This document provides an overview of key concepts in electric current and circuits, including: - Current is defined as the rate of flow of electric charge. It is measured in amperes. - Ohm's law states that the current through a conductor is directly proportional to the voltage applied. - Resistance depends on the material's resistivity, length, and cross-sectional area. It increases with temperature for most materials. - Electrical power is equal to current times voltage or voltage squared over resistance. It is measured in watts. - Components like bulbs connected in series or parallel affect how voltage and current are distributed in a circuit.

1. REVIEW LECTURE-7
UNIT-7 (Current
Electricity)
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2.  The rate of flow of charge in a circuit is defined as current.
𝐼 =
𝑄
𝑡
=
𝑛𝑒
𝑡
𝑜𝑟 𝑄 = 𝐼𝑡
 Electric current is a scalar quantity.
 S.I unit of current is ampere
 1 ampere = 6.25 x 1018 electrons / second.
 In metals electric conduction takes place due to flow of free
electrons only. But in gases and electrolytes, electric conduction
takes place due to flow of both positive and negative ions.
ELECTRIC CURRENT

3. Drift velocity (vd)
 “It is the average velocity attained by free electrons on
applying external electric field.”
 In conductors, 𝑣𝑑 = 10−3
𝑚𝑠−1
 𝑣𝑑 =
𝐼
nAe
(where n is number charges per unit volume)

4.  “If the physical conditions of the conductor (temperature
etc.) remain constant, then current flowing through a
conductor is directly proportional to potential difference
across its ends.
V = I R
Ohmic substances:
 “The substances which obeys Ohm’s law is called ohmic
substances.
 Metals are ohmic substances.”
OHM’S LAW

5. Non-ohmic substances:
 “Those substances which do not obey Ohm’s law are called
non-ohmic substances.”
 Tungsten filament, semi-conductor diode are non-ohmic
substances.

6. Slope of I-V graph is equal to conductance.
(If V along x-axis and I along y-axis)
 Slope of I-V graph is equal to resistance.
(If I along x-axis and V along y-axis)

7.  𝑅 =
𝑉
𝐼
 𝑅 = 𝜌
𝐿
𝐴
 Resistance of conductor depends on following factors:
(i)R  L (ii) R  1/A
(iii) R  T (except semi-conductors)
(iv) Nature of material
 If length of wire becomes double by stretching of wire
then area becomes half (keeping volume constant) and
resistance of wire becomes four times. 𝓁′
= 𝑛𝓁 ⇒ 𝑅′
=
𝑛2
𝑅 𝑛 > 1
 Conductance is reciprocal of resistance.
 Resistivity is a resistance of 1 m cube of a substance.
𝜌 =
𝑅𝐴
ELECTRICAL RESISTANCE

8.  Unit of  is ohm × meter
 Conductivity is reciprocal of resistivity. 𝜎 =
1
𝜌
Its SI unit is (-m)-1
 Resistivity depends on
(a) nature of material (b) temperature.

9. Temperature coefficient of resistance ():
 𝛼 =
𝑅𝑡−𝑅0
𝑅0𝑡
. 𝑅𝑇 = 𝑅𝑜 1 + 𝛼𝛥𝑇
 Its unit is oC-1.
 For conductors  is positive i.e., on increasing temperature,
resistance increase.
 For semiconductors (Si, Ge)  is negative, i.e., on increasing
temperature resistance decreases.
 If two conductors (e.g. iron and platinum) have equal
resistivity, even then they can be differentiated, by
measuring their temperature co-efficient of resistivity.

10.  Emf of a source is defined as potential difference between
its output terminals when either its internal resistance is
zero or no current is
 Every source of emf has its own resistance called internal
resistance.
 Terminal potential difference is a voltage between output
terminals of a source of emf when current is drawn from it.
  = Vt + Ir 𝑟 =
𝐸−𝑉𝑡
𝐼
=
𝐸−𝑉𝑡
𝑉𝑡
𝑅
 For open circuit (r = 0)  = Vt
 For close circuit (r  0)  > Vt
 Charging the battery  < Vt
 Power delivered will be maximum when R = r , Pmax =E2/r
INTERNAL RESISTANCE OF SUPPLY

11.  P = IV= I2R =
𝑉2
𝑅
(Where P represents the power
dissipation)
 Pdiss = I2R (Series)
 Pdiss = 𝑉2
𝑅
(parallel)
 1 watt = 1J /1s
 H = P x t = I2 R t = V I t = V2 t / R (Joule Thomson’s
Effect)
ELECTRICAL POWER

12. Bulbs in Series:
 Total power consumed
1
𝑃
total
=
1
𝑃1
+
1
𝑃2
+. . .
 If ‘n’ bulbs are identical, 𝑃total =
𝑃
𝑛
 Pconsumed  (Brightness) ∝ 𝑉 ∝ 𝑅 ∝
1
𝑃
rated
i.e. in series combination bulb of lesser wattage will give
more bright light and p.d. appeared across it will be
Combinations of Bulbs

13. Bulbs in Parallel:
 Total power consumed
Ptotal = P1 + P2 + P3 +...+ Pn
 If ‘n’ identical bulbs are in parallel Ptotal = nP
 Pconsumed  (Brightness) ∝ PR ∝ I ∝ 1/R i.e. in parallel
combination, bulb of greater wattage will give more
bright light and more current will pass through it.

14.  “When a power of 1kW is maintained through a circuit
for 1 hour, then energy dissipated is 1kWh.”
1 KWh = 1000 W × 3600 sec = 3.6 × 106 J = 3.6 MJ
 Commercial unit of electrical energy
Energy in KWh =
PowerinWatt ×time of use in hour
1000
Amount of electricity bill=
Power in Watt ×time of use in hour
1000
× cost of unit
Kilowatt-hour

15. If the resistivity of the conductor is 2  10-6 m then its
conductivity is
A. 2  106 -1m-1
B. 5  106 -1m-1
C. 5  10-5 -1m-1
D. 5  105 -1m-1
QUESTION-1

16. A new flashlight cell of emf 1.5 volts gives a current of
15 amps, when connected directly to an ammeter of
resistance 0.04 ohm. The internal resistance of cell is
A) 0.04 ohm
B) 0.06 ohm
C) 10 ohm
D) 0.10 ohm
QUESTION-2

17. The resistance of a bulb filament is 100 Ω at a
temperature of 100°C. If its temperature coefficient of
resistance be 0.005 per °C, its resistance will become
200 Ω at a temperature of
(a) 200°C
(b) 400°C
(c) 300°C
(d) 500°C
QUESTION-3

18. A wire of uniform area of cross section is cut into two
parts of equal lengths. The resistivity of any part
(a) Remain same
(b) Is doubled
(c) Is halved
(d) None of these
QUESTION-4

19. An electrical motor has power 2000 W and the resistance
is 2 Ω. Find potential difference.
A. 6V
B. 63.25V
C. 0.5V
D. 0.09V
QUESTION-5

20. The V-I graph of a conductor at two different
temperatures as shown in Fig. The relation between
resistance will be
A. R1 > R2
B. R1 = R2
C. R2 > R1
D. R1 = 2 R2
QUESTION-6

21. An electron is circulating in a circular path with a
frequency of 50 Hz. What is the associated current?
A. 0.8 × 10–17 A
B. 8 × 10–17A
C. 0.4 × 10–17A
D. 80 × 10–17A
QUESTION-7

22. A certain appliance uses 350 W. If it is allowed to run
continuously for 24 days, how many kilowatt-hours of
energy does it consume?
A. 201.6 kWh
B. 20.16 kWh
C. 2.01 kWh
D. 8.4 kWh
QUESTION-8