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:

Transformer relations: \[ \frac{V_s}{V_p}=\frac{N_s}{N_p},\quad P_{in}=P_{out} \] Step-up transformer increases voltage but reduces current.

Answer 1

Step 1: Principle and construction:
A transformer uses mutual induction between two coils wound on a common laminated iron core.

In the diagram, the primary coil has \(N_p\) turns and is joined to the AC source of voltage $V_p$.
The secondary coil has \(N_s\) turns, with $N_s > N_p$ for a step-up transformer, and is joined to the load at voltage $V_s$.
The iron core is closed and laminated so that the flux stays inside it.

Step 2: Emf per turn:
In an ideal transformer the flux $\Phi$ through each turn is the same for both coils.
So the emf induced in one turn is $-\frac{d\Phi}{dt}$ for both coils.

Step 3: Emf of the full coils:
The primary has $N_p$ turns, so $e_p = N_p \times \left(-\frac{d\Phi}{dt}\right)$.
The secondary has $N_s$ turns, so $e_s = N_s \times \left(-\frac{d\Phi}{dt}\right)$.

Step 4: Relate the emf to the terminal voltages:
The resistance of the coils is negligible, so the terminal voltages equal these emfs in magnitude.
That gives $V_p = N_p k$ and $V_s = N_s k$, where $k = \left|\frac{d\Phi}{dt}\right|$ is the same for both.
Dividing the second by the first:
\[ \frac{V_s}{V_p} = \frac{N_s k}{N_p k} = \frac{N_s}{N_p} \]

Final Answer:
The ratio of secondary voltage to primary voltage equals the ratio of the number of turns: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$.