Characteristics of alternating currents Peak current, I

Peak-to-peak current, I

Period, T = 20 ms

Frequency, f = 1 / T = 50 Hz

Angular Frequency, ω = 2 π f = 314 rad s

Instantaneous current: the current at a particular instant.

Since this A.C. signal can be described by the equation: I = I0 sin (ω t)

or V = V0 sin (ω t)

the instantaneous current I or voltage V at time t is given by

Root-mean-square current of an alternating current is defined as that steady {NOT

Max (Instantaneous) Power, P

The

{Values of I & V may be either R.M.S. or peak but not instantaneous values; N

Power Loss during Transmission of Electrical Power Power Generated at power station P

where I: current in the transmission, Vi: Voltage at which power is transmitted

Thus to reduce power loss, for a given amt of power generated, electricity is transmitted at high voltage V

Rectification with a diode If a single diode is connected to an A.C. circuit as shown, a

The graphs for the input and output voltages, and the output current, are shown below.

In the regions A and C, the diode is forward biased, allowing current to flow. When the input voltage becomes negative, the diode prevents the current flow, because it is reverse biased.

_{0}= 3 APeak-to-peak current, I

_{p-p}= 6 APeriod, T = 20 ms

Frequency, f = 1 / T = 50 Hz

Angular Frequency, ω = 2 π f = 314 rad s

^{-1}Instantaneous current: the current at a particular instant.

Since this A.C. signal can be described by the equation: I = I0 sin (ω t)

or V = V0 sin (ω t)

the instantaneous current I or voltage V at time t is given by

*I*or_{0}sin (ωt)*V*. Note: Both the period and amplitude of a sinusoidal A.C should be_{0}sin (ωt)**constant**.Root-mean-square current of an alternating current is defined as that steady {NOT

*direct*} current that produces the same heating effect {ie I^{2}R} as the alternating current in a given resistor.(Instantaneous) sinusoidal current: I = I

I_{0}sinωt , { Similarly, V = V_{0}sinωt }_{rms}= I_{o}/ √2, V_{rms}= v_{o}/ √2, {for**sinusoidal**ac only} Relationship between Peak, & RMS values of PD & Current: V

_{0}= I_{0}R , V_{rms}= I_{rms}R**Mean/Ave Power, P**= ½ x Maximum Instantaneous Power =_{ave}= I_{rms}^{2}R = V_{rms}^{2}/ R = I_{rms}/ V_{rms}**½ I**{for sinusoidal AC}_{0}V_{0}Max (Instantaneous) Power, P

_{max}= I_{0}V_{0}= I_{0}^{2}RThe

**root-mean-square**(R.M.S.) value, I_{rms}, of an A.C. is the magnitude of the direct current that produces the same**average**heating effect as the alternating current in a given resistance whereas peak value is the maximum current of an AC.Ideal transformer: V

{Mean power in the primary coil = Mean power in the secondary coil )_{p}I_{p}= V_{s}I_{s}→ N_{S}/ N_{P}= V_{S}/ V_{P}= I_{P}/ I_{S}{Values of I & V may be either R.M.S. or peak but not instantaneous values; N

_{S}/ N_{P}: turns ratio}Power Loss during Transmission of Electrical Power Power Generated at power station P

_{gen}= V_{i}I,where I: current in the transmission, Vi: Voltage at which power is transmitted

I = P

Power Loss in Transmission Cables, P_{gen}/ V_{i}_{L}= I^{2}R_{C}= (P_{gen}/ V_{i})^{2}R_{C};R_{C}= cable resistanceThus to reduce power loss, for a given amt of power generated, electricity is transmitted at high voltage V

_{i}{ie low current}. {V_{i}is NOT the pd across the cables}Rectification with a diode If a single diode is connected to an A.C. circuit as shown, a

**half-wave rectification**occurs.The graphs for the input and output voltages, and the output current, are shown below.

In the regions A and C, the diode is forward biased, allowing current to flow. When the input voltage becomes negative, the diode prevents the current flow, because it is reverse biased.