is the same on either side, as is the magnetic field strength, and so `(Deltaphi)/(Deltat)` is the same on either side. The input primary voltage V_P is also related to changing flux by

`V_P = -N_P(Deltaphi)/(Deltat)` .....(ii)

Lenz's law tells us that the primary coil opposes the change in flux caused by the input voltage V_P hence the minus sign is given.

Dividing both the equation, we get

`V_S/V_P = N_S/N_P`

`V_S = V_P(N_S/N_P)`