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Alternating Currents

Sinusoidal Current

  • Period, T is the time for one complete cycle of the alternating current (a.c.)
  • Frequency, f is the number of oscillations per unit time; f = 1/T
  • Peak value, I0/V0 is the highest point on the graph
  • Instantaneous current/voltage, I/V is the current/voltage at a particular instant
    • I = I0sinωt
    • V = V0sinωt
    • where ω = 2πf
  • The root-mean-squared (r.m.s) value, Irms/Vrms is the value of steady current/voltage that produces the same power in a resistor as the alternating current/voltage
    • Irms = I0/√2
    • Vrms = V0/√2
Mean Power in an a.c. Supply
  • For a sinusoidal alternating current, the peak power is twice the average power
    • P = IV and using Irms and Vrms
    • P = I0/√2 × V0/√2 = ½IV
  • For example:
    1. An alternating voltage is represented by the equation: V = 220sin(120πt)
      For this alternating voltage, determine

      1. peak voltage
        • Simply using the equation, the peak voltage, V = 220 V
      2. the r.m.s voltage
        • VrmsV0/√2220/√2
        • Vrms = 156 v
      3. the frequency
        • The quantity is sin() is equal to ωt:
          • ∴ ω = 120π
        • Also, ω = 2πf, so:
          • f = 120π/
          • ∴ f = 60 Hz
Transformer
  • A Transformer is a device used to increase or decrease the current or voltage of an alternating current
  • An Ideal Transformer has no power loss in the transformer
    • Input power = Output power
  • The p.d VP across the primary coil causes an alternating current IP to flow, producing a magnetic field in the soft iron core
  • The secondary coil is thus in a changing magnetic field, and an alternating current IS is induced in it, producing an alternating e.m.f. VS across the secondary coil
  • Step-up Transformer: The primary coil has fewer turns than the secondary coil, hence the output voltage is greater (current decreases by the same factor)
  • Step-down Transformer: The primary coil has greater turns than the secondary coil, hence the output voltage is lower (current increases by the same factor)
  • Transformer Relationships:
    • IPVP = ISVS
    • NS/NP = VS/VP = IP/IS
    • or simply use ratios
Phase Difference in VP/VS and IP/IS
  • The alternating current in the primary coil is not in phase with the alternating e.m.f. induced in the secondary coil:
    • Current in the primary coil gives rise to the magnetic field
    • The magnetic field in the core is in phase with the current in the primary coil
    • The magnetic flux cuts the secondary coil inducing the e.m.f. in the secondary coil
    • The e.m.f. induced is proportional to the rate of change of field, so not in phase
  • VP and VS have a phase difference of 90° with IS, IP and Φ
Eddy Currents
  • If a metallic conductor moves in a magnetic field, an e.m.f. is induced which will make free electrons in the metal move, causing electric current — eddy currents
  • The eddy currents will oppose change in flux linkage of the conductor by Lenz’s law and energy of motion will be dissipated as heat
Energy Loss in a Practical Transformer
  • Some power is lost due to resistance in the coils of transformers causing them to heat up
  • Some power is lost as the magnetic flux flows back and forth. To minimize this, a soft magnetic material is used where magnetic flux direction can change easily
  • Losses also occur in the core due to eddy currents: induced currents flows through the iron core and dissipate energy due to its resistance. Currents can be reduced y making the core out of thin laminated sheets; flux can easily flow but eddy currents cannot
Transmission of Electrical Energy
  • Electricity transmission lines have resistance, therefore, energy will be lost through heating in the wires
  • Electricity transmitted at high voltage a.c. supply:
    • High Voltage: For same power, current is smaller, so less heating and voltage loss in cables/wires
    • A.c. Supply: This can change the output voltage efficiently using transformers
Half-wave Rectification
  • For one half of the time, the voltage is 0, this means that the power available from a half-wave rectified supply is reduced
Full-wave Rectification
  • The four diodes are known as a bridge diode
  • When current is flowing for the first half of period
  • When current is flowing for the second half of period
Smoothing
  • In order to produce steady d.c. from ‘bumpy’ d.c. that results from rectification, a smoothing capacitor is required
  • The capacitor charges and maintains the voltage as a.c. voltage rises, (first half of the wave)
  • As the wave slopes downward, the capacitor begins to discharge in order to maintain the voltage

  • A small capacitor discharges more rapidly than a large capacitor and gives rise to a greater ripple in the output
  • If the load resistor is small, the capacitor will also discharge rapidly
  • CR is the time constant of a capacitor resistor: It is the time taken for a charge to fall 1/e times the original value
    • The value should be much greater than the time period of a.c. supply so the capacitor does not have sufficient time to discharge significantly
  • In general, the greater the value R × C, the smoother the rectified a.c.
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