Current and Electricity

“I am somewhat exhausted; I wonder how a battery feels when it

pours electricity into a non-conductor?”
― Arthur Conan Doyle, The Adventure of the Dying Detective
 Electrical Forces : Origin and Form

• Eelctrical Forces are experienced by all charged particles.

• From lightning in the sky to the glowing of a lamp, from what holds atoms
together as molecules to the nerve impulses that travel along your nervous
system, electricity is all around us.
• It’s universal like the Forces of Gravitation.
• Charges or Charged Particles exert forces on each other similar to the masses
in case of Gravitation.

Charges could be of two types : Positive or Negative. Experiments in

electrostatics show that -
Like charges repel and opposite charges attract.
The Charges : Facts and Trivia
Electric Current : The flow of charges.

Circuit : a closed system of

wires or pipes through
which electricity or liquid
can flow.

Analogues :

Water : Charge
Valve : Switch
Constrictions : Resistance
Pump : Cell (Battery)
 Current : Rate of Charge Flow

 As a physical quantity, current is the rate at which charge flows past a point on a
circuit. As depicted in the diagram below, the current in a circuit can be determined if
the quantity of charge Q passing through a cross section of a wire in a time t can be
measured. The current is simply the ratio of the quantity of charge and time.

I=
The rate of electrical flow is measured in Amperes (S.I.) (named After the French
Scientist Andre Marie Ampere). 1 Ampere is a rate of flow equal to 1 Coulomb of
charge per second.
1A=
Current : Rate of Charge Flow

Ben Franklin, envisioned positive charges as

the carriers of charge. As such, an early
convention for the direction of an electric
current was established to be in the direction
that positive charges would move. The
convention has stuck and is still used today.
 Potential and Potential Difference : The Cause for
the flow of charges
• Charges flow only when they are pushed or driven. A sustained current
requires a suitable pumping device to provide a difference in electric
potential - voltage. An electrical pump is some sort of voltage source.
• Batteries and electric generators do work to pull negative charges away
from positive ones.
• The work done by whatever means in separating the opposite charges is
available at the terminals of the battery or generator. These different
values of energy per charge create a difference in potential (voltage).
• This voltage provides the electrical pressure to move electrons through
a circuit joined to these terminals.
Quantifying the Potential Difference (Voltage)

 
The work done in moving a unit
positive charge between two points in a
circuit is called the potential difference
or voltage between the two points.
V=
Unit (S.I.): Joules/Coulomb = Volts (V)
(named after Italian Scientist Alessandro Volta)
1 V=
Potential Difference and Electric Potential

The Potential Difference across

the bulb is given by VP-VQ.
Current will pass from left to right
in the figure.

The Potential at P is given by VP

and the Potential at Q is given by
VQ.

What is meant by the Potential ?

Electric Potential

An electric (electrostatic) potential is the amount of

work needed to move a unit positive charge from a
reference point to a specific point inside the field without
producing any acceleration.

Typically, the reference point is a point at Infinity.

The Potential at Infinity is 0. (Something to think about)
The Reference Point

 - : Work Done to bring a unit positive charge from

infinity to P. and - : Work done to bring a unit positive
charge from infinity to Q. Hence Potential difference
between P and Q
( - )-( - ) = -
Potential Difference =Voltage across the entire circuit= Electromotive force

In a battery, a series of chemical reactions occur in

which electrons are transferred from one terminal to
another. There is a potential difference (voltage)
between these poles.

The maximum potential difference a power source

can have is called the electromotive force or
(EMF), e. The term isn't actually a force, simply the
amount of energy per charge (J/C or V).
 Instruments to Measure Current and Voltage

Series! Parallel! High or Low Resistance!!!

Current is measured by an instrument

called ammeter. Voltage is measured by an instrument called
An ammeter should have very low voltmeter.
resistance. An Voltmeter should have very high resistance.
It is always connected in series with the It is always connected in parallel with the
electrical circuit. electrical circuit.
Components of an Electrical Circuit
A Complete Circuit
 Some Quick Numerical Problems

1. 1 microampere = How many electrons.

(a) 6.25 x 109 electrons
(b) 6.25 x 106 electrons
(c) 6.25 x 1012 electrons
(d) 6.25 electrons

2. Calculate the amount of charge that would flow in 1 hour through the elements of
an electric bulb drawing a current of 0.4 A.

3. A current of 4 A exists in a 10 resistor for 4 minutes. Find the charge and the
number of electrons that pass through any cross-section of the resistor in this time.
 Solutions to the problems
 1.Q=±ne (n = number of electrons, e = charge of 1 electron = 1.6 x 10-19 C)
t can be taken as 1 s. I = . I = 1 x 10-6 A (microampere). ∴ Q = 1 x 10-6 C
∴ n = = = 6.25 x 1012.

2. Current, I=0.4 A, Time t=1hr=1x60x60=3600 s. Charge, Q= I x t

Hence Q = 0.4x3600 coulomb = 1440 coulombs.

3. We know the relation, Q=I x t, Given t= 4 minutes = 4 x 60 = 240 seconds.

Total charge, Q=4 x 240 = 960 C. Also, number of electrons = n = .
∴ n = = = 6 x 1021.
Ohm’s Law : Voltage , Current and something
called Resistance
 The relationship among voltage, current, and resistance is summarized by a statement
called Ohm's law.
German Scientist Georg Simon Ohm did a series of experiment to find out that

The potential difference (voltage) across an ideal conductor is proportional to the

current through it.  The constant of proportionality is called the "resistance", R.
Ohm's Law is given by:
 More about Resistance

 Resistanceis a measure of the opposition to current flow in an electrical circuit. Resistance

is measured in Ohms, symbolized by the Greek letter Omega (Ω).
Ohm's law tells us that a
potential difference of 1 volt
established across a circuit that
has a resistance of 1 ohm will
The proportionality between Voltage and Current for a given conductor
produce a current means that
of 1 ampere.
we'll get twice the current for twice the voltage. The greater the voltage, the greater the
current. But, if the resistance of a circuit is doubled, the current will be half what it
would be otherwise. The greater the resistance, the smaller the current. Ohm's law
makes good sense.
The Ohm’s law triangle
A generic circuit verifying Ohm’s Law
 Graphical Representation of Ohm’s Law

 
Let us compare the two equations :

We can compare the two and say I is y and

V is x where is the slope of the straight
line.
Finding the Resistance Graphically
 Factors affecting the Resistance of a conductor

 Resistanceis proportional to length. This means more the length of the conductor,
more the resistance.

Resistance is inversely proportional to the cross sectional area of the conductor. This
means more the cross sectional area, lesser is the resistance.

Combining the two we can write

(ρ (rho) = electrical resistivity of the material. S.I. unit is Ohm-m (-m)

Graphical Representation of and A dependence of
 
R
Categorization of Materials based on Resistance

Conductors : They have free electrons and they allow the passage of electricity
through their body with minimum resistance.

Insulators : They do not have free electrons and they are incapable of passing
electricity through their body. They offer maximum resistance to the flow of current.

Semi-conductors : Their response to the flow of current is somewhat in between the

conductors and insulators. With temperature their resistance to the flow of the current
changes and they become closer to being conductors.

Home-work : Find examples of each and at least three applications for each of
the categories.
 Temperature also affects the Resistance in different
materials
Resistance increases with the temperature of the wire.

In ideal conductors (metals) the hotter wire has a larger resistance because of increased
vibration of the atoms. When a material gets hotter the atoms in their arrangement vibrate
more. This makes it difficult for the electrons to move without interaction with an atom and
increases resistance.

In semi-conductors the case is opposite as with temperature the resistance of the material
decreases.
 Sample problem 1

Question: A torch bulb has a resistance of 1 Ω when cold. It draws a current of 0.2 A

from a source of 2 V and glows. Calculate :
(i) the resistance of the bulb when glowing and
(ii) explain the reason for the difference in resistance.

Answer:
(i) When the bulb glows:
V = I R (Ohm's law)
R = V/I = 2/0.2 =10 Ω
(ii) Resistance of the filament of the bulb increases with increase in temperature.
Hence when it glows its resistances is greater than when it is cold.
 Sample problem 2

 Question: Calculate the resistance of 1 km long copper wire of radius 1 mm. (Resistivity of

copper = 1.72 x 1 0-8 -m)
 Answer:
L = 1 km = 1000 m
R = 1 mm = 1 x 10-3 m
ρ = 1.72 x 10-8   -m

Area of cross section = πr2  = 3.14 x 10-3 x 10-3 =  3.14 x 10-6

= (1.72 x 10-8)  = 5.5 

 Sample Problem 3
Question: When a potential difference of 2 V is applied across the ends of a wire of 5 m
length, a current of 1 A is found to flow through it. Calculate: 
(i) the resistance per unit length of the wire.  
(ii) the resistance of 2 m length of this wire.
(iii) the resistance across the ends of the wire if it is doubled on itself. (Visualise!)
 Answer: (i) V = IR (Ohm's law) R= = = 2 Ω
Resistance per unit length: Ω/m = 0.4 Ω/m
(ii) Resistance of 2 m length of the wire = 0.4 Ω/m x 2 m = 0.8 Ω
Sample Problem 3 continued

 (iii)When the wire is doubled on itself:

(a) the area of cross-section is doubled. If A is the original cross sectional area, now
it is A1 = 2A.
(b) The length becomes half i.e. 1 =  
Resistance R1 = =
= =
(∵ R = where R = 2 Ω)
∴ R1 = x 2 = 0.5 Ω.
Resistance for a system (combination) of Resistors

What are the differences that we can find or predict to happen in these two circuits ?
The analogy of a water tank and it’s power to
rotate a blender (schematic)

Series arrangement for the flow of water

Parallel arrangement for the flow of water

Series and Parallel configuration in electrical
circuits

Series arrangement for the electrical circuit

Parallel arrangement for the electrical circuit

A Series of Resistors: Important Features

1. Electric current has but a single

pathway through the circuit. This
means that the current passing through
the resistance of each electrical device
along the pathway is the same.

2. This current is resisted by the

resistance of the first device, the
resistance of the second, and that of
the third also, so the total resistance to
current in the circuit is the sum of the
individual resistances along the circuit
path.
 A Series of Resistors: Important Features
3. The current in the circuit is numerically equal to the voltage supplied by the source
divided by the total resistance of the circuit. This is in accordance with the Ohm's law.

4. The total voltage impressed across a series circuit divides among the individual electrical
devices in the circuit so that the sum of the "voltage drops" across the resistance of each
individual device is equal to the total voltage supplied by the source. This characteristic
follows from the fact that the amount of energy supplied to the total current is equal to the
sum of energies supplied to each device.

5. The voltage drop across each device is proportional to its resistance-Ohm's law applies
separately to each device. This follows from the fact that more energy is dissipated when a
current passes through a large resistance than when the same current passes through a small
resistance.
 To find the equivalent resistance for a series
configuration

   eq. I
   eq. II

𝑉 1=𝐼𝑅 1 𝑉 2=𝐼𝑅 2 𝑉 3=𝐼𝑅 3 

      eq. III

𝑉 =𝐼 (𝑅 1+𝑅 2+𝑅 3) eq. IV

 

  = Equivalent resistance for the series configuration

Resistors in Parallel: Important Features

1. Each device connects the same two

points A and B of the circuit. The
voltage is therefore the same across
each device.

2. The total current in the circuit

divides among the parallel branches.
Since the voltage across each branch
is the same, the amount of current in
each branch is inversely proportional
to the resistance of the branch-Ohm's
law applies separately to each branch.
Resistors in Parallel: Important Features

3. The total current in the circuit equals the sum of the currents in its parallel
branches. This sum equals the current in the battery or other voltage source.

4. As the number of parallel branches is increased, the overall resistance of the circuit
is decreased. Overall resistance is lowered with each added path between any two
points of the circuit. This means the overall resistance of the circuit is less than the
resistance of anyone of the branches.
To find the equivalent resistance for a parallel
configuration

   eq. I
 eq. II

 eq. III

= + + = + + )  eq. IV

+ +

  = Equivalent resistance for the parallel configuration

 One Example:

Determine the following quantities

for each of the two circuits shown
below…
1. the equivalent resistance
2. the total current from the
power supply
3. the current through each resistor
4. the voltage drop across each
resistor
 Solution to the example for (a) : Series
Combination
 1. Resistances in series add up.

(20 Ω = 100 Ω
2. Total current is determined by the voltage of the power supply and the equivalent
resistance of the circuit.
I= = =
3. Current is constant through resistors in series. = 1.25 A
4. The voltage drops can be found using Ohm's law.
= 1.25 A x 20 Ω = 25 V
= 1.25 A x 30 Ω = 37.5 V = (25 + 37.5 + 62.5) V = 125 V
= 1.25 A x 50 Ω = 62.5 V
 Solution to the example for (a) : Parallel
Combination
 1.Resistances in parallel is given by the sum-of-inverse rule.
+ + + + = ∴ Rp = = 12.5
2. Total current is determined by the voltage of the power supply and the equivalent
resistance of the circuit. I = = =
3. Current for each resistor will be found by using the Ohm’s Law.
I1 = = = 6.25 A

I2 = = = 1.25 A = (6.25 + 1.25 + 2.50) A = 10 A

I3 = = = 2.50 A
4. T = 125 V
Heating Effect of Current
Energy is supplied to the circuit from the battery.
The electrical appliance, plugged in the circuit, uses up the energy.
But certain amount of energy is wasted: In the form of Heat Energy.
Why
The current in the conductor is due to the drifting of electrons inside a conductor in a
direction opposite to the flow of electrons.

During their drifting they collide with their atoms vibrating about their mean position
and lose some of kinetic energy to the vibrating atoms which increases the amplitude
of the vibration of the atoms.

This increases the thermal energy of the metal with the corresponding rise in
temperature of the conductor.
 The mathematical expression for heat dissipated in
the circuit
 Let
the Potential Difference in a circuit be V and the current flowing through the resistance
R be I. Therefore in time t seconds charge Q is moved in the circuit such that .

The work done to move the charge Q through the potential difference V is given by W =
QV.

Power supplied to the circuit is given by the energy supplied (W Joules) in time t seconds.
∴ P = = = V I (∵ )

Hence energy supplied to the circuit = W = P t = V I t = VIt.

This energy is spent or dissipated as the heat energy (H).
∴ H = VIt = (IR)It = I2Rt. (∵ V=IR).
 Joules law of heating
  H = VIt = = I2Rt. (∵ V=IR).
This is known as Joules law of Heating.
(i) The amount of heat produced in a current conducting wire, is proportional to the square of
the amount of current that is flowing through the wire, when the electrical resistance of the
wire and the time of current flowing are constant.
I2 (when R and t are constants)
(ii) The amount of heat produced is proportional to the electrical resistance of the wire when
the current in the wire and the time of current flowing are constant.
R (when I and t are constants)
(iii) The heat generated due to the flow of current is proportional to the time of current
flowing, when the electrical resistance and the amount of current is constant.
t (when R and I are constants)
Some electrical appliances using the Joule Heating
 Power: Definition and a small recap

 The work done to move the charge Q through the potential difference V is given by W = QV.

Power supplied to the circuit is given by the energy supplied (W Joules) in time t seconds.
∴ P = = = V I (∵ ) = V x = ( ∵ )
The SI unit of Power is Watt. 1 Watt = 1 Volt x 1 Ampere.
In practice we use kW for Power (1 kW = 1000 Watt)
Energy = Power x Time
∴ kWh is the unit of electrical energy consumed.
1kWh = 1000 Watt x 3600 seconds
= 3.6 x 106 watt-second
= 3.6 x 106 Joules.
 Some numerical problems on Joules Law of
Heating and Electric Power

Q1: How much heat is produced by 2 kW electric heater when it is operated for 30 minutes.

Solution:
P = VI and W = VIt = P x t.

Rating of the electric heater is the power rating for the heater.
∴ P = 2kW = 2000 Watt.
Time of run = 30 minutes = 1800 seconds.

∴ W = 2000 x 1800 J
= 3600000 J = 3.6 x 106 J.
 Some numerical problems on Joules Law of
Heating and Electric Power
Q2: An electric iron is marked 250 V, 500 W. What current does it take if connected to the
correct voltage? What is hot resistance? If the iron is used for one hour daily for 30 days in
a month, what will be the monthly bill at Rs 4.50 per unit? How does the cost get affected if
the voltage is only 200 V instead of 250 V?   In a month total working hours = 1 x 30 = 30 hours.
  Wattage of the iron = 500 Watt = 0.5 kW.
Solution:
Wattage rating of an iron, ∴ Total Power consumed = 30 x 0.5 = 15 kWh
P = 500 watts, Voltage rating, V = 250 V ∴ Total Cost for the month = Rs. 4.5 x 15 = Rs. 67.5
Current drawn when connected to rated
voltage of 250 volts, If the Voltage is 200 V, then, Power drawn is given by
I = = = 2 Amp. P = VI = V x = = = 320 Watt = 0.320 kW.
Hot Resistance = R = = = 125 Ω. ∴ Total Power consumed = 30 x 0.320 = 9.6 kWh
∴ Total Cost for the month = Rs. 4.5 x 9.6 = Rs. 43.2