4. Electricity
I. Choose the best answer
(a) Rate of change of charge is electrical power.
(b) Rate of change of charge is current.
(c) Rate of change in energy is current.
(d) Rate of change of current is a charge.
2. SI unit of resistance is ______.
(a) mho (b) joule
(c) ohm (d) ohmmeter.
3. In a simple circuit, why does the bulb glow when you close the switch?
(a) The switch produces electricity.
(b) Closing the switch completes the circuit.
(c) Closing the switch breaks the circuit.
(d) The bulb is getting charged
4. Kilowatt-hour is the unit of _______.
(a) resistivity
(b) conductivity
(c) electrical energy
(d) electrical power.
II. Fill in the blanks
1. When a circuit is open, Current cannot pass through it.
2. The ratio of the potential difference to the current is known as Ohm’s law.
3. The wiring in a house consists of domestic electric circuits.
4.The power of an electric device is a product of electric current and the potential difference.
5. LED stands for Light Emitting Diode.
III. State whether the following statements are true or false: If false correct the statement.
1. Ohm’s law states the relationship between power and voltage.
Answer: False.
Correct Statement: Ohm’s law states the relationship between potential difference and current.
2. MCB is used to protect household electrical appliances.
Answer: True.
3. The SI unit for electric current is the coulomb.
Answer: False.
Correct Statement: The SI unit of electric current is Ampere.
4. One unit of electrical energy consumed is equal to 1000 kilowatt-hour.
Answer: True.
5. The effective resistance of three resistors connected in series is lesser than the lowest of the individual resistances.
Answer: False.
Correct Statement: The effective resistance of three resistors connected in series is greater than the highest of the individual resistance.
IV. Match the items in column-I to the items in column-II:
V. Assertion and Reason Type Questions
Mark the correct choice as
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) If the assertion is true, but the reason is false.
(d) If the assertion is false, but the reason is true.
1.Assertion: Electric appliances with a metallic body have three wire connections.
Reason: Three – pin connections reduce heating of the connecting wires
Answer:
(c) If the assertion is true, but the reason is false.
Correct Reason: Three – pin connections to protect the electrical shocking. Because appliances carry high electric current.
2. Assertion: In a simple battery circuit the point of highest potential is the positive terminal of the battery.
Reason: The current flows towards the point of the highest potential.
Answer:
(c) If the assertion is true, but the reason is false.
Correct Reason: The current flows towards the points of the lower potential.
3. Assertion: LED bulbs are far better than incandescent bulbs.
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.
VI. Very Short Answer Questions.
1. Define the unit of current.
The SI unit of electric current is ampere (A). 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.
2. What happens to the resistance, when the conductor is made thicker?
As the resistance is inversely proportional to the area, (R α 1/A) thick wires will cause low resistance .
3. Why is tungsten metal used in bulbs, but not infuse wires?
Because tungsten has a high melting point it is used in blubs. It cannot be used in fuse wires because in fuse wires the material must have a low melting point.
4. Name any two devices, which are working on the heating effect of the electric current.
(i) Electric heater
(ii) Fuse wire
VII. Short Answer Questions.
1. Define electric potential and potential difference.
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.
Electric 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.
2. What is the role of the earth wire in domestic circuits?
This 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 the metallic electric appliance. Thus, the earth wire serves as a protective conductor, which saves us from electric shocks.
3. State Ohm’s law.
According to Ohm’s law, at a constant temperature, the steady current ‘I’ flowing through a conductor is directly proportional to the potential difference ‘V’ between the two ends of the conductor.
I ∝ V
V = IR.
4. Distinguish between the resistivity and conductivity of a conductor.
5. What connection is used in domestic appliances and why?
Parallel connection is used in domestic appliances. When any disconnection of one circuit in our home, does not affect the other circuit.
VIII. Long Answer Questions.
1. With the help of a circuit diagram derive the formula for the resultant resistance of three resistances connected: (a) in series and (b) in parallel
(a) Resistors in Series:
A series circuit connects the components one after the other to form a ‘single-loop’. A series circuit has only one loop through which current can pass. If the circuit is interrupted at any point in the loop, no current can pass through the circuit and hence no electric appliances connected in the circuit will work. Series circuits are commonly used in devices such as flashlights. Thus, if
(i) Resistors are connected end to end so that the same current passes through each of them, then they are said to be connected in series.
(ii) Let, three resistances R1, R2 and R3 be connected in series.
(iii) Let the current flowing through them be I.
(iv) According to Ohm’s Law, the potential differences V1, V2 and V3 across R1 , R2 and R3 respectively, are given by:
V1 = IR1 …. (1)
V2 = IR2 …. (2)
V3 = IR3 …. (3)
The sum of the potential differences across the ends of each resistor is given by:
V = V1 + V2 + V3
using equations (1), (2) and (3), we get
V = IR1 + IR2 + IR3 …. (4)
(v) The effective resistor is a single resistor, which can replace the resistors effectively, so as to allow the same current through the electric circuit.
(vi) Let, the effective resistance of the series-combination of the resistors, be RS. Then,
V = IRS ….(5)
Combining equations (4) and (5)
IRS = IR1 + IR2 + IR3
RS = R1 + R2 + R3
Thus, we can understand that
(vii) When a number of resistors are connected in series, their equivalent resistance or effective resistance is equal to the sum of the individual resistances.
(viii) When V resistors of equal resistance R are connected in series, the equivalent resistance is ‘nR’. i.e., RS = nR
(ix) The equivalent resistance in a series combination is greater than the highest of the individual resistances.
(b) Resistors in Parallel:
A parallel circuit has two or more loops through which current can pass. If the circuit is disconnected in one of the loops, the current can still pass through the other loop(s). The wiring in a house consists of parallel circuits.
(i) Consider that three resistors R1, R2 and R3 are connected across two common points A and B.
(ii) The potential difference across each resistance is the same and equal to the potential difference between A and B. This is measured using the voltmeter.
(iii) The current I arriving at A divides into three branches I1, I2 and I3 passing through R1, R2 and R3 respectively.
According to the Ohm’s law, we have,
I1 = V/ R1 -----------(1)
I2 = V/R2 ------------(2)
I3 = V/R3 -------------(3)
The total current through the circuit is given by
I = I1 + I2 + I3
Using equations (1), (2) and (3), we get
I = V/R1 + V/R2+ V/R3----------------- (4)
Let the effective resistance of the parallel combination of resistors be RP Then,
I = V/ Rp --------------------------------(5)
Combining equations (4) and (5), we have
V/Rp = V/R1 + V/R2 + V/R3
1/Rp = 1/ R1 + 1/R2 + 1/R3
(iv) Thus, when a number of resistors are connected in parallel, the sum of the reciprocals of the individual resistances is equal to the reciprocal of the effective or equivalent resistance. When ‘n’ resistors of equal.
resistances R are connected in parallel, the equivalent resistance is R/n.
i.e. 1/Rp = 1/R + 1/R + 1/R …… + 1/R = n/R
Hence, Rp = R/n.
(vi) The equivalent resistance in a parallel combination is less than the lowest of the individual resistances.
2.(a) What is meant by electric current?
(b) Name and define its unit.
(c) Which instrument is used to measure the electric current? How should it be connected in a circuit?
(a) Electric current is the rate of flow of charges in a conductor.
(b) The unit of current is ampere. 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. Ammeter is always connected in series.
3.(a) State Joule’s law of heating.
(b) An alloy of nickel and chromium is used as the heating element. Why?
(c) How does a fuse wire protect electrical appliances?
(a) The heat produced in the resistor is H = W = VQ
We know that the relation between the charge and current is Q = It.
Using this, we get H = VIt
From Ohm’s Law, V = IR. Hence, we have H = I2Rt
This is known as Joule’s law of heating.
Joule’s law of heating states that the heat produced in any resistor is:
Directly proportional to the square of the current passing through the resistor. H = VIt
Directly proportional to the resistance of the resistor. V = IR
Directly proportional to the time for which the current is passing through the resistor. H = I2Rt
(b) An alloy of nickel and chromium is used as the heating element. Because:
it has high resistivity
it has a high melting point
it is not easily oxidized.
(c) (i) The fuse wire is connected in series, in an electric circuit.
(ii) When a large current passes through the circuit, the fuse wire melts due to Joule’s heating effect and hence the circuit gets disconnected.
(iii) Therefore, the circuit and the electric appliances are saved from any damage.
(iv) The fuse wire is made up of a material whose melting point is relatively low.
4.Explain about domestic electric circuits, (circuit diagram not required)
In our homes, electricity is distributed through the domestic electric circuits wired by the electricians. The first stage of the 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. 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). The function of the fuse wire or an MCB is to protect the household electrical appliances from overloading due to excess current.
An MCB is a switching device, which can be activated automatically as well as manually. It has a spring attached to the switch, which is attracted by an electromagnet when an excess current passes through the circuit. Hence, the circuit is broken and the protection of the appliance is ensured.
The electricity is brought to houses by 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’. The electricity supplied to your 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. After the electricity meter, these wires enter into the main switch, which is used to discontinue the electricity supply whenever required. 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 5 A 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 15 A rating, which is used to two insulated wires. Out of these two wires, run electric appliances with a high power rating, such as air-conditioners, refrigerators, electric iron and heaters. It should be noted that all the circuits in a house are connected in parallel, so that the disconnection of one circuit does not affect the other circuit. One more advantage of the parallel connection of circuits is that each electric appliance gets an equal voltage.
5. (a) What are the advantages of LED TV on normal TV?
(b) List the merits of the LED bulb.
(a) (i) It has a 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) (i) As there is no filament, there is no loss of energy in the form of heat. It is cooler than the incandescent bulb.
(ii) In comparison with the fluorescent light, the LED bulbs have a significantly low power requirement.
(iii) It is not harmful to the environment.
(iv) A wide range of colours is possible here.
(v) It is cost-efficient and energy-efficient.
(vi) Mercury and other toxic materials are not required.
IX. Numerical Problems:
1. An electric iron consumes energy at the rate of 420 W when heating is at the maximum rate and 180 W when heating is at the minimum rate. The applied voltage is 220 V. What is the current in each case?
Solution:
Case: 1
Power (P) = 420W
Applied Voltage (V) = 220V
Case: 2
Power (P) = 180 W
Applied Voltage (V) = 2.20 V
2. A 100-watt electric bulb is used for 5 hours daily and four 60 watt bulbs are used for 5 hours daily. Calculate the energy consumed (in kWh) in the month of January.
Answer:
Energy used by 100 W bulb is E = P × t
= 100 × 5 = 500 Wh
Energy used by four 60 W bulbs E = 4 × 60 × 5 = 1200 Wh
Total energy per day = 500 + 1200 = 1700 Wh
= 1.7 kWh
Number of days in January = 31 days.
Energy consumed in January = 31 × 1.7 = 52.7 kWh.
3. A torch bulb is rated at 3 V and 600 mA. Calculate it’s?
(a) power
(b) resistance
(c) energy consumed if it is used for 4 hours.
4. A piece of wire having a resistance R is cut into five equal parts.
(a) How will the resistance of each part of the wire change compare 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 the ratio of the effective resistance in series connection to that of the parallel connection?
Solution:
XI. HOTS Questions
1. Two resistors when connected in parallel give the resultant resistance of 2 ohms, but when connected in series the effective resistance becomes 9 ohms. Calculate the value of each resistance.
Solution:
Resultant resistance of parallel combination RP = 2 Ω
Resultant resistance of series combination RS = 9 Ω
2. How many electrons are passing per second in a circuit in which there is a current of 5 A?
Answer:
3. A piece of wire of resistance 10 ohms is drawn out so that its length is increased to three times its original length. Calculate the new resistance.
Solution:
Resistance (R) = Resistivity (r) × Length (L) / Area (A)
When the length increases by three times, the cross section will reduce by three times. Hence the length will be 3L while area = A/3
