AC generators in power stations use electromagnetic induction to convert kinetic into electrical energy. Electromagnetic induction is then used in transformers to transmit the electricity efficiently to where it is needed.
Key Concepts
A transformer consists of two coils - primary and secondary. The primary coil produces a changing magnetic field due to the alternating current supplied within. The secondary coil cuts the changing magnetic flux and a changing EMF is induced, causing a current to flow. The strength of the magnetic field is increased by the presence of the solf iron core.
The process of transmitting electrical power from a generator to a town is optimised to reduce thermal energy losses.
The main culprit is usually thermal energy loss in power lines, due to the size of the current. To reduce this current and hence the wasted energy:
- A step-up transformer is used before the power lines (to increase voltage an reduce current)
- A step-down transformer is used after the power lines (to reduce voltage to an acceptable size for household use)
Use quizzes to practise application of theory.
START QUIZ!
A straight conductor of length \(x\) travels through a perpendicular magnetic field of flux density \(y\) with constant speed \(z\).
The induced EMF between the ends of the conductor is:
\(\varepsilon = BLv =yxz\)
An AC generator induces 12 V at 100 revolutions per second.
The freqency is increased to 200 revolutions per second. What are the new output voltage and time period?
The initial time period is \({1\over 100}=0.01 \text{ s}=10\text { ms}\)
Doubling the frequency doubles the induced EMF and halves the time period.
A coil with \(N\) turns and area \(A\) rotates in a uniform field \(B\).
When the induced EMF is maximum, the flux enclosed is:
The EMF, which is proportional to the rate of change of flux, is greatest when flux enclosed is zero. This is because there is the largest possible change in flux.
A coil with \(N\) turns and area \(A\) rotates in a uniform field \(B\).
When the induced EMF is zero, the flux linkage is:
Rate of change of flux is minimum when the flux enclosed is maximum.
\(N\Phi=BAN\cos \theta\) is maximised when \(\cos \theta=1\)
A coil with \(N\) turns and area \(A\) rotates clockwise in a uniform field \(B\).
At the position shown the EMF is:
The clockwise rotation means that the top and bottom of the coil were recently moving parallel to the field lines when no EMF was induced.
The EMF could now be increasing or decreasing (but will be increasing in magnitude) depending on which direction is taken to be positive.
A coil with \(N\) turns and area \(A\) rotates in a uniform field \(B\).
At what angle \(\theta\) will the EMF be half its maximum value?
The EMF induced is proportional to \(\sin\theta\). When \(\sin \theta={1\over 2}\), \(\theta = 30°\).
NB: We can check that \(\sin\theta\) is appropriate here by noticing that the angle is between the perpendicular and the field lines.
A coil with \(N\) turns and area \(A\) rotates in a magnetic field \(B\) with frequency \(f\). The mean value of the induced EMF is:
\(BAN\sqrt2πf\)
Note that the average requested was the 'mean'. Since EMF is sinusoidal (\(\varepsilon = BANω\sinθ\)), the mean is 0.
A coil with \(N\) turns and area \(A\) rotates in a magnetic field \(B\) with frequency \(f\). The rms value of the induced EMF is:
\(BANπf\over \sqrt2\)
\(BAN\sqrt2πf\)
\(\varepsilon_\text{rms} = {\varepsilon_0\over \sqrt2} ={BAN2πf\over \sqrt2}=BAN\sqrt2 \pi f\)
Light bulb A is connected to an AC supply of peak voltage 6 V. An identical bulb B is connected to a 6 V battery.
What is the ratio \(\text{power dissipated in A} \over \text{power dissipated in B}\)?
\(P_\text{A}={{V_\text{rms}}^2\over R} ={{V_0}^2\over {\sqrt2}^2R}={1\over 2}{{V_0}^2\over R}={1\over2}P_\text{B}\)
Light bulb A is connected to an AC supply of rms voltage 6 V. An identical bulb B is connected to a 6 V battery.
What is the ratio \(\text{power dissipated in A} \over \text{power dissipated in B}\)?
\(P_\text{A}={{V_\text{rms}}^2\over R}=P_\text{B}\)
Identical components are connected in two different ways to produce two different transformers, A and B.
What is the ratio \(V_\text{rms} \text{ across load A} \over V_\text{rms} \text{ across load B}\)?
Let's call the EMF of the supply \(\varepsilon\):
\(V_\text{load A}={\varepsilon\over 2}\)
\(V_\text{load B}=2{\varepsilon}\)
The rms voltage of the supply in the diagram is 6 V and the load resistor has resistance 1 kΩ.
The peak current through the load is:
\(N_s/N_p = 2\Rightarrow V_\text{rms}=12 \text{ V}\)
\(I_\text{rms} = {12\over1000} = 12 \text{ mA}\)
\(I_0=\sqrt2 I_\text{rms}\)
A power station delivers 2 MW at a potential difference of 50 kV through cables of resistance 4 Ω.
The current through the cables is:
\(P=IV \Rightarrow I={P\over V}={2\times 10^6\over 50\times 10^3}=40\text{ A}\)
NB: We cannot use \(V = IR\) as the potential difference given is across the power station (and not across the cables).
How much of Transformers have you understood?
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