Question

A step down transformer reduces 220 V to 11 V. The primary coil draws 5 A current and secondary coil supplies 90 A efficiency of transform will be.
(A) 4.4%
(B) 20%
(C) 33%
(D) 90%

Answer 1

The efficiency of the transformer is defined as the ratio of useful output power to the input power. The input and output power are measured in the same unit. Its unit is either in Watts (W) or KW. Transformer efficiency is denoted by $\eta$. We know that the efficiency of a transformer is given by
$\eta = \dfrac{{{\text{output power}}}}{{{\text{Input power}}}} \times 100$
On substituting the values we can easily calculate the efficiency of the transformer.

Complete step by step solution:
According to question initial data,
We are given with voltage across the secondary coil, ${E_s} = 11{\text{V}}$
The current flowing through the secondary coil, ${I_s} = 90{\text{A}}$.
Voltage across the primary coil ${E_p} = 220{\text{V}}$
The current through the primary coil is ${I_P} = 5{\text{A}}$
Also we know that the efficiency of the transformer is given by
$\eta = \dfrac{{{\text{output power}}}}{{{\text{Input power}}}} \times 100$
$\eta = \dfrac{{{E_s}{I_s}}}{{{E_P}{I_P}}} \times 100$
On substituting the values we get
$\eta = \dfrac{{11 \times 90}}{{220 \times 5}} \times 100$
$\eta = 90\%$
$\therefore$ Efficiency of the transformer is 90%

Hence the option (d) is correct.

Note: Generally there are two types of transformer. A transformer designed to increase the voltage from primary to secondary is called a step-up transformer. A transformer designed to reduce the voltage from primary to secondary is called a step-down transformer. These types of transformer are used in order to minimize heat loss during the transmission of electricity.