Transformers
Transformers ($ V_s/V_p = N_s/N_p $)
A transformer is a static electrical device that transfers electrical energy from one circuit to another through electromagnetic induction. Its main function is to change the voltage of an alternating current (AC) supply, either increasing it (step-up transformer) or decreasing it (step-down transformer), without changing its frequency. Transformers are essential for the efficient transmission and distribution of electrical power over long distances.
Principle of a Transformer
A transformer works on the principle of mutual induction. When an alternating voltage is applied to one coil (the primary coil), it produces a changing magnetic flux in the core. This changing magnetic flux links with the other coil (the secondary coil) and, according to Faraday's Law of electromagnetic induction, induces an alternating EMF across it.
Simplified diagram of a transformer.
Construction
A simple transformer consists of:
- Core: A laminated soft iron core provides a path of low reluctance for the magnetic flux. Using soft iron and a laminated structure minimises energy losses due to hysteresis and eddy currents in the core. The core can be of various shapes (e.g., shell type, core type).
- Primary Coil: A coil of insulated wire wound on the core, connected to the AC voltage source (input). Let $N_p$ be the number of turns in the primary coil.
- Secondary Coil: Another coil of insulated wire wound on the same core, connected to the load circuit (output). Let $N_s$ be the number of turns in the secondary coil. There is no direct electrical connection between the primary and secondary coils; energy transfer is through the magnetic field in the core.
Working (Ideal Transformer)
Let an alternating voltage $V_p$ be applied across the primary coil. This alternating voltage drives an alternating current through the primary coil.
According to the law of self-induction, the applied voltage $V_p$ is almost equal to the back EMF induced in the primary coil (especially when the primary coil resistance is negligible).
Let $\Phi$ be the instantaneous magnetic flux linked with each turn of the primary and secondary coils (assuming that almost all the flux produced by the primary passes through the secondary - perfect flux linkage, and the core has negligible resistance).
The instantaneous induced EMF in the primary coil ($e_p$) due to the changing flux is, by Faraday's Law:
$ e_p = -N_p \frac{d\Phi}{dt} $
In an ideal transformer with negligible resistance in the primary coil, the applied primary voltage $v_p$ is equal and opposite to the induced back EMF $e_p$:
$ v_p = -e_p = N_p \frac{d\Phi}{dt} \quad \ldots (1) $
The same changing magnetic flux $\Phi$ also links with the secondary coil. The instantaneous induced EMF in the secondary coil ($e_s$) is:
$ e_s = -N_s \frac{d\Phi}{dt} \quad \ldots (2) $
This induced EMF $e_s$ across the secondary coil terminals is the output voltage $v_s$ supplied to the load (assuming the secondary coil resistance is negligible).
$ v_s = e_s = -N_s \frac{d\Phi}{dt} $
Divide equation (2) by equation (1):
$ \frac{v_s}{v_p} = \frac{-N_s (d\Phi/dt)}{N_p (d\Phi/dt)} $
Assuming $d\Phi/dt \neq 0$ (i.e., there is a changing flux, which is true for AC):
$ \frac{v_s}{v_p} = \frac{N_s}{N_p} $
This relationship holds for instantaneous voltages. It also holds for the magnitudes of peak voltages ($V_{s0}/V_{p0}$) and RMS voltages ($V_{s, rms}/V_{p, rms}$). Let's use $V_p$ and $V_s$ to denote RMS voltages.
$ \frac{V_s}{V_p} = \frac{N_s}{N_p} $
This equation is called the transformer equation. The ratio $\frac{N_s}{N_p}$ is called the turns ratio or transformation ratio.
Step-up and Step-down Transformers
- Step-up transformer: If $N_s > N_p$, then $V_s > V_p$. The secondary voltage is greater than the primary voltage. Used to increase AC voltage.
- Step-down transformer: If $N_s < N_p$, then $V_s < V_p$. The secondary voltage is less than the primary voltage. Used to decrease AC voltage.
Currents in an Ideal Transformer
In an ideal transformer, there is no energy loss. The power delivered to the primary circuit is equal to the power delivered by the secondary circuit to the load.
Power in primary = $V_p I_p$ (assuming pure resistive or unity power factor load, or using apparent power)
Power in secondary = $V_s I_s$ (assuming power factor is determined by load)
$ V_p I_p = V_s I_s $
Rearranging this equation:
$ \frac{I_s}{I_p} = \frac{V_p}{V_s} $
Using the transformer equation $\frac{V_s}{V_p} = \frac{N_s}{N_p}$:
$ \frac{I_s}{I_p} = \frac{N_p}{N_s} $
This means that in a transformer, the ratio of currents is inversely proportional to the ratio of turns.
- In a step-up transformer ($N_s > N_p$, $V_s > V_p$), the secondary current is less than the primary current ($I_s < I_p$). The voltage is stepped up, but the current is stepped down.
- In a step-down transformer ($N_s < N_p$, $V_s < V_p$), the secondary current is greater than the primary current ($I_s > I_p$). The voltage is stepped down, but the current is stepped up.
The quantity that is essentially conserved in an ideal transformer (when transferring energy) is the power, and the frequency of the AC voltage remains unchanged.
Energy Losses in a Real Transformer
Real transformers are not ideal and have some energy losses:
- Joule Loss (Copper Loss): Energy is lost as heat due to the resistance of the copper wires in the primary and secondary coils ($I^2R$ loss). This loss can be minimised by using thick wires with low resistance, especially in coils carrying high current.
- Eddy Current Loss: The changing magnetic flux in the iron core induces eddy currents in the core material, leading to power dissipation as heat ($I_{eddy}^2 R_{core}$). This loss is minimised by using a laminated core, where thin sheets of iron are insulated from each other, increasing the resistance to eddy current flow.
- Hysteresis Loss: Energy is lost due to the repeated magnetisation and demagnetisation of the iron core as the AC current changes direction. This loss is proportional to the area of the hysteresis loop of the core material. Using soft iron (which has a narrow hysteresis loop) for the core minimises this loss.
- Flux Leakage: Not all the magnetic flux produced by the primary coil links with the secondary coil. Some flux lines may leak out into the surrounding air. This reduces the mutual inductance and hence the induced EMF in the secondary. This loss is minimised by winding the primary and secondary coils closely together on the core.
Despite these losses, modern transformers are highly efficient, typically 95% or more, especially large power transformers.
Uses of Transformers
Transformers are indispensable components in the power grid:
- Power Transmission: At power generating stations, step-up transformers are used to increase the voltage of the generated AC power to very high levels (e.g., from 11 kV to 132 kV or 400 kV) for transmission over long distances. Transmitting power at high voltage significantly reduces the current ($P=VI$), and therefore reduces the power loss ($I^2R$) in the transmission lines.
- Power Distribution: Along the transmission route, step-down transformers are used at substations to reduce the voltage to lower levels (e.g., from 400 kV to 33 kV or 11 kV) for local distribution. Further step-down transformers (e.g., at neighbourhood poles) reduce the voltage to the standard domestic supply voltage (220 V in India) for household use.
- In Electronic Appliances: Small step-down transformers are used in power adapters and electronic devices to reduce the mains voltage to the lower voltages required by the electronic circuits.
- Isolation Transformers: Transformers with a 1:1 turns ratio are used to isolate circuits from the mains supply for safety purposes.
Example 1. A step-down transformer is used to reduce the mains 220 V supply to 12 V. The primary coil has 1100 turns. (a) Calculate the number of turns required in the secondary coil. (b) If the transformer is ideal and supplies a current of 2 A to the load, what is the current in the primary coil?
Answer:
Given:
Primary voltage, $V_p = 220 \, V$ (RMS)
Secondary voltage, $V_s = 12 \, V$ (RMS)
Number of turns in primary coil, $N_p = 1100$
(a) We use the transformer equation $\frac{V_s}{V_p} = \frac{N_s}{N_p}$. We need to find the number of turns in the secondary coil, $N_s$.
Rearranging the formula: $ N_s = N_p \times \frac{V_s}{V_p} $
Substitute the given values:
$ N_s = 1100 \times \frac{12 \, V}{220 \, V} $
$ N_s = 1100 \times \frac{12}{220} = 1100 \times \frac{6}{110} = 10 \times 6 = 60 $
The number of turns required in the secondary coil is 60 turns.
(b) Assuming the transformer is ideal, the power input equals the power output: $V_p I_p = V_s I_s$. We need to find the current in the primary coil, $I_p$. The current in the secondary coil, $I_s = 2 \, A$.
Rearranging the formula: $ I_p = I_s \times \frac{V_s}{V_p} $
Substitute the given values:
$ I_p = 2 \, A \times \frac{12 \, V}{220 \, V} $
$ I_p = 2 \, A \times \frac{12}{220} = 2 \, A \times \frac{6}{110} = 2 \, A \times \frac{3}{55} = \frac{6}{55} \, A $
$ I_p \approx 0.109 \, A $
Alternatively, using the turns ratio relationship $\frac{I_s}{I_p} = \frac{N_p}{N_s}$: $I_p = I_s \times \frac{N_s}{N_p}$
$ I_p = 2 \, A \times \frac{60}{1100} = 2 \, A \times \frac{6}{110} = \frac{12}{110} \, A = \frac{6}{55} \, A $
The current in the primary coil is approximately 0.109 Amperes. This is less than the secondary current, as expected for a step-down transformer (voltage is stepped down, current is stepped up from the primary perspective, or stepped down from the secondary perspective relative to primary current).