**Summary:**

- When an alternating voltage V
_{p}is applied to the primary coil, an a.c. current flows in the primary coil. - This a.c. current sets up a changing magnetic field in the iron core. The flux through the primary coil is linked to the flux in the secondary coil through the iron core.
- This alternating magnetic field induces an e.m.f. in both the coils. (See Below)
- The output voltage in the secondary coil V
_{s}will give rise to an a.c. current in the coil itself. Thus, electrical energy can be delivered to any devise at the output. - Since there is no flux leakage, the rate of flux change $\frac{d \phi}{dt}$ at the primary coil and the secondary coil must be the same. Since, $\frac{V_{s}}{V_{p}} = \frac{\epsilon_{s}}{\epsilon_{p}}$, $\frac{V_{s}}{V_{p}} = \frac{- N_{s} \frac{d \phi}{dt}}{- N_{p} \frac{d \phi}{dt}} = \frac{N_{s}}{N_{p}}$.
- In addition, an ideal transformer is 100% efficient. Hence, power output from the secondary coil = power supplied to the primary coil. $V_{s} l_{s} \, = \, V_{p} l_{p}$

**Note: **Transformer works by the principle of magnetic induction. When an a.c. flows through the primary coil, it sets up a changing flux which in turn induces an e.m.f.

**Detailed Walkthrough Of The E.M.F. Induced In Both Coils:**

**At the primary coil:**

The changing magnetic field induces an e.m.f. E_{p} in the primary coil (due to self-inductance) This induced e.m.f. opposes the applied voltage, V_{p}

$V_{p} – \mid E_{p} \mid = I_{p} r_{p}$, where

- $r_{p}$ = the primary coil resistance
- $I_{p}$ = the current in the primary coil

Induced e.m.f., E_{p}(due to self-inductance) = $- \, \frac{d \left( N_{p} \phi \right) }{dt} = \, – \, N_{p} \frac{d \phi}{dt}$, where

- $N_{p}$ = the number of turns in the primary coil,
- $\phi$ = the magnetic flux in the iron core linking the coils
- $\frac{d \phi}{dt}$ = the rate of change of magnetic flux in the iron core.

For an ideal transformer, r_{p} = 0, hence $V_{p} = \mid E_{p} \mid $

**At the secondary coil,**

When connected to a load, the output voltage V_{s} (Secondary voltage) is given by:

$V_{s} = \mid E_{s} \mid – I_{s} r_{s}$,where

- $r_{s}$ = the secondary coil resistance,
- $I_{s}$ = the current in the secondary coil

Induced e.m.f. in the secondary coil, E_{s} (through mutual induction) = $- \, \frac{d \left( N_{S} \phi \right)}{dt} = N_{S} \frac{d \phi}{dt}$, where

- $N_{s}$ = the number of turns in the secondary coil,
- $\phi$ = the magnetic flux in the iron core linking the coils
- $\frac{d \phi}{dt}$ = the rate of change of magnetic flux in the iron core.

For an ideal transformer, r_{s} = 0, hence V_{s} = |E_{s}|

**Important:**

$\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}$

For step-up transformer, N_{s} > N_{p}

For step-down transformer, N_{s} < N_{p}