Question

A transformer is used to light 140W, a 24V lamp from 240V AC Mains. The current input to the mains is 0.7A. The efficiency of the transformer is nearest to
(A) $48\%$
(B) $63.8\%$
(C) $83.3\%$
(D) $90\%$

Answer 1

Hint
Using the quantities given in the question we will calculate the input power of the transformer using a formula of power. Then will use the formula of efficiency i.e. $\eta = \dfrac{{output\,power}}{{input\,power}} \times 100$ , and put the values of output and input powers and get the percentage of efficiency.

Complete step by step answer
Transformer is a voltage control device that converts or changes the level of voltage between circuits and it is used widely in the distribution and transmission of alternating current power.
Given that primary voltage ${V_P}$ is 240V, output power ${P_{out}}$ is 140W, primary current ${I_P}$ is 0.7A.
Now we calculate the input power in the transformer i.e. ${P_{in}}$
We know that power $P = I \times V$ , where Iis the current and V is the voltage.
So ${P_{in}} = {I_P} \times {V_P}$
Now as we know that efficiency is denoted by $\eta$ and we calculate efficiency as
$\eta = \dfrac{{output\,power}}{{input\,power}} \times 100$ , we put the value of output and input power in this equation.
So , $\eta = \dfrac{{{P_{out}}}}{{{P_{in}}}} \times 100 = \frac{{140W}}{{{I_P} \times {V_P}}} \times 100$ , now we calculate ${P_{in}} = {I_P} \times {V_P} = 0.7 \times 240W$ =168W.
Now we put the value of ${P_{in}}$ in the equation of efficiency :-
$\eta = \dfrac{{140W}}{{16.8W}} \times 100$ , now we calculate the efficiency i.e. $\eta = \dfrac{{140W}}{{168W}} \times 100 = 83.33\%$.
We get $\eta = 83.33\%$ , so the correct option is (C).

Note
If a system has high efficiency then the system is called better . we can solve this question by another method. In that method we use the formula of power $P = \dfrac{{{V^2}}}{R}$ and calculate the internal resistance of the lamp. Then we calculate the individual power for both the voltages and apply the efficiency formula.