Alternating Current

Question 1

What to take note about sine squared graph?

Answer 1

Smooth oscillating curve (Not mcdonalds graph with sharp turns)
* as long as there is any sinusoidal curve, squaring gives squared sinusoidal graph, never a McDo graph

Question 2

Define root mean square (r.m.s.) value of alternating current (or voltage)

Answer 2

rms value of alternating current/voltage is that value of steady direct current/voltage that would produce thermal energy at same rate in resistor

Question 3

What are the three steps to deduce average current/voltage/etc using rms?

Answer 3

1. Find period and square ‘every’ single value
2. Find mean of squared value which is area under graph for 1 period divided by 1 period
   *drawing squared variable against time graph (eg V² - t) can help with calculation
3. Take square root of squared value to get rms value

Question 4

What are the formulae for SINUSOIDAL alternating currents?

Answer 4

I = (I peak) sinωt OR (I peak )cosωt

I rms = I peak/√2

V rms = V peak/√2

P rms = 0.5P peak

Question 5

What to take note about sinusoidal alternating currents?

Answer 5

When question involves household appliances, transformers or when waveform is not stated,
ASSUME that alternating current varies sinusoidally with time

Question 6

How do transformers operate?

Answer 6

(based on principle of EMI)
a) when alternating voltage Vp applied to primary coil, alternating current Ip flow in primary coil
b) Ip sets up alternating mag flux in iron core
c) changing mag flux linkage induce alternating output voltage Vs in secondary coil, thus induce alternating current Is in secondary coil

Question 7

Give formula for transformers

Answer 7

Ns/Np = Vs/Vp = Ip/Is

Question 8

Describe step-up and step-down transformers

Answer 8

• ideal transformers mean no energy loss
• output voltage can be increased/decreased by changing turns ratio in primary and secondary coil
• for step up, Ns > Np, so Vs>Vp
• for step down, Ns<Np, so Vs<Vp

Question 9

Explain how energy is lost in transformers

Answer 9

1. Resistance of windings
   - copper wire used for winding hv resistance, cause heat loss (P loss = I²R)
2. Flux leakage
   - all flux produced by primary coil may not b linked to secondary coil (leaked out) if design of core is poor eg cracks
3. Eddy currents
   - changing B field not oni induce current in secondary coil but oso eddy current in iron coil itself, flow in little circles in iron core
   - eddy current cause heat loss (I²R)
   - can b reduced by use laminated soft iron core comprising of thin sheet soft iron insulated fr one another
4. Hysteresis
   - magnetisat n of core is repeatedly reversed by alternating B field
   - this repeating core magnetisat n process expends energy, appears as heat
   - heat generated can be minimised by use magnetic material easily magnetised, demagnetised (eg soft iron)

Question 10

Describe half-wave rectification

Answer 10

• simplest rectifier circuit is diode connected in series w ac input
• since diode allow current in oni 1 direct n, oni half of incoming ac input reach output

Question 11

Describe full-wave rectification

Answer 11

• four diodes can b arranged in bridge network form full wave rectifying circuit
• bridge network allow AC flow thru resistor as direct current
• voltage output hv same mag as AC voltage but in 1 direct n oni
  *McDo graph