CBSE Class 10 Science Electricity Assignment Set 09

Answer: Electric charge is a basic property of matter that causes it to experience a force in an electric field. Its SI unit is the coulomb (C).
1. Charges are of two kinds, positive and negative; like charges repel and unlike charges attract.
2. Charge is quantised: it exists in integral multiples of the charge of an electron (1.6 × 10⁻¹⁹ C).
3. Charge is conserved: it can neither be created nor destroyed, only transferred.

Teacher's Note:
1. Charge is the cause of all electrical effects.
2. Learn the SI unit and one numerical value to support the answer.

Answer: (a) An ammeter measures current; it is connected in series in the circuit.
(b) A voltmeter measures potential difference; it is connected in parallel across the component.

Teacher's Note:
1. The ammeter has a very low resistance and the voltmeter has a very high resistance.
2. Their connections in the circuit are different.

Answer: Resistance is the property of a conductor to oppose the flow of charges through it, given by R = V/I. Its SI unit is the ohm (Ω).
Resistivity of a material is the resistance of a conductor of that material of unit length and unit area of cross-section, ρ = RA/l. Its SI unit is the ohm metre (Ω m).

Teacher's Note:
1. Resistance depends on the dimensions of the conductor but resistivity depends only on the material and temperature.
2. Resistivity is also called specific resistance.

Answer: Ohm's law: the potential difference V across the ends of a conductor is directly proportional to the current I flowing through it, provided its temperature remains the same. So V = IR, where R is the resistance.
Verification: Connect a cell or battery, a key, an ammeter in series and a nichrome wire. Connect a voltmeter across the wire.
Vary the current using a rheostat and note the readings of current I and potential difference V.
The ratio V/I stays constant, and the graph of V against I is a straight line passing through the origin.

Teacher's Note:
1. Ohm's law is an experimental law and holds only for ohmic conductors at constant temperature.
2. The slope of the V-I graph gives the resistance.

Answer: The resistance of a conductor depends on
1. its length: R is directly proportional to length l,
2. its area of cross-section: R is inversely proportional to the area A,
3. the nature (material) of the conductor, and
4. its temperature.
Together, R = ρl/A.

Teacher's Note:
1. A longer or thinner wire has more resistance.
2. Resistance of metals increases when the temperature rises.

Answer: Joule's law: the heat produced in a resistor is directly proportional to the square of the current, to the resistance and to the time for which the current flows: H = I²Rt.
Derivation: If a charge Q moves across a potential difference V, the work done is W = VQ. Since Q = It, W = VIt. Using V = IR, W = I²Rt.
The electrical energy used up in the resistor is converted to heat, so H = I²Rt.

Teacher's Note:
1. The heating effect of current is used in heaters, geysers and electric irons.
2. The same energy can also be written as H = VIt or V²t/R.

Answer: Electric power is the rate at which electrical energy is consumed or work is done: P = W/t = VI. Electric energy is the total work done: E = P × t.
1 watt is the power consumed when a current of 1 ampere flows under a potential difference of 1 volt (1 W = 1 V × 1 A).
1 joule is the work done (or energy transferred) when a charge of 1 coulomb moves across a potential difference of 1 volt.

Teacher's Note:
1. Power is measured in watts and energy in joules or kilowatt-hours.
2. Energy is obtained by multiplying power by time.

Answer: 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.

Teacher's Note:
1. The kilowatt-hour is the commercial unit of electrical energy.
2. One unit of electricity on a bill means 1 kWh.

Answer: An ammeter has a very small resistance, so when connected in series it measures the whole current without changing it.
A voltmeter has a very large resistance, so when connected in parallel it measures the potential difference without drawing current from the component.

Teacher's Note:
1. Wrong connections can damage the meters or give wrong readings.
2. The difference lies in the resistance of the two meters.

Answer: Alloys have a higher resistivity than their constituent metals. They also do not oxidise (burn) readily at high temperatures, so they last long and produce large heat.

Teacher's Note:
1. Nichrome is the usual alloy used in heaters and toasters.
2. Pure metals would oxidise quickly and be less effective.

Answer: In series, the same current flows through all the resistors, so the total length of the path increases and the resistances add up: R = R1 + R2 + R3.
In parallel, the current divides among the branches, and each branch gives an extra path, which increases the effective area for the current. Hence 1/R = 1/R1 + 1/R2 + 1/R3, and the equivalent resistance is less than the smallest resistance.

Teacher's Note:
1. Series is like increasing the length of the wire and parallel is like increasing its thickness.
2. The equivalent resistance in parallel is always smaller than the smallest individual resistance.

Answer: (a) Series: the same current I flows through each resistor, and the total potential difference V = V1 + V2 + V3. Using V = IR for each, IR = IR1 + IR2 + IR3, so R = R1 + R2 + R3.
(b) Parallel: the potential difference V is the same across each resistor, and the total current I = I1 + I2 + I3. So V/R = V/R1 + V/R2 + V/R3, which gives 1/R = 1/R1 + 1/R2 + 1/R3.

Teacher's Note:
1. Remember what stays the same in each combination: the current in series and the voltage in parallel.
2. Draw the circuit diagram in the exam as part of the answer.

Answer: (i) Connect two resistors in parallel: 1/R = 1/9 + 1/9 = 2/9, so R = 4.5 Ω. Connect the third in series with this combination: 4.5 + 9 = 13.5 Ω.
(ii) Connect two resistors in series, giving 18 Ω. Connect the third in parallel with this combination: R = (18 × 9)/(18 + 9) = 162/27 = 6 Ω.

Teacher's Note:
1. Try the parallel pair and series combination first.
2. Always check the final value by calculation.

Answer: The graph of V (on the x-axis) against I (on the y-axis) is a straight line passing through the origin, which shows that I is proportional to V (Ohm's law).
R = V/I = 0.5/0.1 = 5 Ω; the ratio is the same for every pair of readings.
Therefore the resistance of the resistor is 5 Ω.

Teacher's Note:
1. The slope of the graph gives the resistance when V is on the vertical axis.
2. Use any pair of readings from the table and check with a second pair.

Answer: For the 100 W lamp: I1 = P/V = 100/220 = 0.45 A.
For the 60 W lamp: I2 = 60/220 = 0.27 A.
Total current from the line = I1 + I2 = 160/220 = 0.73 A (approximately).

Teacher's Note:
1. In a parallel circuit, the total current is the sum of the currents in each lamp.
2. Use P = VI to find the current in each lamp.

Answer: The resistance required R = V/I = 220/5 = 44 Ω.
For n identical resistors in parallel, R = 176/n.
So 176/n = 44, giving n = 4.
Four resistors of 176 Ω are required.

Teacher's Note:
1. First find the total resistance needed from Ohm's law.
2. Then divide the resistance of one resistor by that value.

Answer: A diameter divides the circle into two equal semicircles, each of resistance 20/2 = 10 Ω.
These two semicircular parts are in parallel between the ends of the diameter.
1/R = 1/10 + 1/10 = 2/10, so R = 5 Ω.

Teacher's Note:
1. The two halves of the wire provide two parallel paths between the same points.
2. The equivalent resistance is one quarter of the resistance of the whole wire.