Question

A transformer connected to $220V$ mains is used to light a lamp of rating $100W$ and $110V$. If the primary current is $0.5A$, the efficiency of the transformer is (approximately)
A. $60\%$
B. $35\%$
C. $50\%$
D. $90\%$
E. $44\%$

Answer 1

The question talks about the performance of a transformer which either steps down the voltage, steps up the voltage depending on which side of the source, it could be the primary source or secondary source.

Complete step by step answer:
The efficiency of a transformer is equal the ratio of power output to power input multiplied by 100%
Power is the amount of energy transferred per unit time. The unit of Power is Watt.
Power in a transformer can be defined as the product of Voltage with the current flowing through it.
$P = IV$, where P is power, I is current and V is voltage
For the main source or primary source, power is P = $0.5A \times 220V = 110W$
For the secondary source, power is $100W$
The primary source produces the input power while the secondary source produces the output power, as we were able to deduce from the question that the lamp is the secondary source because after current was initiated from the primary mains source to move up to produce a secondary one.
Mathematically it can be represented as
$Efficiency\% = \dfrac{{Powe{r_{output}}}}{{Powe{r_{input}}}} \times 100 \\\ \eta = {\text{ }}\dfrac{{100}}{{110}} \times 100 = 90.9\% \\\$
which is approximately equal to 90%
Therefore, the correct option is Option D.

Note: Sometimes the power output and power input can be calculated from other parameters such as resistance, voltage and current, the formulas for calculating are-
$P = {I^2}R \\\ P = {V^2}/R \\\$
From the two equations above we can easily get the power input and the power output, and then taking their ratio gives the efficiency of the transformer.