A transformer having efficiency of 90% is working on $200\, V$ and $3 \,kW$ power supply. If the current in the secondary coil is $6\, A$, the voltage across the secondary coil and the current in the primary coil respectively are

Question

In an a.c. circuit, voltage and current are given by: \[ V = 100 \sin(100t) \, \text{V} \quad \text{and} \quad I = 100 \sin\left(100t + \frac{\pi}{3}\right) \, \text{mA}. \] The average power dissipated in one cycle is:

Answer 1

To solve this problem, we need to find the voltage across the secondary coil and the current in the primary coil of the transformer. We know the following information from the problem:

Efficiency of the transformer = 90%
Primary voltage $ V_p = 200\,\text{V} $
Power supply $ P = 3\,\text{kW} = 3000\,\text{W} $
Secondary current $ I_s = 6\,\text{A} $

The efficiency of the transformer is given by the formula:

\text{Efficiency} = \frac{\text{Output Power}}{\text{Input Power} } \times 100\%

Given that the efficiency is 90%, we can write:

0.9 = \frac{V_s \times I_s}{V_p \times I_p}

Where:

$ V_s $ is the secondary voltage.
$ I_p $ is the primary current.

Using the input power: V_p \times I_p = 3000 \, \text{W}, we have:

I_p = \frac{3000}{200} = 15 \, \text{A}

Now, substituting into the efficiency equation:

0.9 = \frac{V_s \times 6}{200 \times 15}

This simplifies to:

0.9 = \frac{V_s \times 6}{3000}

V_s \times 6 = 2700

V_s = \frac{2700}{6} = 450 \, \text{V}

Thus, the voltage across the secondary coil is $ 450 \, \text{V} $ and the current in the primary coil is $ 15 \, \text{A} $.

The correct option is:
450 V, 15 A

Answer 2