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Assistant

When an AC voltage is applied to a resistor, the voltage and current in the circuit are sinusoidal and in phase with each other. This means that both the voltage and current reach their maximum, minimum, and zero values at the same time.

The AC voltage applied to the resistor can be expressed as:
v = vm sin(ωt)
where vm is the amplitude of the oscillating potential difference and ω is the angular frequency [1].

Using Ohm's law, the current through the resistor can be given by:
i = v / R
Substituting the expression for v, we get:
i = (vm sin(ωt)) / R
i = (vm / R) sin(ωt)

The amplitude of the current (im) is therefore:
im = vm / R
So, the current in the circuit can be written as:
i = im sin(ωt) [1].

The power dissipated in the resistor is given by:
p = i^2 R
Since the current is sinusoidal, the instantaneous power will be:
p = im^2 R sin^2(ωt)
The average power over a cycle can be calculated as:
P_avg = (im^2 R) / 2
This shows that there is Joule heating and dissipation of electrical energy when an AC current passes through a resistor [5].

In summary:

The voltage and current are in phase.
The current varies sinusoidally with time.
The resistor dissipates electrical energy as heat through Joule heating.

References: [1]leph107.pdf, page 2 [5] leph107.pdf, page 3

User
What happens when you apply AC voltage to a resistor?

💁‍♀️ Sources

1. leph107.pdf, page 2

Physics
7.2 AC VOLTAGE APPLIED TO A RESISTOR
Figure 7.1 shows a resistor connected to a source e of
ac voltage. The symbol for an ac source in a circuit
diagram is . We consider a source which produces
sinusoidally varying potential difference across its
terminals. Let this potential difference, also called ac
voltage, be given by

NICOLA TESLA (1856 – 1943)

(7.1)
v = vm sin ω t
where vm is the amplitude of the oscillating potential
difference and w is its angular frequency.
Nicola Tesla (1856 –
1943) Serbian-American
scientist, inventor and
genius. He conceived the
idea of the rotating
magnetic field, which is the
basis of practically all
alternating
current
machinery, and which
helped usher in the age of
electric power. He also
invented among other
things the induction motor,
the polyphase system of ac
power, and the high
frequency induction coil
(the Tesla coil) used in radio
and television sets and
other electronic equipment.
The SI unit of magnetic field
is named in his honour.

FIGURE 7.1 AC voltage applied to a resistor.

To find the value of current through the resistor, we
apply Kirchhoff’s loop rule ∑ ε (t ) = 0 (refer to Section
3.13), to the circuit shown in Fig. 7.1 to get

vm sin ω t = i R
vm
sin ω t
R
Since R is a constant, we can write this equation as

or i =

i = i m sin ω t
where the current amplitude im is given by

vm
(7.3)
R
Equation (7.3) is Ohm’s law, which for resistors, works equally
well for both ac and dc voltages. The voltage across a pure resistor
and the current through it, given by Eqs. (7.1) and (7.2) are
plotted as a function of time in Fig. 7.2. Note, in particular that
both v and i reach zero, minimum and maximum values at the
same time. Clearly, the voltage and current are in phase with
each other.
We see that, like the applied voltage, the current varies
sinusoidally and has corresponding positive and negative values
during each cycle. Thus, the sum of the instantaneous current
values over one complete cycle is zero, and the average current
is zero. The fact that the average current is zero, however, does
im =

FIGURE 7.2 In a pure
resistor, the voltage and
current are in phase. The
minima, zero and maxima
occur at the same
respective times.

178

(7.2)

Rationalised 2023-24

2. leph107.pdf, page 8

Physics
7.5

AC VOLTAGE APPLIED TO A CAPACITOR

Figure 7.7 shows an ac source e generating ac voltage v = vm sin wt
connected to a capacitor only, a purely capacitive ac circuit.
When a capacitor is connected to a voltage source
in a dc circuit, current will flow for the short time
required to charge the capacitor. As charge
accumulates on the capacitor plates, the voltage
across them increases, opposing the current. That is,
a capacitor in a dc circuit will limit or oppose the
current as it charges. When the capacitor is fully
charged, the current in the circuit falls to zero.
When the capacitor is connected to an ac source,
as in Fig. 7.7, it limits or regulates the current, but
FIGURE 7.7 An ac source
does not completely prevent the flow of charge. The
connected to a capacitor.
capacitor is alternately charged and discharged as
the current reverses each half cycle. Let q be the
charge on the capacitor at any time t. The instantaneous voltage v across
the capacitor is
q
(7.15)
C
From the Kirchhoff’s loop rule, the voltage across the source and the
capacitor are equal,
v=

vm sin ω t =

q
C

To find the current, we use the relation i =
i =

dq
dt

d
(vm C sin ω t ) = ω C vm cos(ω t )
dt

π
Using the relation, cos(ω t ) = sin  ω t +  , we have

2

π

i = im sin  ω t + 
(7.16)

2
where the amplitude of the oscillating current is im = wCvm. We can rewrite
it as
vm
(1/ ω C )
Comparing it to im= vm/R for a purely resistive circuit, we find that
(1/wC) plays the role of resistance. It is called capacitive reactance and
is denoted by Xc,
im =

Xc= 1/wC

(7.17)

so that the amplitude of the current is

184

im =

vm
XC

(7.18)

Rationalised 2023-24

3. leph107.pdf, page 23

Alternating Current
with the source frequency and the attractive force would average to
zero. Thus, the ac ampere must be defined in terms of some property
that is independent of the direction of the current. Joule heating
is such a property, and there is one ampere of rms value of
alternating current in a circuit if the current produces the same
average heating effect as one ampere of dc current would produce
under the same conditions.
5.

In an ac circuit, while adding voltages across different elements, one
should take care of their phases properly. For example, if V R and VC
are voltages across R and C, respectively in an RC circuit, then the
total voltage across RC combination is VRC = VR2 + VC2

and not

VR + VC since V C is p/2 out of phase of V R.
6.

Though in a phasor diagram, voltage and current are represented by
vectors, these quantities are not really vectors themselves. They are
scalar quantities. It so happens that the amplitudes and phases of
harmonically varying scalars combine mathematically in the same
way as do the projections of rotating vectors of corresponding
magnitudes and directions. The ‘rotating vectors’ that represent
harmonically varying scalar quantities are introduced only to provide
us with a simple way of adding these quantities using a rule that
we already know as the law of vector addition.

There are no power losses associated with pure capacitances and pure
inductances in an ac circuit. The only element that dissipates energy
in an ac circuit is the resistive element.

In a RLC circuit, resonance phenomenon occur when XL = X C or

ω0 =

1
. For resonance to occur, the presence of both L and C
LC

elements in the circuit is a must. With only one of these (L or C )
elements, there is no possibility of voltage cancellation and hence,
no resonance is possible.
9.

The power factor in a RLC circuit is a measure of how close the
circuit is to expending the maximum power.

10. In generators and motors, the roles of input and output are
reversed. In a motor, electric energy is the input and mechanical
energy is the output. In a generator, mechanical energy is the
input and electric energy is the output. Both devices simply
transfor m energy from one form to another.
11. A transformer (step-up) changes a low-voltage into a high-voltage.
This does not violate the law of conservation of energy. The
current is reduced by the same proportion.

199

Rationalised 2023-24

4. leph103.pdf, page 13

Current
Electricity
Using Ohm’s law V = IR, we get
P = I 2 R = V 2/R

(3.33)

as the power loss (“ohmic loss”) in a conductor of resistance R carrying a
current I. It is this power which heats up, for example, the coil of an
electric bulb to incandescence, radiating out heat and light.
Where does the power come from? As we have
reasoned before, we need an external source to keep
a steady current through the conductor. It is clearly
this source which must supply this power. In the
simple circuit shown with a cell (Fig.3.11), it is the
chemical energy of the cell which supplies this power
for as long as it can.
The expressions for power, Eqs. (3.32) and (3.33),
show the dependence of the power dissipated in a
resistor R on the current through it and the voltage
FIGURE 3.11 Heat is produced in the
across it.
resistor R which is connected across
Equation (3.33) has an important application to
the terminals of a cell. The energy
dissipated in the resistor R comes from
power transmission. Electrical power is transmitted
the chemical energy of the electrolyte.
from power stations to homes and factories, which
may be hundreds of miles away, via transmission
cables. One obviously wants to minimise the power
loss in the transmission cables connecting the power stations to homes
and factories. We shall see now how this can be achieved. Consider a
device R, to which a power P is to be delivered via transmission cables
having a resistance Rc to be dissipated by it finally. If V is the voltage
across R and I the current through it, then
P=VI
(3.34)
The connecting wires from the power station to the device has a finite
resistance Rc. The power dissipated in the connecting wires, which is
wasted is Pc with
Pc = I 2 Rc
P 2 Rc
(3.35)
V2
from Eq. (3.32). Thus, to drive a device of power P, the power wasted in the
connecting wires is inversely proportional to V 2. The transmission cables
from power stations are hundreds of miles long and their resistance Rc is
considerable. To reduce Pc, these wires carry current at enormous values
of V and this is the reason for the high voltage danger signs on transmission
lines — a common sight as we move away from populated areas. Using
electricity at such voltages is not safe and hence at the other end, a device
called a transformer lowers the voltage to a value suitable for use.
=

3.10 CELLS, EMF, INTERNAL RESISTANCE
We have already mentioned that a simple device to maintain a steady
current in an electric circuit is the electrolytic cell. Basically a cell has
two electrodes, called the positive (P) and the negative (N), as shown in

Rationalised 2023-24

5. leph107.pdf, page 3

Alternating Current
not mean that the average power consumed is zero and
that there is no dissipation of electrical energy. As you
know, Joule heating is given by i 2R and depends on i 2
(which is always positive whether i is positive or negative)
and not on i. Thus, there is Joule heating and
dissipation of electrical energy when an
ac current passes through a resistor.
The instantaneous power dissipated in the resistor is

p = i 2 R = i m2 R sin 2 ω t
The average value of p over a cycle is*

(7.4)

[7.5(b)]
p = i m2 R < sin2 ωt >
2
Using the trigonometric identity, sin w t =
1/2 (1– cos 2wt ), we have < sin2 wt > = (1/2) (1– < cos 2wt >)
and since < cos2wt > = 0**, we have,
< sin2 ω t > =

1
2

Thus,
1 2
im R
[7.5(c)]
2
To express ac power in the same form as dc power
(P = I2R), a special value of current is defined and used.
It is called, root mean square (rms) or effective current
(Fig. 7.3) and is denoted by Irms or I.

Geor ge
Westinghouse
(1846 – 1914) A leading
proponent of the use of
alternating current over
direct current. Thus,
he came into conflict
with Thomas Alva Edison,
an advocate of direct
current. Westinghouse
was convinced that the
technology of alternating
current was the key to
the electrical future.
He founded the famous
Company named after him
and enlisted the services
of Nicola Tesla and
other inventors in the
development of alternating
current
motors
and
apparatus
for
the
transmission of high
tension current, pioneering
in large scale lighting.

GEORGE WESTINGHOUSE (1846 – 1914)

[7.5(a)]
p = < i 2 R > = < i m2 R sin 2 ω t >
where the bar over a letter (here, p) denotes its average
value and <......> denotes taking average of the quantity
inside the bracket. Since, i 2m and R are constants,

FIGURE 7.3 The rms current I is related to the
peak current im by I = i m / 2 = 0.707 im.
T

* The average value of a function F (t ) over a period T is given by F (t ) =

1  sin 2ω t  T
1
=
[sin 2ω T − 0] = 0
2ω  0 2ω T

** < cos 2ω t > = T ∫ cos 2ω t dt = T 
0

Rationalised 2023-24

1
F (t ) dt
T ∫0

179

6. leph103.pdf, page 3

Current
Electricity
number of electrons travelling in any direction will be equal to the number
of electrons travelling in the opposite direction. So, there will be no net
electric current.
Let us now see what happens to such a
piece of conductor if an electric field is applied.
To focus our thoughts, imagine the conductor
in the shape of a cylinder of radius R (Fig. 3.1).
Suppose we now take two thin circular discs FIGURE 3.1 Charges +Q and –Q put at the ends
of a dielectric of the same radius and put
of a metallic cylinder. The electrons will drift
positive charge +Q distributed over one disc
because of the electric field created to
and similarly –Q at the other disc. We attach
neutralise the charges. The current thus
will stop after a while unless the charges +Q
the two discs on the two flat surfaces of the
and –Q are continuously replenished.
cylinder. An electric field will be created and
is directed from the positive towards the
negative charge. The electrons will be accelerated due to this field towards
+Q. They will thus move to neutralise the charges. The electrons, as long
as they are moving, will constitute an electric current. Hence in the
situation considered, there will be a current for a very short while and no
current thereafter.
We can also imagine a mechanism where the ends of the cylinder are
supplied with fresh charges to make up for any charges neutralised by
electrons moving inside the conductor. In that case, there will be a steady
electric field in the body of the conductor. This will result in a continuous
current rather than a current for a short period of time. Mechanisms,
which maintain a steady electric field are cells or batteries that we shall
study later in this chapter. In the next sections, we shall study the steady
current that results from a steady electric field in conductors.

3.4 OHM’S LAW
A basic law regarding flow of currents was discovered by G.S. Ohm in
1828, long before the physical mechanism responsible for flow of currents
was discovered. Imagine a conductor through which a current I is flowing
and let V be the potential difference between the ends of the conductor.
Then Ohm’s law states that
VµI
or, V = R I

(3.3)

where the constant of proportionality R is called the resistance of the
conductor. The SI units of resistance is ohm, and is denoted by the symbol
W. The resistance R not only depends on the material of the conductor
but also on the dimensions of the conductor. The dependence of R on the
dimensions of the conductor can easily be determined as follows.
Consider a conductor satisfying Eq. (3.3) to be in the form of a slab of
length l and cross sectional area A [Fig. 3.2(a)]. Imagine placing two such
identical slabs side by side [Fig. 3.2(b)], so that the length of the
combination is 2l. The current flowing through the combination is the
same as that flowing through either of the slabs. If V is the potential
difference across the ends of the first slab, then V is also the potential
difference across the ends of the second slab since the second slab is

Rationalised 2023-24

FIGURE 3.2
Illustrating the
relation R = rl/A for
a rectangular slab
of length l and area
of cross-section A.

7. leph107.pdf, page 1

Chapter Seven

ALTERNATING
CURRENT

7.1 INTRODUCTION
We have so far considered direct current (dc) sources and circuits with dc
sources. These currents do not change direction with time. But voltages
and currents that vary with time are very common. The electric mains
supply in our homes and offices is a voltage that varies like a sine function
with time. Such a voltage is called alternating voltage (ac voltage) and
the current driven by it in a circuit is called the alternating current (ac
current)*. Today, most of the electrical devices we use require ac voltage.
This is mainly because most of the electrical energy sold by power
companies is transmitted and distributed as alternating current. The main
reason for preferring use of ac voltage over dc voltage is that ac voltages
can be easily and efficiently converted from one voltage to the other by
means of transformers. Further, electrical energy can also be transmitted
economically over long distances. AC circuits exhibit characteristics which
are exploited in many devices of daily use. For example, whenever we
tune our radio to a favourite station, we are taking advantage of a special
property of ac circuits – one of many that you will study in this chapter.
* The phrases ac voltage and ac current are contradictory and redundant,
respectively, since they mean, literally, alternating current voltage and alternating
current current. Still, the abbreviation ac to designate an electrical quantity
displaying simple harmonic time dependance has become so universally accepted
that we follow others in its use. Further, voltage – another phrase commonly
used means potential difference between two points.

Rationalised 2023-24

8. leph107.pdf, page 24

Physics
EXERCISES
7.1

7.2

7.3
7.4
7.5
7.6
7.7

7.8

A 100 W resistor is connected to a 220 V, 50 Hz ac supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle?
(a) The peak voltage of an ac supply is 300 V. What is the rms voltage?
(b) The rms value of current in an ac circuit is 10 A. What is the
peak current?
A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine
the rms value of the current in the circuit.
A 60 mF capacitor is connected to a 110 V, 60 Hz ac supply. Determine
the rms value of the current in the circuit.
In Exercises 7.3 and 7.4, what is the net power absorbed by each
circuit over a complete cycle. Explain your answer.
A charged 30 mF capacitor is connected to a 27 mH inductor. What is
the angular frequency of free oscillations of the circuit?
A series LCR circuit with R = 20 W, L = 1.5 H and C = 35 mF is connected
to a variable-frequency 200 V ac supply. When the frequency of the
supply equals the natural frequency of the circuit, what is the average
power transferred to the circuit in one complete cycle?
Figure 7.17 shows a series LCR circuit connected to a variable
frequency 230 V source. L = 5.0 H, C = 80mF, R = 40 W.

FIGURE 7.17

(a) Determine the source frequency which drives the circuit in
resonance.
(b) Obtain the impedance of the circuit and the amplitude of current
at the resonating frequency.
(c) Determine the rms potential drops across the three elements of
the circuit. Show that the potential drop across the LC
combination is zero at the resonating frequency.

200

Rationalised 2023-24

9. leph107.pdf, page 4

Physics
It is defined by
I = i2 =

i
1 2
im = m
2
2

= 0.707 im

(7.6)

In terms of I, the average power, denoted by P is
1 2
im R = I 2 R
2
Similarly, we define the rms voltage or effective voltage by

P = p=

= 0.707 vm
2
From Eq. (7.3), we have
v m = i mR

(7.7)

(7.8)

R
2
2
or, V = IR
(7.9)
Equation (7.9) gives the relation between ac current and ac voltage
and is similar to that in the dc case. This shows the advantage of
introducing the concept of rms values. In terms of rms values, the equation
for power [Eq. (7.7)] and relation between current and voltage in ac circuits
are essentially the same as those for the dc case.
It is customary to measure and specify rms values for ac quantities. For
example, the household line voltage of 220 V is an rms value with a peak
voltage of
or,

vm = 2 V = (1.414)(220 V) = 311 V
In fact, the I or rms current is the equivalent dc current that would
produce the same average power loss as the alternating current. Equation
(7.7) can also be written as
P = V2 / R = I V (since V = I R )
Example 7.1 A light bulb is rated at 100W for a 220 V supply. Find
(a) the resistance of the bulb; (b) the peak voltage of the source; and
(c) the rms current through the bulb.
Solution
(a) We are given P = 100 W and V = 220 V. The resistance of the
bulb is
V 2 (220 V )
=
= 484 Ω
P
100 W
2

180

EXAMPLE 7.1

(b) The peak voltage of the source is

v m = 2V = 311 V
(c) Since, P = I V
I

P
V

100 W
220 V

0.454A

Rationalised 2023-24

10. leph107.pdf, page 14

Physics
Example 7.6 A resistor of 200 W and a capacitor of 15.0 mF are
connected in series to a 220 V, 50 Hz ac source. (a) Calculate the
current in the circuit; (b) Calculate the voltage (rms) across the
resistor and the capacitor. Is the algebraic sum of these voltages
more than the source voltage? If yes, resolve the paradox.
Solution
Given

R = 200 Ω, C = 15.0 µF = 15.0 × 10−6 F
V = 220 V, ν = 50 Hz
(a)
In order to calculate the current, we need the impedance of
the circuit. It is

Z = R 2 + X C2 = R 2 + (2π ν C )−2
= (200 Ω )2 + (2 × 3.14 × 50 × 15.0 × 10−6 F)−2
= (200 Ω)2 + (212.3 Ω)2
= 291.67 Ω

Therefore, the current in the circuit is
I =

(b)

V
220 V
=
= 0.755 A
Z 291.5 Ω

Since the current is the same throughout the circuit, we have
V R = I R = (0.755 A)(200 Ω) = 151 V
VC = I X C = (0.755 A)(212.3 Ω ) = 160.3 V

EXAMPLE 7.6

The algebraic sum of the two voltages, VR and VC is 311.3 V which
is more than the source voltage of 220 V. How to resolve this
paradox? As you have learnt in the text, the two voltages are not
in the same phase. Therefore, they cannot be added like ordinary
numbers. The two voltages are out of phase by ninety degrees.
Therefore, the total of these voltages must be obtained using the
Pythagorean theorem:
V R +C = VR2 + VC2

= 220 V
Thus, if the phase difference between two voltages is properly taken
into account, the total voltage across the resistor and the capacitor
is equal to the voltage of the source.

7.7 POWER IN AC CIRCUIT: THE POWER FACTOR
We have seen that a voltage v = vm sinwt applied to a series RLC circuit
drives a current in the circuit given by i = im sin(wt + f) where
im =

190

vm
Z

 X − XL 
and φ = tan −1  C


R

Therefore, the instantaneous power p supplied by the source is

Rationalised 2023-24

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`final_search_query`

`references`

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]

`raw_output_text[0]`

The AC voltage applied to the resistor can be expressed as:
v = vm sin(ωt)
where vm is the amplitude of the oscillating potential difference and ω is the angular frequency.

Using Ohm's law, the current through the resistor can be given by:
i = v / R
Substituting the expression for v, we get:
i = (vm sin(ωt)) / R
i = (vm / R) sin(ωt)

The amplitude of the current (im) is therefore:
im = vm / R
So, the current in the circuit can be written as:
i = im sin(ωt).

In summary:
- The voltage and current are in phase.
- The current varies sinusoidally with time.
- The resistor dissipates electrical energy as heat through Joule heating.

References:leph107.pdf, page 2 leph107.pdf, page 3

`output_text[0]`

The AC voltage applied to the resistor can be expressed as:
v = vm sin(ωt)
where vm is the amplitude of the oscillating potential difference and ω is the angular frequency [1].

Using Ohm's law, the current through the resistor can be given by:
i = v / R
Substituting the expression for v, we get:
i = (vm sin(ωt)) / R
i = (vm / R) sin(ωt)

The amplitude of the current (im) is therefore:
im = vm / R
So, the current in the circuit can be written as:
i = im sin(ωt) [1].

In summary:
- The voltage and current are in phase.
- The current varies sinusoidally with time.
- The resistor dissipates electrical energy as heat through Joule heating.

References: [1]leph107.pdf, page 2 [5] leph107.pdf, page 3

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