#### 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, I
_{0}/V_{0}is the highest point on the graph - Instantaneous current/voltage, I/V is the current/voltage at a particular instant
- I = I
_{0}sinωt - V = V
_{0}sinωt - where ω = 2πf

- I = I
- 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}

- I

**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

- P = IV and using I
- For example:
- An alternating voltage is represented by the equation: V = 220sin(120πt)

For this alternating voltage, determine- peak voltage
- Simply using the equation, the peak voltage, V = 220 V

- the r.m.s voltage
- V
_{rms}=^{V0}/_{√2}=^{220}/_{√2} - V
_{rms}= 156 v

- V
- the frequency
- The quantity is sin() is equal to ωt:
- ∴ ω = 120π

- Also, ω = 2πf, so:
- f =
^{120π}/_{2π} - ∴ f = 60 Hz

- f =

- The quantity is sin() is equal to ωt:

- peak voltage

- An alternating voltage is represented by the equation: V = 220sin(120πt)

**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

- 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
_{P}V_{P}= I_{S}V_{S} ^{NS}/_{NP}=^{VS}/_{VP}=^{IP}/_{IS}- or simply use ratios

- I

**Phase Difference in V**_{P}/V_{S} and I_{P}/I_{S}/Φ

_{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.