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Question
The primary coil having NP turns of an ideal transformer is supplied with an alternating voltage VP. Obtain an expression for the voltage VS induced in its secondary coil having NS turns.
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Solution
For the simple transformer shown in the figure, the output voltage VS depends almost entirely on the input voltage VP 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 VS to be
`V_S = -N_S((Deltaphi)/(Deltat))` .....(i)
Where NS 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 VP 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 VP 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)`
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