PHYSICS A LEVEL(FORM SIX) NOTES – AC THEORY(1)

AC THEORY

When a battery is connected to a circuit the current flows steadily in one direction, this is called a Direct current (d.c).

The use of Direct currents is limited to a few applications e.g charging of batteries, electroplating etc.

Most of electrical energy is generated and used in the form of alternating current due to many reasons including,

i) Alternating voltages can be changed in value very easily by means of transformers.

ii) A.c motors are simpler in construction and cheaper than d.c motors.

ALTERNATING VOLTAGE AND CURRENT

i) Alternating voltage

An alternating voltage is one whose magnitude changes with time and direction reverses periodically. The instantaneous value (i.e value at any time t) of an alternating voltage is given by,

where,

E = Value of the Alternating voltage at time t

E0 = Maximum value of the Alternating voltage

ω = Angular frequency of supply
From

,where f is the frequency of the alternating voltage, If T is the time period of alternating voltage then

The voltage varies from zero to a positive peak (+E0) then back via zero to negative peak (-E0) and so on.
In time period T, the wave completely cycle.

ii) (ii)Alternating current

This is one whose magnitude changes with time and direction reverses periodically.

The Instantaneous value (

value at any time t) of sinusoidally varying alternating current is given by

where

I = value of alternating current at time t

I0 = maximum value (Amplitude) of alternating current.

ω = Angular frequency of supply.

Figure below shows the waveform of alternating current.

Current varies sinusoidally with time. The current increases gradually from zero to a positive peak (+Io), then back via zero to a negative peak (–Io) and so on.

In time period T, the wave completes a cycle.

An Alternating voltage and current also be represented as a cosine function of time.

Both these representations give the same result as is given by the one containing sine functions.

MEASUREMENT OF ALTERNATING CURRENT

Since the average value of sinusoidal alternating current is zero, an ordinary (DC) ammeter or galvanometer will not show any deflection when connected in an AC circuit.

Due to inertia it will not be possible for the needle to oscillate with the frequency of the current.

Therefore to measure AC we use hot wire instruments because the heating effect of current is independent of the direction of current.

MEAN OR AVERAGE VALUE OF ALTERNATING CURRENT

The mean value or average value of alternating current over one complete circle is zero.

It is because the area of positive half cycle is exactly equal to the area of the negative half cycle.

However, we can find the average or mean value of alternating current over any half cycle.

Half Cycle Average Value of a.c
This is that value of steady current (d.c) which would send the same amount of charge through a circuit for half the time period of a.c

as it sent by the a.c through the same circuit in the same time.It is represented by Im or

The instantaneous value of alternating current is given by,

Suppose current I remains constant for a small time

. Then small amount of charge sent by alternating current in a small time

is given by

If Q is the total charge sent by the positive half cycle of a.c (

If the Im is the half cycle average means values of positive half cycle of a.c then by definition

From equation (i) and (ii)

Hence half cycle average value of a.c is 0.637 times the peak value of a.c
For positive ½ cycle Im

+0.637 I0

For negative half cycle Im

-0.637 I0

Obviously, average value of a.c over a complete cycle is zero.

MEAN OR AVERAGE VALUE OF ALTERNATIVE VOLTAGE
Half Cycle Average value of alternating

This is that value of steady

, (

which would send the same amount charge through a circuit for half the time period of alternating

as is sent by the alternating

through the same circuit in the same time.It is denoted by

The instantaneous value of alternating e. m. f is given by

Suppose this alternating

is applied to a circuit of resistance R. Then by ohm’s law the instantaneous value of alternating current is

If this current remains constant a small time

, then small amount of charge send by alternating

in small time

is given by

If Q is the total charge sent by positive half cycle of a alternating

then

is the half cycle average or mean value of the positive half cycle of alternating

then by definition

From equation (i) and equation (ii)

Therefore, half cycle average value of alternating

is 0.637 times the peak value of alternating

.
For positive half cycle

For negative half cycle

A d.c voltmeter or ammeter reads average (or d.c) value. Therefore, they can be used to measure alternating voltage on current.

It is because the average value of alternating voltage or current over a complete cycle is zero.

We use a.c meters to measure alternating voltage/current.

ROOT MEAN SQUARE VALUE OF ALTERNATING CURRENT

The average value cannot be used to specify an alternating current (or voltage).

It is because its value is zero over one cycle and cannot be used for power calculations.

Therefore we must search for more suitable criterion to measure the effectiveness of an alternating current or voltage.

The obvious choice would be to measure it in terms of direct current that would do work (or produce heat) at the same average rate as a.c under similar conditions.

This equivalent direct current is called the root mean square (

) or effective value of alternating current.

Effective or value of Alternating Current

The root mean square (r.m.s) of alternating current is that steady current (d.c) which when flowing through a given resistance for given time produces the same amount of heat as produced by the alternating current when flowing through the same resistance for the same time.

– It is also called virtual value of a.c

– It is denoted by called

For example, when we say that

or effective value of an alternating current is 5A, it means that the alternating current will do the work (or produce heat) at the same rate as 5A direct current under similar conditions.

RELATION BETWEEN R.M.S. VALUE AND PEAK VALUE OF A.C

Let the alternating current be represented by

If this alternating current flows through a resistance R for a small time

, then small amount of heat produced is given.

In one complete cycle

for time 0 to T of alternating current the total amount of heat produced in R is given by

If Irms is the virtual or

value of the alternating current, then heat produced in R in the same time

(0 to T) is given by

From equation (i) and (ii), we have

Hence the

value of effective value or virtual value of alternating current is 0.707 times the peak value of alternating current.

The

value is the same whether calculated for
half cycle or full cycle.

Alternative Method
Let the alternating current be represented by

If this current is passed through a resistance R, then power delivered at any instant is given by

Because the current is squared, power is always positive since the value of

varies

between 0 and 1

Average power delivered

is the virtual or

value of the alternating current then by definition

Power delivery P

From equation (I) and (ii)

ROOT MEAN SQUARE VALUE OF ALTERNATING E.M.F

The root mean square (r.m.s) value of alternating voltage is that steady voltage (d.c voltage when applied to a given resistance for a given time produces the same amount of heat as is produced by the alternative

when applied to the same resistance for the same time.

It is also called virtual value of alternating e.m.f and is donated by

The instantaneous value of alternating

is given by

When this alternating voltage is applied to a resistance R, small amount of heat produced in a small time

But

Then,

In one complete cycle

for time 0 to T the total amount of heat produced in resistance R is

If Erms is the

value of the alternating

, then heat produced in the same resistance R for the same time T is

From the equation (i) and equation (ii)

Hence the

value of alternating

is 0.707 times the peak value of the alternating

Therefore,

value is the same whether calculated for half cycle or full cycle.

Importance of r.m.s values

An alternating voltage or current always specified in terms of

values.

T
hus an alternating current of 10A is the one which has the same heat effect as 10A d.c under similar conditions. The following points may be noted carefully.

i) The domestic a.c supply is 230V, 50H. It is the

or effectively value. It means that the alternative voltage available has the same effect as 230 V d.c under similar conditions.

The equation of this alternative voltage is

ii) When we say that alternating current in a circuit is 5A, we specifying the

value.

– It means that the alternating current flowing in the circuit has the same heating effect as 5A d.c under similar conditions.

iii) A.C ammeters and voltmeters record

values of current and voltage respectively.

The alternating voltage/current can be measured by utilizing the heating effect of electric current.

Such Instruments are called hot wire instruments and measure the

value of the voltage/current since r.m.s value is the same for half cycle or complete cycle.

WORKED EXAMPLES

1. An a.c main supply is given to be 220V what is the average

during a positive half cycle?

Solution

Average

during positive half is given by

2. An alternating current I is given by

Find

(i) The maximum value

(ii) Frequency

(iii) Time period

(iv) The instantaneous value when t= 3ms

Solution

Comparing the given equation of the alternating current with the standard form

(i) Maximum value,

(ii) Frequency f

f = 50 HZ

(iii) Time period T

(iv)

When

3. Calculate the

value of the current shown in figure below

Solution

4. An a.c voltmeter records 50V when connected across the terminals of sinusoidal power source with frequency 50Hz. Write down the equation for the instantaneous voltage provided by the source.

Solution

An a.c voltmeter records

values

Now

Here

5. An alternating voltage of 50 Hz has maximum value of 200

volts. At what time measured from a positive maximum value will the instantaneous voltage be 141.4 volts.

Solution

This equation is valid when time is measured from the instant the voltage is zeroi.e point O. Since the time is measured from the positive maximum value at point A.

The above equation is modified to

Let the value of voltage become 141.4 volts t second after passing then the maximum positive value.

6. An alternating current of frequency 60 Hz has a maximum value of 120 A.

i) Write down the equation for the instantaneous value.

ii) Recording time from the instant the current is zero and becoming positive. Find the instantaneous value after 1/360 second.

iii) Time taken to reach 96 A for the first time.

Solution

i) The instantaneous value of alternating current is given by

ii) Since point 0 has been taken as the reference the current equation is

When

iii) To reach the current 96A for the first time

7. Find the r.m.s value of the current after current I shown in figure below

Solution

Consider the current variation over one complete cycle.

A.C CIRCUIT

An A.C circuit is the closed path followed by alternating current.
When a sinusoidal alternating voltage is applied in a circuit, the resulting alternating current is also sinusoidal and has the same frequency as that of applied voltage.

However there is generally a phase difference between the applied voltage and the resulting current.

As we shall see, this phase difference is introduced due to the presence of inductance (L) and capacitance (C) in circuit.

While discussing A.C circuits our main points of interest are;

(i) Phase difference between the applied voltage and circuit current
(ii) Phasor diagram. It is the diagram representation of the phase difference between the applied voltage and the result circuit current.
(iii) Wave diagram
(iv) Power consumed.

A.C CIRCUIT CONTAINING RESISTANCE ONLY

When an alternating voltage is applied across a pure resistance, then from electrons

current flow in one direction for the first half cycle of the supply and then flow in the opposite direction during the next half cycle, thus constitute alternate current in the circuit.

Consider a pure resistor of resistance R connected across an alternating source of

Suppose the instantaneous value of the alternating

is given by

If I is the circuit current at that instant, then by ohm’s law

The value of I will be maximum Io when

Therefore equation (ii) becomes

From

Since sin

t = 1 then I = Io

1) Phase Angle

It is clear from equation (i) and (iii) the applied

and circuit current are in phase with each other

they pass through their zero values at the same instant and attain their peak value both positive and negative peaks at the same instant.

This is indicated in the wave diagram shown in figure below.

The

diagram shown in figure below also reveals that current is in phase with the applied voltage.

Hence in an a.c circuit, current through R is in phase with voltage across R.

This means that current in R varies in step with voltage across R. If voltage across R is maximum current in R is also maximum, if voltage across R is zero, current in R is also zero and so on.

2) Power Absorbed

In a.c circuit, voltage and current vary from instant to instant. Therefore power at any instant is equal to the product of voltage and current at that instant.

Instantaneous power P

Since power varies from instant to instant the average power over a complete cycle is to be considered.

This is found by integrating equation………..(iv) with respect to time for 1 cycle and dividing by the time of 1 cycle. The time per one cycle is T.

Average power P

Therefore, average power absorbed by a resistor in an a.c circuit is equal to the product of virtual voltage ( Erms) across it and virtual current ( Irms) through it.

Obviously, this power is supplied by the source of alternating

Since

WORKED EXAMPLES

1. An a.c circuit consists of a pure resistance of 10Ω and is connected across an a.c supply of 230V, 50 Hz.

Calculate

i) Circuit current

ii) Power dissipated and

iii) Equations for voltage and current

Solution

Erms = 230V R = 10Ω f = 50HZ

i) Circuit current

ii) Power dissipated P

= 230 x 23

∴ P = 5290 W

iii)Equations for voltage and current

x 230

= 325. 27V

= 2 x 23

∴

= 32.52A

ω = 2

= 2

ω = 314

∴ The equations of voltage and current

E = 325.27 sin 314t and I = 3.52 sin 314t

2. In a pure resistive circuit, the instantaneous voltage and current are given

E = 250 sin 314t

I = 10 sin 314t

Determine

i) Peak power

ii) Average power

Solution

In a pure resistive circuit

i) Peak power =

= 250 x 10

∴ Peak power = 2500W

ii) Average power P

P =

∴ P = 1250 W

3. Calculate the resistance and peak current in a 1000 W hair dryer connected to 120V, 60Hz supply. What happens if it is connected to 240V line?

Solution

Peak current Io

Resistance of hair dryer R

When connected to 240V line, the average power delivered would be

This would undoubtedly melt the heating element or the coils of the motor.

4. A voltage E = 60sin 314t is applied across a 20 Ω resistor. What will;

i) An a.c ammeter

ii) Ordinary moving coil ammeter in series with resistor read?

Solution

i) E = 60 sin 314t

An a.c ammeter will read the r.m.s value.

Therefore a.c a meter will read 2.12A

ii) An ordinary moving coil ammeter will read average value of alternating current. Since the average value of a.c over one cycle is zero, this meter will record zero reading.

5. What is the peak value of an alternating current which produces three times the heat per second as a direct current of 2A in a resistor R?

Solution

Heat per second by 2A