Question

A transformer having the efficiency of 90% is working on 200V and 3kW power supply. If the current in the secondary coil is 6A, the voltage across the secondary coil and the current in the primary coil respectively are:
A. 300V, 15A
B. 450V, 15A
C. 450V, 13.5A
D. 600V, 15A

Answer 1

Hint: A transformer has a primary coil and a secondary coil. The ratio of the power of the secondary coil to the power of primary coil is called efficiency of the transformer i.e. $\eta =\dfrac{{{P}_{s}}}{{{P}_{p}}}$. With formula find the power of the secondary coil and then find the voltage by using the formula P=Vi. You can also find the current in the primary coil by using P=Vi.

Formula used:
$\eta =\dfrac{{{P}_{s}}}{{{P}_{p}}}$
P=Vi

Complete step-by-step answer:
A transformer is a device that can increase or decrease the voltage of the circuit as per requirement. It consists of a primary coil and secondary coil. The primary coil acts as an input and the secondary coil acts as an output.
Both the coils have some power associated with them. The power of the input or primary coil passed to the secondary coil. However, there is some leakage of this power and hence the output power (${{P}_{s}}$) is less than the input power (${{P}_{p}}$).
Therefore, we define the efficiency of the transformer which is equal to $\eta =\dfrac{{{P}_{s}}}{{{P}_{p}}}$ …. (i).
It is given that the efficiency of the transformer is 90%, which means the ratio is equal to 0.9. And the input power is given to 3kW. This means ${{P}_{p}}=3kW=3000W$
Substitute the values of $\eta $ and input power in equation (i).
$\Rightarrow 0.9=\dfrac{{{P}_{s}}}{3000}$
$\Rightarrow {{P}_{s}}=0.9\times 3000=2700W$.
Therefore, the power of the secondary coil is 2700W.
Let the voltage of the secondary coil and current flowing in the secondary coil be ${{V}_{s}}$ and ${{i}_{s}}$ respectively.
Power is equal to the product of the voltage and the current in the coil.
This gives us that ${{P}_{s}}={{V}_{s}}{{i}_{s}}$
It is given that ${{i}_{s}}=6A$ and we found that ${{P}_{s}}=2700W$.
This gives that,
$2700=6{{V}_{s}}$
${{V}_{s}}=\dfrac{2700}{6}=450V$
This means that the voltage across the secondary coil is 450V.
It is given that the voltage across the primary coil is 200V and the power in this coil is 3000W.
Let the voltage across the primary coil be ${{V}_{p}}$ and the current in this coil be ${{i}_{p}}$.
Therefore, ${{P}_{p}}={{V}_{p}}{{i}_{p}}$.
Substitute the values of ${{V}_{p}}$ and ${{P}_{p}}$
$\Rightarrow 3000=200{{i}_{p}}$
$\Rightarrow {{i}_{p}}=\dfrac{3000}{200}=15A$
This means the current in the primary coil is 15A.
Hence, the correct option is B.

Note: The ratio of the voltages across the primary and secondary coils depends on the number of turns in both coils. i.e. $\dfrac{{{V}_{s}}}{{{V}_{p}}}=\dfrac{{{N}_{s}}}{{{N}_{p}}}$.
Here, ${{N}_{s}}$ and ${{N}_{p}}$ are the number of turns in the secondary and the primary coils respectively.
Accordingly, transformers are used in two ways. One is to get a voltage greater than the voltage of the primary coil (i.e. when ${{N}_{s}}$>${{N}_{p}}$). This type of transformer is called a set-up transformer.
Other is get the voltage of the secondary coil smaller than the voltage of the primary coil (i.e. ${{N}_{s}}$<${{N}_{p}}$). This type of transformer is called a set-down transformer.