Animated Solution for Physics - Electromagnetic Induction: A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of 2.2 kW. If the current in the secondary coil is 10 A, then the input voltage and current in the primary coil are

## The Magic of Transformers: Decoding Voltage and Current

Transformers are the unsung heroes of our modern electrical grid. They allow us to transmit massive amounts of power across vast distances with minimal loss, simply by manipulating the delicate balance between voltage and current. In this problem, we are going to step into the shoes of an electrical engineer and decode the primary circuit of a transformer based purely on its output.

Imagine a transformer as a magnetic bridge connecting two separate electrical circuits. On the input side, known as the primary coil, we have $N_1=300$ turns of wire wrapped around an iron core. On the output side, the secondary coil, we have $N_2=150$ turns.

We are given that the secondary side is working hard, delivering an output power of $P_2=2.2\text{ kW}$ (which is $2200\text{ W}$) to a load, and a current of $I_2=10\text{ A}$ is flowing through it.

Finding Secondary Voltage

Before we can figure out what is happening on the primary side, we need a complete picture of the secondary side. We know the power and the current, but what about the voltage?

Electrical power is simply the product of voltage and current:

By rearranging this fundamental equation, we can isolate the secondary voltage $V_2$:

Substituting our known values:

So, the secondary coil is operating at $220\text{ V}$.

Unlocking the Primary Voltage

The true magic of a transformer lies in its turns ratio. The ratio of the voltages across the coils is exactly equal to the ratio of the number of turns in those coils. This is because the changing magnetic flux links both coils equally.

We want to find the primary voltage $V_1$, so let's rearrange the equation:

Now, we plug in the numbers:

Notice that because the primary coil has twice as many turns as the secondary coil, its voltage is twice as high. This makes it a step-down transformer, stepping the voltage down from $440\text{ V}$ to $220\text{ V}$.

Finding the Primary Current

Finally, we need to find the primary current $I_1$. In physics, we love conservation laws. For an ideal transformer (which we assume unless told otherwise), the power put into the primary coil must exactly equal the power taken out of the secondary coil. Energy cannot be created or destroyed!

Since power is voltage times current, we can write:

We already know the total power is $2200\text{ W}$, and we just calculated the primary voltage to be $440\text{ V}$. Let's solve for $I_1$:

Alternatively, you could use the inverse turns ratio for current: $\frac{I_1}{I_2}=\frac{N_2}{N_1}$, which gives $I_1=10\times \frac{150}{300}=5\text{ A}$. Both paths lead to the same beautiful truth.

By systematically applying the principles of power and the transformer turns ratio, we have completely solved the circuit. The input voltage is $440\text{ V}$ and the input current is $5\text{ A}$. This perfectly matches option (a).