Question

A transformer is used to light a \[120W\], \[24V\] lamp from \[240V\]a.c. mains. The current in the main cable is \[0.6A\]. The efficiency of the transformer is:
A) \[48\% \]
B) \[63.8\% \]
C) \[83.3\% \]
D) \[90\% \]

Answer 1

Hint: As given in question we have to calculate efficiency of so we will first calculate input power that is need and output has given in question as \[120W\], and then we will take out percentage of output and input through the formula\[\dfrac{O}{I} \times 100\].

Complete step by step answer:
As in question we are given with output power as, \[120W\]
And we are also given with input voltage of lamp as, \[24V\]
And we are also given with current of ,main cable as, \[0.6A\]
And a.c. mains voltage as, \[240V\]
As to calculate input power we know the formula of input power is
\[P = V \times I\], so substituting the values of voltage and current from question as
\[P = 240 \times 0.6\]
\[P = 140\], hence input voltage is calculated.
So to calculate efficiency we know
\[E = \dfrac{{120}}{{140}} \times 100\], where values of output and input power are substituted from above.
\[E = 0.833 \times 100\]
\[E = 83.3\% \]

Hence, The correct option is C.

Additional information: Efficiency of anything is known as to check that the system is doing the effective job or not and we can also increase the efficiency of the system by changing some parts of the system or by changing the inputs that we are giving the system through any source. In energy efficiency it means to make the same energy by using less energy and use remaining energy for some other work.

Note: Here we can also change the system’s efficiency by changing the voltage and current source so that output voltage will decrease and our efficiency will increase. Efficiency also means to make a system work in less time than before and effective output by system in less time.