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

Concept: A transformer works on the principle of mutual induction. A changing current in the primary coil produces a changing magnetic flux in the core, which induces an emf in the secondary coil. A step-up transformer increases voltage by increasing the number of turns in the secondary coil.

Step 1: State transformer voltage ratio. For an ideal transformer, \[ \frac{V_s}{V_p} = \frac{N_s}{N_p}. \] This shows that voltage is directly proportional to the number of turns. 

Step 2: Explain step-up transformer. In a step-up transformer: \[ N_s > N_p \Rightarrow V_s > V_p. \] Thus, it increases voltage while decreasing current, keeping power constant. 

Step 3: Given data. \[ N_p = 100,\quad N_s = 5000 \] \[ P = 3.3\,kW = 3300\,W,\quad V_p = 220\,V \] 

Step 4: Find primary current. \[ P = V_p I_p \] \[ I_p = \frac{P}{V_p} \] \[ I_p = \frac{3300}{220} \] \[ I_p = 15\,A \] \[ \boxed{I_p = 15\,A} \] 

Step 5: Find secondary voltage. \[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \] \[ V_s = 220 \times \frac{5000}{100} \] \[ V_s = 220 \times 50 \] \[ V_s = 11000\,V \] \[ \boxed{V_s = 1.1\times10^4\,V} \]