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, I_rms/V_rms is the value of steady current/voltage that produces the same power in a resistor as the alternating current/voltage
  • I_rms = ^I₀/_√2
  • V_rms = ^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 I_rms and V_rms
  • 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
        • V_rms = ^V₀/_√2 = ²²⁰/_√2
        • V_rms = 156 v
     3. the frequency
        • The quantity is sin() is equal to ωt:
          • ∴ ω = 120π
        • Also, ω = 2πf, so:
          • f = ^120π/_2π
          • ∴ 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 V_P across the primary coil causes an alternating current I_P 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 I_S is induced in it, producing an alternating e.m.f. V_S 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:
  • I_PV_P = I_SV_S
  • ^N_S/_NP = ^V_S/_VP = ^I_P/_IS
  • or simply use ratios

Phase Difference in V_P/V_S and I_P/I_S/Φ

• 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
• V_P and V_S have a phase difference of 90° with I_S, I_P 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.