Generators, Transformers and Rwanda's Grid – Electromagnetic Induction (Senior 6 Physics) | Ganzaa

Electromagnetic Induction

Lessons

Senior 6 · Physics

Electromagnetic Induction

4 lessons

Lesson 3 of 4

Generators, Transformers and Rwanda's Grid

25 min

From a turbine to a socket

REG's list of power plants gives Nyabarongo I, a hydropower station commissioned in 2014, an installed capacity of 28 MW. Inside it, water turns turbines, and each turbine turns a generator: coils and magnets moving past one another, exactly as in Faraday's experiments but on a giant scale. Before that energy reaches a phone charger in Huye or a welding workshop in Rubavu, its voltage is raised to over a hundred thousand volts and lowered again several times by transformers. This lesson explains both machines, and why the grid is built this way.

The AC generator

Figure 3.1: A coil turning in a magnetic field, and the emf it produces.

A coil of $N$ turns and area $A$ turns at constant angular velocity $\omega$ in a uniform field $B$ . Let the normal to the coil make an angle $\theta = \omega t$ with the field. The flux through each turn is

\Phi = BA\cos\omega t

and by Faraday's law

\varepsilon = -N\frac{d\Phi}{dt} = NBA\omega\sin\omega t = \varepsilon_0\sin\omega t

\varepsilon_0 = NBA\omega

The emf is a sine wave of frequency $f = \frac{\omega}{2\pi}$ , the rate at which the coil turns. It is zero when the plane of the coil is perpendicular to the field (the flux is greatest but momentarily not changing) and greatest when the plane is parallel to the field (the flux is zero but changing fastest; the sides of the coil cut the field lines at right angles). The slip rings keep each end of the coil connected to the same brush, so the output is alternating current. A DC generator replaces them with a split-ring commutator, which reverses the connections every half-turn.

Definition

A generator (dynamo) converts mechanical energy into electrical energy by rotating a coil in a magnetic field, or a magnet inside a coil. For a coil of $N$ turns and area $A$ rotating at $\omega$ in a field $B$ , the emf is $\varepsilon = NBA\omega\sin\omega t$ , with peak value $\varepsilon_0 = NBA\omega$ .

Worked example 1: the peak emf of a generator

A generator coil of 100 turns, each of area $0.020\,\mathrm{m^2}$ , turns 50 times a second in a field of 0.50 T. Find the peak emf, and write an equation for the emf.

\omega = 2\pi f = 2\pi \times 50 = 314\,\mathrm{rad\,s^{-1}}

\varepsilon_0 = NBA\omega = 100 \times 0.50 \times 0.020 \times 314 = 314\,\mathrm{V}

So $\varepsilon = 314\sin(314t)$ volts. Turning the coil twice as fast would double both the peak emf and the frequency.

The transformer

Figure 3.2: The primary and secondary coils share one core, so the same changing flux passes through every turn of both.

The alternating current in the primary makes an alternating flux $\Phi$ in the core. The core guides almost all of it through the secondary, so every turn of both coils has the same $\frac{d\Phi}{dt}$ . By Faraday's law

V_p \approx N_p\frac{d\Phi}{dt}, \qquad V_s = N_s\frac{d\Phi}{dt} \qquad \Rightarrow \qquad \frac{V_s}{V_p} = \frac{N_s}{N_p}

A step-up transformer has $N_{s} > N_{p}$ ; a step-down transformer has $N_{s} < N_{p}$ . In an ideal transformer no energy is lost, so the power in equals the power out:

V_p I_p = V_s I_s \qquad \Rightarrow \qquad \frac{I_s}{I_p} = \frac{N_p}{N_s}

Stepping the voltage up steps the current down in the same ratio. A transformer needs a CHANGING flux, so it does not work on steady DC.

Energy loss	Cause	How it is reduced
copper (resistive) loss	$I^{2} R$ heating in the windings	thick copper wire of low resistance
eddy currents	currents induced in the core itself (lesson 4)	a core of thin, insulated laminations
hysteresis	energy used to magnetise the core back and forth every cycle	a core of soft iron (or silicon steel), easily magnetised
flux leakage	some flux from the primary misses the secondary	coils wound close together, on one closed core

Large transformers reach efficiencies of about 99%, and are cooled with oil and radiator fins because even 1% of many megawatts is a lot of heat.

Definition

A transformer changes the voltage of an alternating supply by mutual induction between two coils on a common iron core. For an ideal transformer $\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}$ . Its efficiency is $\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\%$ .

Worked example 2: a distribution transformer

A transformer in Kigali steps 15 kV down to 0.4 kV. (a) Find the turns ratio and the number of primary turns if the secondary has 60 turns. (b) Treating it as an ideal single-phase transformer (a simplification: real ones are three-phase), find the currents when it supplies 250 kW.

\text{(a)}\; \frac{N_p}{N_s} = \frac{15\,000}{400} = 37.5, \qquad N_p = 37.5 \times 60 = 2250 \text{ turns}

\text{(b)}\; I_s = \frac{P}{V_s} = \frac{250\,000}{400} = 625\,\mathrm{A}, \qquad I_p = \frac{P}{V_p} = \frac{250\,000}{15\,000} = 16.7\,\mathrm{A}

Worked example 3: efficiency

A transformer steps 230 V down to 12 V to light a 60 W lamp at full brightness. It is 90% efficient. Find the input power and the currents in the two coils.

P_{\text{in}} = \frac{60}{0.90} = 66.7\,\mathrm{W}, \qquad I_p = \frac{66.7}{230} = 0.29\,\mathrm{A}, \qquad I_s = \frac{60}{12} = 5.0\,\mathrm{A}

Why the grid uses high voltage

A power station must deliver power $P = V I$ along lines of resistance $R$ . The power wasted as heat in the lines is

P_{\text{loss}} = I^2R = \left(\frac{P}{V}\right)^{2}R

For the same power, raising the voltage ten times cuts the current ten times and the loss one hundred times. That is why Rwanda's grid, like every national grid, transmits at high voltage and uses transformers at each end.

Figure 3.3: The voltage steps on Rwanda's grid.

REG reports that Rwanda's high-voltage transmission network operates at 220 kV and 110 kV, and measured 1158 km of lines at the end of June 2023. Its 2024 distribution master plan gives the distribution voltage as 30 kV across the country, except in Kigali and Rwamagana, which operate at 15 kV. Distribution transformers then step down to 0.4 kV (400 V between two lines of the three-phase supply), which gives $\frac{400}{\sqrt{3}} \approx 230\,\mathrm{V}$ between a line and neutral for homes.

Worked example 4: the benefit of high voltage

A power station sends 20 MW along a line of total resistance $5.0\,\Omega$ . Compare the power lost when it is sent at 110 kV and at 30 kV.

At 110 kV:

I = \frac{20 \times 10^{6}}{110 \times 10^{3}} = 182\,\mathrm{A}

P_{\text{loss}} = 182^2 \times 5.0 = 1.65 \times 10^{5}\,\mathrm{W} \;(0.83\%)

At 30 kV:

I = \frac{20 \times 10^{6}}{30 \times 10^{3}} = 667\,\mathrm{A}

P_{\text{loss}} = 667^2 \times 5.0 = 2.2 \times 10^{6}\,\mathrm{W} \;(11\%)

At the lower voltage more than a tenth of the power would be lost as heat.

Experiment 3.1: testing the turns rule

Apparatus: a demountable transformer kit (a laminated C-core with a clamp and plug-in coils of, for example, 120, 240 and 480 turns), a school low-voltage AC power supply set to 2 V to 6 V, two AC voltmeters (or multimeters on AC), leads.

Fit a 240-turn primary and a 120-turn secondary on the core and clamp the core closed.
Connect the primary to the low-voltage AC supply and a voltmeter across each coil. Record $V_{p}$ and $V_{s}$ .
Repeat with other pairs of coils. Compare $\frac{V_s}{V_p}$ with $\frac{N_s}{N_p}$ .
Unclamp the core and lift one half slightly: $V_{s}$ falls because flux leaks out of the gap.

Warning

Safety. Use ONLY the school's low-voltage AC supply, never the mains, and never make a step-up arrangement that could give more than about 12 V at the secondary. Switch off before changing coils or moving the core; the core can trap fingers when it is clamped. Coils get warm if left on: switch off between readings.

Warning

Common mistakes. (1) Thinking a step-up transformer creates energy: it raises the voltage but lowers the current, and $P_{\text{out}} \le P_{\text{in}}$ . (2) Using $P_{\text{loss}} = \frac{V^2}{R}$ with the transmission voltage: that $V$ is not across the line resistance; use $I^{2} R$ . (3) Expecting a transformer to work on DC. (4) Mixing up where the generator emf is zero: the flux is greatest there, not zero.

Tip

Voltage follows turns, current goes against them: $\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}$ .

How transformers work (Cognito). Watch how the alternating current in the primary coil makes a changing field in the iron core, and how the turns ratio sets the output voltage.

✦

Practice

A generator coil of 250 turns, each of area $0.010\,\mathrm{m^2}$ , rotates at 3000 rev/min in a field of 0.20 T. Find the peak emf.
At what position of the coil is the emf of a generator greatest? Explain why, although the flux through the coil is then zero.
A transformer steps 230 V down to 11.5 V. Its secondary has 50 turns. How many turns has the primary?
The secondary of the transformer in question 3 supplies 2.0 A. Assuming it is ideal, find the primary current.
A transformer takes 500 W from the supply and delivers 470 W. Find its efficiency and name two ways energy is lost.
A small power station sends 5.0 MW through lines of resistance $4.0\,\Omega$ . Find the power lost if it is transmitted at 30 kV and at 110 kV.
Mugisha says: "A step-up transformer is useful because it gives out more energy than it takes in." Evaluate this statement.

Show answersHide answers

$\omega = 3000 \times \frac{2\pi}{60} = 314\,\mathrm{rad\,s^{-1}}$ ; $\varepsilon_0 = 250 \times 0.20 \times 0.010 \times 314 = 157\,\mathrm{V}$
When the plane of the coil is parallel to the field. The flux is then zero but changing at its fastest rate ( $\varepsilon \propto \sin\omega t$ , the rate of change of $\cos\omega t$ ); the sides of the coil move at right angles to the field lines.
$N_p = N_s\frac{V_p}{V_s} = 50 \times \frac{230}{11.5} = 1000$ turns
$I_p = \frac{V_s I_s}{V_p} = \frac{11.5 \times 2.0}{230} = 0.10\,\mathrm{A}$
$\eta = \frac{470}{500} = 94\%$ . Losses: heating of the windings ( $I^{2} R$ ), eddy currents and hysteresis in the core, flux leakage.
At 30 kV: $I = 167\,\mathrm{A}$ , $P_{\text{loss}} = 167^2 \times 4.0 = 1.1 \times 10^{5}\,\mathrm{W}$ . At 110 kV: $I = 45.5\,\mathrm{A}$ , $P_{\text{loss}} = 8.3 \times 10^{3}\,\mathrm{W}$ , about 13 times less.
Wrong. A transformer cannot create energy: at best $V_{s} I_{s} = V_{p} I_{p}$ . A step-up transformer raises the voltage and lowers the current in the same ratio. It is useful because the smaller current wastes far less power ( $I^{2} R$ ) in transmission lines.

Key points

AC generator: $\varepsilon = NBA\omega\sin\omega t$ , peak $\varepsilon_0 = NBA\omega$ ; slip rings give AC, a commutator gives DC.
Transformer: $\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}$ (ideal); losses from resistance, eddy currents, hysteresis and flux leakage.
High-voltage transmission cuts $I$ , and the loss $I^{2} R$ falls as $\frac{1}{V^2}$ . Rwanda transmits at 220 kV and 110 kV and distributes at 30 kV (15 kV in Kigali and Rwamagana), then 0.4 kV.

Sources and credits

23.8: Electric Generators and 23.10: Transformers, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
13.7: Electric Generators and Back Emf, from University Physics Volume 2 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Rwanda's grid (facts quoted): Transmission and Power Plant list, Rwanda Energy Group (REG); Rwanda Electricity Distribution Master Plan, June 2024 revision, REG.
Photo: Substation Power Transformer 1.jpg by Yoshieslunchbox, Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: GCSE Physics - How Transformers Work, Cognito, embedded from YouTube.

The experiment, the three diagrams, the worked examples and the practice exercises are written by Ganzaa.

1From a turbine to a socket

2The AC generator

3The transformer

4Why the grid uses high voltage

5Experiment 3.1: testing the turns rule

From a turbine to a socket

The AC generator

The transformer

Why the grid uses high voltage

Experiment 3.1: testing the turns rule