The transformer raises or lowers the AC voltage through mutual induction. It consists of two coils wound on the same core. The alternating current passing through the primary creates a time-varying magnetic flux through the core. This changing flux induces an alternating emf in the secondary.

Let $N_p$ and $N_s$ be number of turns in primary and secondary. The *transformer turn ratio* is defined as $N_s/N_p$. If $V_p$ is applied (input) voltage to primary and $V_s$ voltage across secondary (output voltage) then Faraday's law gives
\begin{align}
\frac{V_p}{V_s}=\frac{N_p}{N_s}
\end{align}

A *step up transformer* increases the voltage and have $N_s>N_p$. A *step down transformer* decreases the voltage and have $ N_s < N_p$. In power distribution, step-up transformers is used at the power station and step-down transformers are used at the user end.

In an *ideal transformer* there is no loss of power i.e., $P_\mathrm{out}=P_\mathrm{in}$ which gives
\begin{align}
V_p I_p=V_s I_s
\end{align}
where $I_p$ and $I_s$ are current in primary and secondary. The *efficiency* of a transformer is defined as the ratio of output power to the input power. The efficiency of an ideal transformer is 1. The efficiency is less than 1 for real transformer due to losses in transformer. The losses occurs due to heating effect, flux leakage, eddy currents, hysteresis and humming.

The frequency of output voltage/current is same as that of input. The secondary voltage is out of phase with the primary.

A paragraph type problem was asked in IIT JEE 2013. There are two questions related to a paragraph. This problem clears doubts related to the use of transformer in power distribution.

**Paragraph:**
A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 20 km away from the power plant for consumers' usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers' end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with a power factor unity. All the currents and voltages mentioned are rms values.

**Problem 1: **
If the direct transmission method with a cable of resistance 0.4 Ohm/km is used, the power dissipation (in %) during transmission is

- 20
- 30
- 40
- 50

**Solution:**
The current through the cable is given by $i=P/V=600\times 10^3/4000=150$ A. The resistance of the cable is $R=0.4\times 20=8 $ Ohm. The power loss in the cable is $i^2 R=180 $ kW which amounts to 30% of 600 kW.

**Problem 2: ** In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1:10. If the power to the consumers has to be supplied at 200 V, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is

- 200:1
- 150:1
- 100:1
- 50:1

**Solution: **
For an ideal transformer with power factor unity, $V_s/V_p=N_s/N_p$, where $V$ is the voltage and $N$ is the number of turns (subscript $s$ stands for secondary and $p$ for primary). For step-up transformer, $V_p=4000$ V and $N_p/N_s=1/10$, which gives $V_s=40000$ V. For step-down transformer, $V_p=40000$ V and $V_s=200$ V, which gives $N_p/N_s=200$.

**Question 1:** The core of a transformer is laminated so that

**Question 2:** 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 3:** A transformer has 50 turns in the primary and 100 in the secondary. If the primary is connected to a 220 V DC supply, what will be the voltage across the secondary?