Samacheer Class 10 Science - Electricity

Reason: Three pin connections reduce heating of the connecting wires
Answer: c) if the assertion is true, but the reason is false.

Reason: The current flows towards the point of the highest potential
Answer: c) if the assertion is true, but the reason is false.

Reason: LED bulbs consume less power than incandescent bulbs.
Answer: a) if both the assertion and the reason are true and the reason is the correct explanation of the assertion.

The current flowing through a conductor is said to be one ampere, when a charge of one coulomb flows across any cross section of a conductor, in one second. Hence,
$1 ampere = 1 coulomb / 1 second$

(i) Decreases : The resistance decreases as the conductor is made thicker.
(ii) Reason: Resistance is inversely proportional to area of cross section A.
$i. e., R \alpha 1/A$
$here, A = πr2$
Where, r is the radius which determines the thickness.

Tungsten has high melting point, it can bear high heat for glowing. But in fuse wire, the wire used in it should melt. So a metal (wire) which has low melting point should be used in a fuse wire, but not tungsten wire.

Electric iron, and electric toaster, or electric oven, and electric heater.

Electric potential : The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against the electric force.
Potential difference : The electric potential difference between two points is defined as the amount of work done in moving a unit positive charge from one point to another point against the electric force.

The earth wire provides a low resistance path to the electric current. The earth wire sends the current from the body of the appliance to the Earth, whenever a live wire accidentally touches the body of metallic electric appliance. Thus, the earth wire serves as a protective conductor, which saves us from electric shocks.

Ohm’s law states that, at a constant temperature, the steady current 'I' flowing through a conductor is directly proportional to the potential difference 'V', between two ends of the conductor. I \propto V
$I = (1/R) V$
$V =IR$

Resistivity
(i) Electrical resistivity of a material is defined as the resistance of a conductor of unit length and unit area of cross section.
(ii) Its unit is ohm metre.
(iii) Electrical resistivity of a conductor is a measure of the resisting power of a specified material to the passage of an electric current. It is a constant for a given material.
Conductivity
(i) The reciprocal of electrical resistivity of a material is called its electrical conductivity.
(ii) It's unit is mho metre-1.
(iii) Electrical conductivity of a conductor is a measure of its ability to pass the current through it.

(i) All the circuits in a house are connected in parallel, so that the disconnection of one circuit does not affect the other circuit.
(ii) One more advantage of the parallel connection of circuits is that each electric appliance gets an equal voltage.

Resistors in series

$3. i.e., V = V1+ V2 + V3$
$Here, V = IR$
$V1 = IR1$
$V2 = IR2$
$V3=IR3$
$∴IR = IR1 + IR2 + IR3$
$IR = I [R1 + R2 + R3]$
$i.e., RS = R1 + R2 + R3$

$Ie., RS = nR$

Resistors in parallel

$3. i.e., I = I1 + I2 + I3$
$Here, I = V / R$
$I1 = V / R1$
$I2 = V / R2$
$I3 = V / R3$
$∴V/R = V/R1 + V/R2 + V/R3$
$V(1/R) = V(1/R1 + 1/R2 + 1/R3)$
$i.e., 1/RP = 1/R1 + 1/R2 + 1/R3$

$i.e., 1/RP = n/R$
or
$RP = R/n$

Refer to textbook.

a) Electric current is defined as the rate of flow of charges in a conductor. If a net charge 'Q' passes through any cross section of a conductor in time 't', then the current flowing through the conductor
$is I = Q / t.$
b) The current flowing through a conductor is said to be one ampere, when a charge of one coulomb flows across any cross section of a conductor, in one second. Hence,
$1 ampere = 1 coulomb / 1 second$
c) Ammeter is used to measure electric current. It should be connected in series in a circuit.

a) Joule's law of heating states that the heat produced in any resistor is:
(i) Directly proportional to the square of the current passing through the resistor.
(ii) Directly proportional to the resistance of the resistor.
(iii) Directly proportional to the time for which the current is passing through the resistor.
b) Alloy of nickel and chromium have the following properties:
(i) It has high resistivity,
(ii) It has a high melting point,
(iii) It is not easily oxidized.
So it is used as the heating element.
c) The fuse wire is connected in series, in an electric circuit. When a large current passes through the circuit, the fuse wire melts due to Joule's law of heating and hence the circuit gets disconnected. Therefore, the circuit and the electric appliances are saved from any damage.

(i) The first stage of domestic circuit is to bring the power supply to the main-box from a distribution panel, such as a transformer. The important components of the main - box are (i) a fuse box and (ii) a meter.
(ii) The meter is used to record the consumption of electrical energy. The fuse box contains either a fuse wire or a miniature circuit breaker (MCB).
(iii) The function of the fuse wire or a MCB is to protect the house hold electrical appliances from overloading due to excess current.
(iv) The electricity is brought to houses by two insulated wires, out of these two wires, one wire has a red insulation and is called the 'live wire'. The other wire has a black insulation and is called the 'neutral wire'.
(v) The electricity supplied to house is actually an alternating current having an electric potential of 220 V. Both, the live wire and the neutral wire enter into a box where the main fuse is connected with the live wire.
(vi) After the electricity meter, these wires enter into the main switch, which is used to discontinue the electricity supply whenever required.
(vii) After the main switch, these wires are connected to live wires of two separate circuits. Out of these two circuits, one circuit is of a 5A rating, which is used to run the electric appliances with a lower power rating, such as tube lights, bulbs and fans. The other circuit is of a 15A rating, which is used to run electric appliances with a high power rating. Such as A/C. refrigerators, electric iron and heaters.

a) Advantages of LED TV :
(i) It has brighter picture quality.
(ii) It is thinner in size
(iii) It uses less power and consumes very less energy.
(iv) Its life span is more.
(v) It is more reliable.
b) Merits of LED bulb :
(i) As there is no filament, there is no loss of energy in the form of heat. It is cooler than the incandescent bulbs.
(ii) In comparison with the fluorescent light, the LED bulbs have significantly low power requirement.
(iii) It is not harmful to the environment.

Given
In the first case
$The power consumed by an electric iron P = 420 W$
Applied voltage
$i.e potential difference, V = 220 V$
$To find : Current, I = ? (in each case)$
Solution
$P = V I => I = P/ V = 420/220 = 21/11$
$When heating at maximum rate, I = 21/11 ampere.$
$(or) I = 1.909A.$
$Heating at minimum rate, the rate at which energy is consumed, P = 180 W$
$Applied voltage, V = 220 V$
$I = P/V$
$I = 180/220$
$9/11 = 0.818 A$
When heating at minimum rate,
$1= 9/11 A (or) I = 0.818 A.$

$Energy consumed by Electric bulb of 100 W = ?$
$Time, t = 5 hr$
$For month January = 31 days$
$Number of 60W bulbs = 4$
$60W used for a time = 5 hr$
$To find : Total Energy consumed (in kWh) = ?$
Solution
Energy consumed by 100W bulb
$= Power of the bulb x time$
$= 100 x 31 x 5 = 15500$
$Energy consumed by 100 W = 15.5 kwh$
$Energy consumed by four 60 W bulb = 4x 60 x5x31 = 37,200 wh$
$Four 60 W bulb = 37.2 kwh$
$Total energy consumed = Energy consumed by 100 W + Energy consumed by four 60 W$
$= 15.5 + 37.2$
$Total energy consumed = 52.7 kwh$

a) power
b) resistance
c) energy consumed if it is used for 4 hour.
Given
$The Potential of torch bulb V = 3V.$
$Current I = 600ma$
$= 600 x 10-3 = 0.6A$
To find :
$a) Power, P = ?$
$b) Resistance, R = ?$
$c) Energy consumed = ?$
Solution
$a) P = VI =3x0.6= 1.8W$
$V = I R => R = V/I$
$b) Resistance, R = 3/06 = 50$
c) Energy consumed for 4hr
$= power of the bulb x time = 1.8 x 4$
$Energy, E = 7.2 wh$

a) How will the resistance of each part of the wire change compared with the original resistance?
b) If the five parts of the wire are placed in parallel, how will the resistance of the combination change?
c) What will be ratio of the effective resistance in series connection to that of the parallel connection?
Given
a) Resistance of a piece of wire is proportional to its length i.e R \propto l
$Each piece has a resistance = R$
Solution
Wire is cut into five equal parts
$Resistance of each part = R/5$
$Resistance = R/5$
b) All the five parts are connected in parallel. For parallel connection, the effective resistance is
$1/Rp = 1/R1 + 1/R2$
$1/Rp = 5/R + 5/R + 5/R + 5/R + 5/R$
$1/Rp = 25/R$
$Rp = R/25$
c) The ratio of the effective resistance in series to parallel connection.
$1/Rp = 25/R$
$Rs/Rp = 25/1$
$Ratio of Rs and Rp = 25:1$
$Rs:Rp = 25:1$
XI. HOTS:

$Effective resistance in parallel, Rp = 2Ω$
$Effective resistance in series, R. =9 Ω$
$R12 - 9R1 + 18 = 0$
$(R1 - 3) (R1 - 6) = 0.$
$R1 = 3 (or) 6 Ω$
$When R1 = 3 Ω, R2 = 9-3=6 Ω$
$When R1 = 6 Ω, R2= 9-6=3 Ω$

Given
$Current I = 5 A$
To find :
$Number of electrons passing per second, for 5 A, n = ?$
Solution
$I = Q/t$
$Charge of an electron, e = 1.6 x 10-19 coulomb$
$Then 1 coulomb = 1 / 1.6x10-19 electrons$
$i.e., 1 coulomb = 6.25 x 1018 electrons$
For 1 ampere current, no. of electrons will be 6.25 x 1018 electrons. For 5 ampere current, it is = 5 x 6.25 x 1018 electrons
$= 3.125 x 1019 electrons$
Method 2:
No. of electrons passing
$per second, n = ?$
$I = Q/t$
Where e -> Charge of electron which is equal to 1.6 x 10-19 C
$Q is ‘n‘ no of charges i.e Q = ne$
$Here t = 1S$
$I = ne/t$
$n = [ I x t ]/ e = [5/1.6] x 1019$
$Number of Electrons, n = 3.125 x 1019 electrons$

$Resistance of wire, R = 10 Ω$
$Length is increased to thrice = 3 L$
$To find : New resistance = ?$
Solution
$Resistance (R) = [ resistivity (ρ) * length (L) ] / area (A)$
$R = ρL/A$
$10 = ρL/A$
When length is increased by three times (3L) the area of cross section is reduced by three times (A/3).
$New length = 3L$
$New Area = A/3$
$New resistance R = [ ρ.3L ] / [A/3] = 9 (ρL/A)$
$R = 9.R$
$9 x 10 = 90 Ω$
$R= 90 Ω$
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