the presentation explain about electric current , conventional current its direction ,ohms law and series and parallel combination of resistances

1. Electric current and ohms law
Presented By
Vasudev Shrivastava
P.G.T.(Physics)
J.N.V.Nowgong District
Chhatarpur (M.P.)

3. Electric Current
 Electric current is the continuous flow of electric
charge
 Two types of current are direct and alternating
 Direct current (DC) is when the charge flows in one
direction
 Alternating current (AC) is when the flow of electric
charge regularly reverses direction

4. Electric Current• The rate flow of electric charge is known as
electric current The S.I . Unit of electric
current is ampere .Generally electric
current denoted by I
• An example of a direct current is a flashlight
and most battery-operated devices
• Alternating current is in your home and
school
• Current is defined as the direction in which
the positive charges would flow

5. Conductors, semiconductor and
Insulators• Metals such as copper, and silver are
good electrical conductors An
electrical conductor is a material
through which charge can flow easily
• An electrical insulator is a material
through which charge cannot flow easily
Wood, plastic, rubber and air are good
electrical insulators
• Those material whose conductivity lies in
between conductors and insulators are
known as semiconductor such as
germanium ,silicon which are commonly
use for fabricating diode ,transistor IC

6. Super conductor
 A superconductor is a material that has almost zero
resistance when it is cooled to low temperatures
 Superconductors are trying to reduce the resistance to
zero
 The best superconductor has been cooled to 138K

7. Electric circuit
 The regular
arrangement to flow
electric current is
known as electric
circuit.
 A simple electric
circuit consist a power
source , a switch and
a electric devices on
which electric current
is flowing.

8. While the switch is open:
 Free electrons (conducting electrons) are always moving
in random motion.
 The random speeds are at an order of
106 m/s.
 There is no net movement of charge across a cross
section of a wire.

9. http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/imgele/micohm.gif
What occurs in a wire when the
circuit switch is closed?

10. What occurs in a wire when the
circuit switch is closed?
https://youtu.be/T2M_DC_fEtU
 An electric field is established instantaneously (at
almost the speed of light, 3x108 m/s).
 Free electrons, while still randomly moving,
immediately begin drifting due to the electric field,
resulting in a net flow of charge.
 Average drift velocity is about 0.01cm/s.

11. Closing the switch establishes a potential difference
(voltage) and an electric field in the circuit.
 Electrons
flow in a net
direction
away from
the (-)
terminal.
 The electric
current flow
from higher
potential to
lower
potential
High
Potential
Low
Potential

12. Conventional Current
 By tradition,
direction in
which “positive
charges” would
flow.
 Direction is
opposite of
electron flow.

13. Conventional current

14. Resistance Resistance is the opposition to the flow of charges in a
material or The property of matter which control electric
current .
 The SI unit of resistance is the ohm
 Resistance of a conductor is directly propositional to the
length of conductor
 Resistance of a conductor is inversely propositional to the
area of cross section of conductor .
 The resistance of a conductor is also depends nature of
substance . Two different material have same dimension
but they have different resistances
 As temperature increases the resistance of conductors
increases since the electrons collide more often to each
others .

15. Voltage
 In order for a charge to flow in a conducting wire, the
wire must be connected in a complete loop that
includes a source of electrical energy
 A flashlight will not work if there is no battery

16. Potential Difference
 Reminder potential energy is related to position
 Charges flow from a high to low potential energy
 Potential difference is the difference in electrical
potential energy between two places in an electric field
 Potential difference is measured in joules per coulomb
or volts .Potential difference is also called voltage
 Potential energy = potential difference x charge
 W= qV so that V= W/q unit of p.d. Joule/coulomb

17. Remember: Electric Potential Energy-
Two Unlike Charges
Higher Potential
Energy
Lower Potential
Energy
+
-
•To cause movement of a charge,
there must be a potential difference.

18. Voltage Sources Three common voltage sources are
batteries, solar cells and generators
 A battery is a device that converts
chemical energy to electrical energy
 In a 9-volt battery the voltage drop is
about 9 volts

19. Ohm’s Law
 Named after German scientist Georg Ohm
 If physical state of a metallic conductor remains
constant ,the current flow within conductor is directly
proportional to potential difference between ends of
conductor.
 Ohm’s Law V = I x R or I = V/R
 Voltage (V) = current(I) x resistance (R)
 When the current is in amperes and the resistance is in
ohms the voltage is in volts

20. Example
 What is the voltage if the resistance is 3 ohms and the
current is 3 amps?
 V = I x R =
3 amps x 3 ohms = 9 volts
 Increasing the voltage increases the current
 Having the same voltage and increasing the resistance
will decrease the current
 Multimeters measure current, voltage and resistance

21. Resistivity
 Resistance of metallic
conductor is directly
proportional to length of
conductor.
 Resistance of metallic
conductor is inversely
proportional to area of
cross section conductor.
 Resistivity is a
characteristics property of
matter it depends upon
nature of subsistence and
temperature

22. Resistivity of different
materials

23.  Relation ship between
electric current I
potential difference V
and resistance R and
power R

24. Verification of ohms law
https://youtu.be/_g3FK8726lg

25. Failures of ohms law

26. Combination of resistance Series combination of
resistances  Parallel combination of
resistance

27. Formulas for series /parallel
Combination of resistance

28. Series and parallel combination of
resistances

29. Numerical problems
Q 01 A current of 0.5 A is drawn by a filament of an electric
bulb for 10minutes. Find the amount of electric charge that
flows through the circuit.
Solution
We are given, I = 0.5 A; t = 10 min = 600. ,we know that
Q=It= 0.5 A × 600 s= 300 C
Q 02 How much work is a done in moving a charge of
2 C across two points having potential difference 12 V?
Solution The amount of charge Q, that flows between two
points at potential difference V (= 12 V) is 2 C. Thus, the
amount of work W, done in moving the charge
W= VQ=12 V × 2 C= 24 J.

30. Q.03 (a) How much current will an electric bulb draw from a 220 V
source, if the resistance of the bulb filament is 1200 Ω? (b) How much
current will an electric heater coil draw from a 220 V source, if the
resistance of the heater coil is 100 Ω?
Solution(a) We are given V = 220 V; R = 1200Ω as per ohms law V= IR then
current I = 220 V/1200 Ω = 0.18 A.(b) We are given, V = 220 V, R = 100 Ω.
Again the current in heater I = 220 V/100 Ω = 2.2 A
Q.04 An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω
resistance are connected to a 6 V battery (Fig. 12.9). Calculate (a) the total
resistance of the circuit, (b) the current through the circuit, and(c) the
potential difference across the electric lamp and conductor.

31. Solution The resistance of electric lamp, R1= 20 Ω, The resistance of the
conductor connected in series, R2 = 4 Ω. Then the total resistance in the circuit
R=R1 + R2Rs=20 Ω + 4 Ω = 24 Ω. The total potential difference across the two
terminals of the battery V = 6 V. Now by Ohm’s law, the current through the circuit
is given by I=V/Rs=6 V/24 Ω= 0.25 A.

32. Applying Ohm’s law to the electric lamp and conductor separately, we get potential
difference across the electric lamp,V1= 20 Ω× 0.25 A=5 V; and, that across the conductor,
V2= 4 Ω× 0.25 A= 1 V. Suppose that we like to replace the series combination of electric
lamp and conductor by a single and equivalent resistor. Its resistance must be such that a
potential difference of 6 V across the battery terminals will cause a current of 0.25 A in
the circuit. The resistance R of this equivalent resistor would be R=V/I= 6 V/ 0.25 A= 24 Ω.
This is the total resistance of the series circuit; it is equal to the sum of the two
resistances.
Q.05 If in Fig. 12.12, R1 = 10 Ω, R2 = 40 Ω, R3 = 30 Ω, R4 = 20 Ω, R5 = 60
Ω, and a 12 V battery is connected to the arrangement. Calculate(a) the
total resistance in the circuit, and (b) the total current flowing in the circuit
Solution Suppose we replace the parallel resistors R1
and R2by an equivalent resistor of resistance, R′.
Similarly we replace the parallel resistors R3, R4 and
R5 by an equivalent single resistor of resistance R′′.
Then using Eq. (12.18), we have1/ R′= 1/10 + 1/40=
5/40; that is R′ = 8 Ω.Similarly,1/ R′′= 1/30 + 1/20 +
1/60 = 6/60;that is, R′′ = 10 Ω .Thus, the total
resistance, R= R′ + R′′ = 18 Ω. To calculate the
current, we use Ohm’s law, and get I = V/R = 12 V/18
Ω = 0.67 A.

33. Summary

34. Home Assignment : to solve all
problems of chapter electricity
from ncert text book.