Question

If the input voltage of a transformer is 2500 volts and output current is 80 ampere. The ratio of the number of turns in the primary coil to that in the secondary coil is 20:1. If efficiency of transformer is 100%, the voltage in the secondary coil is
(A) \[\dfrac{{2500}}{{20}}\,{\text{volt}}\]
(B) \[2500 \times 20\,{\text{volt}}\]
(C) \[\dfrac{{2500}}{{80 \times 20}}\,{\text{volt}}\]
(D) \[\dfrac{{2500 \times 20}}{{80}}\,{\text{volt}}\]

Answer 1

Hint:Use the rule that relates the number of turns in the both coils and voltages across the both coils. We have given the voltage in the primary coil. Therefore, the voltage in the secondary can be calculated using that rule. The solution to this question does not deal with output current and efficiency.

Formula used:
\[\dfrac{{{V_P}}}{{{V_S}}} = \dfrac{{{N_P}}}{{{N_S}}}\]
Here, \[{V_P}\] is the voltage at the primary coil, \[{V_S}\] is the voltage at the secondary coil, \[{N_P}\] is the number of turns in the primary coil and \[{N_S}\] is the number of turns in the secondary coil.

Complete step by step answer:
We have given that the voltage in the primary coil is 2500 volt and the output current is 80 ampere.
We have the rule which states that,
\[\dfrac{{{V_P}}}{{{V_S}}} = \dfrac{{{N_P}}}{{{N_S}}}\]
Here, \[{V_P}\] is the voltage at the primary coil, \[{V_S}\] is the voltage at the secondary coil, \[{N_P}\] is the number of turns in the primary coil and \[{N_S}\] is the number of turns in the secondary coil.
We substitute \[\dfrac{{{N_P}}}{{{N_S}}} = \dfrac{{20}}{1}\] in the above equation.
\[\dfrac{{{V_P}}}{{{V_S}}} = \dfrac{{20}}{1}\]
\[ \Rightarrow {V_S} = \dfrac{{{V_P}}}{{20}}\]
We substitute 2500 V for \[{V_P}\] in the above equation.
\[\therefore{V_S} = \dfrac{{2500}}{{20}}\,{\text{volt}}\]

So, the correct answer is option (A).

Additional information:
We know the efficiency of a transformer is given as,
\[\eta = \dfrac{{{\text{Output power}}}}{{{\text{Input power}}}}\]
We have given that the efficiency of this transformer is 100%. Therefore, we substitute 1.0 for \[\eta \] in the above equation.
\[1 = \dfrac{{{\text{Output power}}}}{{{\text{Input power}}}}\]
\[ \Rightarrow {\text{Input power}} = {\text{Output power}}\]
We know the formula for power is,\[P = IV\]. Therefore, we can write the above equation as,
\[{I_i}{V_P} = {I_o}{V_S}\]
\[ \Rightarrow \dfrac{{{V_P}}}{{{V_S}}} = \dfrac{{{I_o}}}{{{I_i}}}\]
Here, \[{I_i}\] is the input current at the primary coil and \[{I_o}\] is the output current at secondary coil.
Therefore, we can calculate the input current as,
\[\dfrac{{20}}{1} = \dfrac{{80}}{{{I_i}}}\]
\[ \therefore{I_i} = 40\,A\]

Note: To solve these types of questions regarding voltage in the coils of the transformer, the key is to remember the rule which relates the number of turns in the both coils to the voltages in the both coils. We don’t need to use the formula for efficiency if the given efficiency is 100%. The efficiency of 100% implies no loss in the power in the transformer coils.