Question

The primary transformer has 40 turns and works on 100 volt and 100 watt. Then the number of turns in the secondary to step up voltage to 400V and the current in the secondary and primary will be
A. 260, 0.25A, 2A
B. 160, 0.25A, 1A
C. 360, 0.55A, 1A
D. None of these

Answer 1

To solve this question we need to use the concept of transformer. The voltage and current in a secondary of a transformer depend on the number of turns in the secondary of a secondary. Obtain the mathematical expression relating the voltage to the number of turns to find the number of turns and then use the expression relating the number of turns and the current to find the required answer.

Complete answer:
The number turns in the primary of the transformer is ${{N}_{1}}=40$
The primary of the transformer works on voltage ${{V}_{1}}=100V$ and power $P=100\text{Watt}$.
The relation between the power, voltage and the current can be expressed as,
$P=IV$
So, the primary current in the transformer will be,
${{I}_{1}}=\dfrac{P}{{{V}_{1}}}=\dfrac{100}{100}=1A$
Now, let the turns in the secondary of the transformer is ${{N}_{2}}$.
The given transformer in the question is a step-up transformer and the step-up voltage is ${{V}_{2}}=400V$.
The relation between the number of turns, voltage and current in the primary and secondary of a transformer can be given as,
$\dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{I}_{2}}}{{{I}_{1}}}$
Here, ${{I}_{2}}$ is the current in the secondary of the transformer.
So, the number of turns in the secondary can be found out as,
${{N}_{2}}=\dfrac{{{V}_{2}}}{{{V}_{1}}}\times {{N}_{1}}=\dfrac{400}{100}\times 40=160$
So, the primary current of the transformer will be,
${{I}_{2}}=\dfrac{{{N}_{1}}}{{{N}_{2}}}\times {{I}_{1}}=\dfrac{40}{100}\times 1A=0.25A$
So, the number of turns in the secondary will be 160, the secondary current will be 0.25A and the primary current 1A.

So, the correct answer is “Option B”.

Note:
The above answer can be directly found from the given voltage by relating the voltage to the current of the transformer. We can write,
$\begin{aligned} & \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{I}_{2}}}{{{I}_{1}}} \\\ & {{I}_{2}}=\dfrac{{{V}_{1}}}{{{V}_{2}}}\times {{I}_{1}} \\\ & {{I}_{2}}=\dfrac{100}{400}\times 1A=0.25A \\\ \end{aligned}$