Question

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

Answer 1

Hint:-This given problem can be solved by taking the consideration of a step down transformer working and calculation of the efficiency of this transformer.

Complete step-by-step solution:
Step 1: A transformer is an electric device which is used for changing the a.c. voltages. A transformer which is used to increase the a.c. voltages from some given values is known as a step up transformer.
And, a transformer that is used to decrease the a.c. voltages from some values is known as a step down transformer.
As we know that \[\mathop E\nolimits_P \mathop I\nolimits_P = \mathop E\nolimits_S \mathop I\nolimits_S \]
Where \[\mathop E\nolimits_{S = } \]the voltage across secondary coil, \[\mathop I\nolimits_S = \]the current across secondary coil,
\[\mathop E\nolimits_P = \]the voltage across primary coil, and \[\mathop I\nolimits_P = \]the current across primary coil
For a step down transformer, \[\mathop E\nolimits_S < \mathop E\nolimits_P \] and \[\mathop I\nolimits_S > \mathop I\nolimits_P \] i.e., secondary current is stronger when secondary voltage is higher i.e., whatever lose is there in the voltage is the gain in the current in the same ratio. So, from this concept efficiency can be defined.
Step 2: Efficiency of a transformer can be defined as the ratio of output power to the input power.
i.e., \[\eta = \dfrac{{\mathop E\nolimits_S \mathop I\nolimits_S }}{{\mathop E\nolimits_P \mathop I\nolimits_P }} \times 100\]=(Output power/ Input power) \[ \times 100\] ………………. (1)
As it is given in the question, \[\mathop E\nolimits_S = 11\]V, \[\mathop E\nolimits_P = 220\], \[\mathop I\nolimits_P = 5\]A, and \[\mathop I\nolimits_S = 90\]
So, by substituting all the given values in the above equation (1), efficiency of the given step down transformer can be calculated as follows –
\[\Rightarrow \eta = \dfrac{{11 \times 90}}{{220 \times 5}} \times 100\], on further solving this equation, we will get-
\[\Rightarrow \eta = \dfrac{9}{{10}} \times 100\]
i.e., \[\Rightarrow \eta = 90\% \]
So, the efficiency of this step down transformer is \[\eta = 90\% \].

Hence, option (D) is correct.

Note:- A transformer works on the principle of mutual induction, i.e., whenever the amount of magnetic flux linked with a coil changes, an e.m.f. is induced in the neighbouring coil.
It should also be noted that a transformer is essentially an a.c. device. It cannot work on d.c. A transformer changes a.c. voltages/currents. It does not affect the frequency of a.c. When a.c. voltage is raised n times, the corresponding alternating current reduces to \[\dfrac{1}{n}\] times.