Question

Draw a labelled diagram of a step-up transformer. State the principle on which it works and obtain the ratio of secondary voltage to primary voltage in terms of number of turns and currents in the two coils.

Hint:

For an ideal transformer: \[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \] \[ \frac{I_s}{I_p} = \frac{N_p}{N_s} \] A step-up transformer has \[ N_s>N_p \] and therefore \[ V_s>V_p. \]

Answer 1

Concept: A transformer is an electrical device used to increase or decrease alternating voltage without changing its frequency. A transformer consists of two insulated coils wound over a common laminated soft iron core.
• The coil connected to the AC source is called the primary coil.
• The coil connected to the load is called the secondary coil.
• A step-up transformer increases the voltage, therefore the secondary coil contains more turns than the primary coil. Labelled Diagram of a Step-Up Transformer:

\[ N_s>N_p \] Hence the transformer acts as a step-up transformer. Principle of Working: A transformer works on the principle of mutual induction. When an alternating current flows through the primary coil, a continuously changing magnetic flux is produced in the iron core. This changing magnetic flux links the secondary coil and induces an emf in it according to Faraday's law of electromagnetic induction. Thus electrical energy is transferred from the primary coil to the secondary coil through magnetic coupling.

Step 1: Induced emf in the primary coil. Let \[ \phi=\text{magnetic flux through each turn}. \] According to Faraday's law, \[ E_p = N_p\frac{d\phi}{dt}. \] Thus, \[ \boxed{ E_p=N_p\frac{d\phi}{dt} } \]

Step 2: Induced emf in the secondary coil. Similarly, \[ E_s = N_s\frac{d\phi}{dt}. \] Hence, \[ \boxed{ E_s=N_s\frac{d\phi}{dt} } \]

Step 3: Obtain the voltage ratio. Dividing the two equations, \[ \frac{E_s}{E_p} = \frac{N_s}{N_p}. \] For an ideal transformer, \[ V_p=E_p, \qquad V_s=E_s. \] Therefore, \[ \boxed{ \frac{V_s}{V_p} = \frac{N_s}{N_p} } \] This is the transformation ratio in terms of the number of turns.

Step 4: Relation between currents and number of turns. For an ideal transformer, \[ \text{Input Power} = \text{Output Power}. \] Therefore, \[ V_pI_p = V_sI_s. \] Substituting \[ \frac{V_s}{V_p} = \frac{N_s}{N_p}, \] we obtain \[ \frac{I_p}{I_s} = \frac{N_s}{N_p}. \] Hence, \[ \boxed{ \frac{I_s}{I_p} = \frac{N_p}{N_s} } \] or \[ \boxed{ \frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} } \] Final Result: For an ideal transformer, \[ \boxed{ \frac{V_s}{V_p} = \frac{N_s}{N_p} } \] and \[ \boxed{ \frac{I_s}{I_p} = \frac{N_p}{N_s} } \]