Question

Draw a labelled diagram of a step-up transformer. Obtain the ratio of secondary voltage to primary voltage in terms of number of turns in the two coils.

Hint:

To easily remember transformer principles: Voltage simply follows the turns.
More turns inherently mean more voltage (Step-up).
Fewer turns inherently mean less voltage (Step-down).
Current, however, does the exact mathematical opposite to conserve energy.

Answer 1

Concept:
A transformer works on mutual induction.
A changing current in the primary coil produces a changing flux in the core, and this flux induces an emf in the secondary coil.
A step-up transformer has more turns in the secondary coil than in the primary coil.

Step 1: Labelled diagram of a step-up transformer:

Draw a closed laminated soft iron core in the form of a rectangle.
Wind the primary coil of a small number of turns \(N_p\) on one limb.
Wind the secondary coil of a larger number of turns \(N_s\) on the other limb, with \(N_s > N_p\).
Connect an AC source (input voltage \(V_p\)) across the primary terminals.
Connect the load across the secondary terminals (output voltage \(V_s\)).
Label: laminated iron core, primary coil, secondary coil, AC input and output.

Step 2: Assumptions for the ideal transformer:
The coil resistances are negligible and there is no flux leakage.
The same flux \(\Phi\) links every turn of both coils.

Step 3: Emf in the two coils:
By Faraday's law, the emf in the primary coil is \[ e_p = -N_p \frac{d\Phi}{dt} \]
The emf in the secondary coil is \[ e_s = -N_s \frac{d\Phi}{dt} \]

Step 4: Take the ratio:
Divide the second equation by the first. The factor \(\frac{d\Phi}{dt}\) cancels.
\[ \frac{e_s}{e_p} = \frac{N_s}{N_p} \]
With negligible resistance, \(V_p = e_p\) in magnitude and \(V_s = e_s\) in magnitude.
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]
For a step-up transformer \(N_s > N_p\), so \(V_s > V_p\).

Final Answer:
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]