CBSE Class 10 Science Electricity VBQs Set 04

Electric Current and Circuit

Question. The expressions that relate (i) Q, I and t and (ii) Q, V and W respectively are (Here the symbols have their usual meanings)
(a) (i) \( I = \frac{Q}{t} \) (ii) \( W = \frac{V}{Q} \)
(b) (i) \( Q = I \times t \) (ii) \( W = V \times Q \)
(c) (i) \( Q = \frac{I}{t} \) (ii) \( V = \frac{W}{Q} \)
(d) (i) \( I = \frac{Q}{t} \) (ii) \( Q = \frac{V}{W} \)
Answer: (b) (i) Q = I × t (ii) W = V × Q

Question. Define electric current.
Answer: Electric current is defined as the rate of flow of electric charge through any cross-section of a conductor. Mathematically, it is represented as \( I = \frac{Q}{t} \), where \( Q \) is the net charge flowing through the cross-section in time \( t \). Its SI unit is Ampere (A).

Question. Define one ampere.
Answer: One ampere is the current that flows through a conductor when one coulomb of charge passes through any cross-section of the conductor in one second. \( 1\text{ A} = \frac{1\text{ C}}{1\text{ s}} \).

Ohm’s Law

Question. State Ohm’s law.
Answer: Ohm’s law states that the electric current flowing through a metallic conductor is directly proportional to the potential difference applied across its ends, provided its temperature and other physical conditions remain constant. \( V \propto I \text{ or } V = IR \).

Question. (a) State Ohm’s Law. Represent it mathematically. (b) Define 1 ohm. (c) What is the resistance of a conductor through which a current of 0.5 A flows when a potential difference of 2 V is applied across its ends?
Answer: (a) Ohm's law states that at constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends. Mathematically: \( V = IR \).
(b) 1 ohm is the resistance of a conductor such that when a potential difference of 1 volt is applied across its ends, a current of 1 ampere flows through it. \( 1\text{ }\Omega = \frac{1\text{ V}}{1\text{ A}} \).
(c) Given: \( I = 0.5\text{ A} \), \( V = 2\text{ V} \).
Using \( R = \frac{V}{I} \)

\( \implies R = \frac{2}{0.5} \)

\( \implies R = 4\text{ }\Omega \).

Factors on which the Resistance of a Conductor Depends

Question. Assertion (A) : The metals and alloys are good conductors of electricity.
Reason (R) : Bronze is an alloy of copper and tin and it is not a good conductor of electricity.
(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.

Question. Assertion (A) : Alloys are commonly used in electrical heating devices like electric iron and heater.
Reason (R) : Resistivity of an alloy is generally higher than that of its constituent metals but the alloys have low melting points then their constituent metals.
(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.

Question. A cylindrical conductor of length ‘l’ and uniform area of cross section ‘A’ has resistance ‘R’. The area of cross section of another conductor of same material and same resistance but of length ‘2l ’ is
(a) \( \frac{A}{2} \)
(b) \( \frac{3A}{2} \)
(c) \( 2A \)
(d) \( 3A \)
Answer: (c) 2A

Question. How is the resistivity of alloys compared with those of pure metals from which they may have been formed?
Answer: The resistivity of an alloy is generally much higher than that of its constituent pure metals.

Question. (i) List three factors on which the resistance of a conductor depends. (ii) Write the SI unit of resistivity.
Answer: (i) Factors: Length of the conductor, Area of cross-section, and Nature of material (Temperature also affects it). (ii) SI unit of resistivity is Ohm-meter (\( \Omega\text{ m} \)).

Question. (a) List the factors on which the resistance of a uniform cylindrical conductor of a given material depends. (b) The resistance of a wire of 0.01 cm radius is 10 \( \Omega \). If the resistivity of the wire is \( 50 \times 10^{-8}\text{ }\Omega\text{ m} \), find the length of this wire.
Answer: (a) Length and Area of cross-section. (b) Given \( R = 10\text{ }\Omega \), \( \rho = 50 \times 10^{-8}\text{ }\Omega\text{ m} \), \( r = 0.01\text{ cm} = 10^{-4}\text{ m} \).
\( A = \pi r^2 = 3.14 \times 10^{-8}\text{ m}^2 \).
Using \( L = \frac{RA}{\rho} \)

\( \implies L = \frac{10 \times 3.14 \times 10^{-8}}{50 \times 10^{-8}} \)

\( \implies L = \frac{31.4}{50} = 0.628\text{ m} \).

Question. Calculate the resistivity of the material of a wire of length 1 m, radius 0.01 cm and resistance 20 ohms.
Answer: Given: \( L = 1\text{ m} \), \( r = 0.01\text{ cm} = 10^{-4}\text{ m} \), \( R = 20\text{ }\Omega \).
\( A = \pi r^2 = 3.14 \times (10^{-4})^2 = 3.14 \times 10^{-8}\text{ m}^2 \).
\( \rho = \frac{RA}{L} = \frac{20 \times 3.14 \times 10^{-8}}{1} \)

\( \implies \rho = 62.8 \times 10^{-8}\text{ }\Omega\text{ m} = 6.28 \times 10^{-7}\text{ }\Omega\text{ m} \).

Question. A copper wire has diameter 0.5 mm and resistivity \( 1.6 \times 10^{-8}\text{ }\Omega\text{ m} \). Calculate the length of this wire to make it resistance 100 \( \Omega \). How much does the resistance change if the diameter is doubled without changing its length?
Answer: Radius \( r = 0.25\text{ mm} = 2.5 \times 10^{-4}\text{ m} \). \( A = \pi r^2 \approx 1.96 \times 10^{-7}\text{ m}^2 \).
\( L = \frac{RA}{\rho} = \frac{100 \times 1.96 \times 10^{-7}}{1.6 \times 10^{-8}} \approx 1225\text{ m} \).
If diameter is doubled, Area becomes 4 times. Since \( R \propto 1/A \), the new resistance becomes \( 1/4 \) of original. \( R_{new} = 100/4 = 25\text{ }\Omega \).

Question. If the radius of a current carrying conductor is halved, how does current through it change?
Answer: Resistance \( R \propto 1/A \propto 1/r^2 \). If radius is halved, resistance becomes 4 times. From Ohm's law (\( I = V/R \)), the current becomes one-fourth (\( 1/4 \)) of its original value, provided potential difference remains same.

Question. Define resistance of a conductor. State the factors on which resistance of a conductor depends. Name the device which is often used to change the resistance without changing the voltage source in an electric circuit. Calculate the resistance of 50 cm length of wire of cross sectional area 0.01 square mm and of resistivity \( 5 \times 10^{-8}\text{ m} \).
Answer: Resistance is the opposition to the flow of current. Factors: Length, Area, Material. Device: Rheostat. Calculation: \( L = 0.5\text{ m} \), \( A = 0.01\text{ mm}^2 = 10^{-8}\text{ m}^2 \), \( \rho = 5 \times 10^{-8}\text{ }\Omega\text{ m} \).
\( R = \rho \frac{L}{A} = \frac{5 \times 10^{-8} \times 0.5}{10^{-8}} = 2.5\text{ }\Omega \).

Resistance of a System of Resistors

Question. Two LED bulbs of 12 W and 6 W are connected in series. If the current through 12 W bulb is 0.06 A the current through 6 W bulb will be
(a) 0.04 A
(b) 0.06 A
(c) 0.08 A
(d) 0.12 A
Answer: (b) 0.06 A

Question. If a person has five resistors each of value \( \frac{1}{5}\text{ }\Omega \), then the maximum resistance he can obtain by connecting them is
(a) 1 \( \Omega \)
(b) 5 \( \Omega \)
(c) 10 \( \Omega \)
(d) 25 \( \Omega \)
Answer: (a) 1 \( \Omega \)

Question. The maximum resistance which can be made using four resistors each of 2 \( \Omega \) is
(a) 2 \( \Omega \)
(b) 4 \( \Omega \)
(c) 8 \( \Omega \)
(d) 16 \( \Omega \)
Answer: (c) 8 \( \Omega \)

Question. The maximum resistance which can be made using four resistors each of resistance \( \frac{1}{2}\text{ }\Omega \) is
(a) 2 \( \Omega \)
(b) 1 \( \Omega \)
(c) 2.5 \( \Omega \)
(d) 8 \( \Omega \)
Answer: (a) 2 \( \Omega \)

Question. Three resistors of 10 \( \Omega \), 15 \( \Omega \) and 5 \( \Omega \) are connected in parallel. Find their equivalent resistance.
Answer: \( \frac{1}{R_p} = \frac{1}{10} + \frac{1}{15} + \frac{1}{5} = \frac{3+2+6}{30} = \frac{11}{30} \).

\( \implies R_p = \frac{30}{11} \approx 2.73\text{ }\Omega \).

Question. Show how would you join three resistors, each of resistance 9 \( \Omega \) so that the equivalent resistance of the combination is (i) 13.5 \( \Omega \), (ii) 6 \( \Omega \)?
Answer: (i) Connect two resistors in parallel and the third in series with them. \( R_{eq} = \frac{9 \times 9}{9 + 9} + 9 = 4.5 + 9 = 13.5\text{ }\Omega \).
(ii) Connect two resistors in series and the third in parallel with them. \( R_{eq} = \frac{(9 + 9) \times 9}{(9 + 9) + 9} = \frac{18 \times 9}{27} = 6\text{ }\Omega \).

Question. (a) A 6 \( \Omega \) resistance wire is doubled on itself. Calculate the new resistance of the wire. (b) Three 2 \( \Omega \) resistors A, B and C are connected in such a way that the total resistance of the combination is 3 \( \Omega \). Show the arrangement of the three resistors and justify your answer.
Answer: (a) Doubling on itself means length is halved (\( L/2 \)) and area is doubled (\( 2A \)). \( R' = \rho \frac{L/2}{2A} = \frac{1}{4} R = 6/4 = 1.5\text{ }\Omega \). (b) Connect two resistors in series (\( 2+2=4 \)) in parallel with the third? No. Connect two in parallel (\( \frac{2 \times 2}{2 + 2} = 1\text{ }\Omega \)) and the third in series with them. \( R_{eq} = 1 + 2 = 3\text{ }\Omega \).

Heating Effect of Electric Current

Question. The resistance of a resistor is reduced to half of its initial value. If other parameters of the electrical circuit remain unaltered, the amount of heat produced in the resistor will become
(a) four times
(b) two times
(c) half
(d) one fourth
Answer: (b) two times

Question. The resistance of a resistor is reduced to half of its initial value. In doing so, if other parameters of the circuit remain unchanged, the heating effects in the resistor will become
(a) two times
(b) half
(c) one-fourth
(d) four times
Answer: (b) half

Question. (i) State Joule’s law of heating. Express it mathematically when an appliance of resistance R is connected to a source of voltage V and the current I flows through the appliance for a time t.
(ii) A 5 \( \Omega \) resistor is connected across a battery of 6 volts. Calculate the energy that dissipates as heat in 10 s.

Answer: (i) Joule's law of heating states that the heat produced in a resistor is directly proportional to the square of the current for a given resistance, directly proportional to the resistance for a given current, and directly proportional to the time for which the current flows through the resistor.
\( \implies \) Mathematically, \( H = I^2Rt \).
\( \implies \) Since \( I = V/R \), it can also be expressed as \( H = \frac{V^2}{R}t \). (ii) Given: \( R = 5 \text{ }\Omega \), \( V = 6 \text{ V} \), \( t = 10 \text{ s} \).
\( \implies \) \( H = \frac{V^2}{R}t = \frac{6^2}{5} \times 10 = \frac{36}{5} \times 10 = 36 \times 2 = 72 \text{ J} \).

Question. (a) Calculate the resistance of a metal wire of length 2 m and area of cross-section \( 1.55 \times 10^{-6} \text{ m}^2 \). (Resistivity of the metal is \( 2.8 \times 10^{-8} \text{ }\Omega\text{ m} \))
(b) Why are alloys preferred over pure metals to make the heating elements of electrical heating devices?

Answer: (a) Given: \( l = 2 \text{ m} \), \( A = 1.55 \times 10^{-6} \text{ m}^2 \), \( \rho = 2.8 \times 10^{-8} \text{ }\Omega\text{ m} \).
\( \implies \) \( R = \rho \frac{l}{A} = \frac{2.8 \times 10^{-8} \times 2}{1.55 \times 10^{-6}} \approx 3.61 \times 10^{-2} \text{ }\Omega \). (b) Alloys are preferred over pure metals for heating elements because:

• They have higher resistivity than their constituent metals.
• They do not oxidize (burn) readily at high temperatures.
• They have higher melting points.

Question. (a) Write the mathematical expression for Joule’s law of heating.
(b) Compute the heat generated while transferring 96000 coulomb of charge in two hours through a potential difference of 40 V.

Answer: (a) The mathematical expression for Joule's law of heating is \( H = I^2Rt \). (b) Given: \( Q = 96000 \text{ C} \), \( t = 2 \text{ hours} \), \( V = 40 \text{ V} \).
\( \implies \) Heat produced \( H = V \times Q \)
\( \implies \) \( H = 40 \text{ V} \times 96000 \text{ C} = 3,840,000 \text{ J} \) or \( 3.84 \times 10^6 \text{ J} \).

Question. Write Joule’s law of heating.

Answer: Joule's law of heating states that the heat \( H \) produced in a conductor of resistance \( R \) when a current \( I \) flows through it for time \( t \) is given by \( H = I^2Rt \). This means the heat produced is directly proportional to (i) square of the current, (ii) resistance of the conductor, and (iii) time for which the current flows.

Question. Explain the use of an electric fuse. What type of material is used for fuse wire and why?

Answer: An electric fuse is a safety device used to protect electrical circuits and appliances from damage due to excessive current or short-circuiting. It is placed in series with the device. The fuse wire is made of a material having a low melting point and high resistance, such as an alloy of lead and tin. When the current exceeds a certain limit, the heat produced (\( I^2Rt \)) melts the fuse wire, breaking the circuit and preventing damage.

Question. (a) Why is tungsten used for making bulb filaments of incandescent lamps?
(b) Name any two electric devices based on heating effect of electric current.

Answer: (a) Tungsten is used for bulb filaments because it has a very high melting point (\( 3380^\circ \text{C} \)) and high resistivity, which allows it to remain incandescent (emit light) at very high temperatures without melting. (b) Electric heater and electric iron.

Question. A fuse wire melts at 5 A. If it is desired that the fuse wire of same material melt at 10 A, then whether the new fuse wire should be of smaller or larger radius than the earlier one? Give reasons for your answer.

Answer: The new fuse wire should have a larger radius. Heat produced \( H = I^2Rt \). For a higher current rating, the resistance \( R \) of the wire must be lower to prevent the wire from melting at lower currents. Since resistance is inversely proportional to the area of cross-section (\( R \propto 1/r^2 \)), a larger radius (thicker wire) will result in lower resistance, allowing it to carry a higher current (10 A) before reaching its melting point.

Electric Power

Question. An electric kettle consumes 1 kW of electric power when operated at 220 V. The minimum rating of the fuse wire to be used for it is
(a) 1 A
(b) 2 A
(c) 4 A
(d) 5 A
Answer: (d) 5 A

Question. Two bulbs of 100 W and 40 W are connected in series. The current through the 100 W bulb is 1 A. The current through the 40 W bulb will be
(a) 0.4 A
(b) 0.6 A
(c) 0.8 A
(d) 1 A
Answer: (d) 1 A

Question. Write the relation between resistance (R) of filament of a bulb, its power (P) and a constant voltage V applied across it.

Answer: The relation is \( P = \frac{V^2}{R} \) or \( R = \frac{V^2}{P} \).

Question. Power of a lamp is 60 W. Find the energy in joules consumed by it in 1s.

Answer: Energy \( E = P \times t = 60 \text{ W} \times 1 \text{ s} = 60 \text{ J} \).

Question. (a) What is the meaning of electric power of an electrical device? Write its SI unit.
(b) An electric kettle of 2 kW is used for 2 h. Calculate the energy consumed in (i) kilowatt hour and (ii) joules.

Answer: (a) Electric power is the rate at which electrical energy is consumed or dissipated in an electric circuit. Its SI unit is watt (W). (b) (i) Energy in kWh \( = \text{Power (kW)} \times \text{time (h)} = 2 \text{ kW} \times 2 \text{ h} = 4 \text{ kWh} \). (ii) Energy in Joules \( = 4 \times 3.6 \times 10^6 \text{ J} = 14.4 \times 10^6 \text{ J} \).

Question. An electric iron has a rating of 750 W; 200 V. Calculate: (i) the current required. (ii) the resistance of its heating element. (iii) energy consumed by the iron in 2 hours.

Answer: (i) Current \( I = \frac{P}{V} = \frac{750}{200} = 3.75 \text{ A} \). (ii) Resistance \( R = \frac{V}{I} = \frac{200}{3.75} = 53.33 \text{ }\Omega \). (iii) Energy \( E = P \times t = 750 \text{ W} \times 2 \text{ h} = 1500 \text{ Wh} = 1.5 \text{ kWh} \).

Question. An electric bulb is connected to a 220 V generator. The current is 2.5 A. Calculate the power of the bulb.

Answer: Power \( P = V \times I = 220 \text{ V} \times 2.5 \text{ A} = 550 \text{ W} \).

Question. (a) Define electric power and state its SI unit. The commercial unit of electrical energy is known as ‘unit’. Write the relation between this ‘unit’ and joule.
(b) In a house, 2 bulbs of 50 W each are used for 6 hours daily and an electric geyser of 1 kW is used for 1 hour daily. Calculate the total energy consumed in a month of 30 days and its cost at the rate of Rs. 8.00 per kWh.

Answer: (a) Electric power is the rate of consumption of electrical energy. SI unit is watt (W).
\( \implies \) 1 unit = 1 kWh = \( 3.6 \times 10^6 \text{ J} \). (b) Energy for 2 bulbs per day = \( 2 \times 50 \text{ W} \times 6 \text{ h} = 600 \text{ Wh} = 0.6 \text{ kWh} \).
\( \implies \) Energy for geyser per day = \( 1 \text{ kW} \times 1 \text{ h} = 1 \text{ kWh} \).
\( \implies \) Total energy per day = \( 0.6 + 1 = 1.6 \text{ kWh} \).
\( \implies \) Total energy for 30 days = \( 1.6 \times 30 = 48 \text{ kWh} \).
\( \implies \) Total cost = \( 48 \times 8 = \text{Rs. } 384 \).

Question. (a) Define power and state its SI unit. (b) A torch bulb is rated 5V and 500 mA. Calculate its (i) power (ii) resistance (iii) energy consumed when it is lighted for \( 2\frac{1}{2} \) hours.

Answer: (a) Power is the rate of doing work. SI unit is Watt. (b) (i) Power \( P = V \times I = 5 \text{ V} \times 0.5 \text{ A} = 2.5 \text{ W} \). (ii) Resistance \( R = \frac{V}{I} = \frac{5}{0.5} = 10 \text{ }\Omega \). (iii) Energy \( E = P \times t = 2.5 \text{ W} \times 2.5 \text{ h} = 6.25 \text{ Wh} = 0.00625 \text{ kWh} \).

Question. Two identical resistors, each of resistance 15 \( \Omega \), are connected in (i) series, and (ii) parallel, in turn to a battery of 6 V. Calculate the ratio of the power consumed in the combination of resistors in each case.

Answer: (i) Series: \( R_s = 15 + 15 = 30 \text{ }\Omega \). Power \( P_s = \frac{V^2}{R_s} = \frac{6^2}{30} = \frac{36}{30} = 1.2 \text{ W} \). (ii) Parallel: \( R_p = \frac{15}{2} = 7.5 \text{ }\Omega \). Power \( P_p = \frac{V^2}{R_p} = \frac{6^2}{7.5} = \frac{36}{7.5} = 4.8 \text{ W} \).
\( \implies \) Ratio \( P_s : P_p = 1.2 : 4.8 = 1 : 4 \).

Question. (a) An electric bulb is rated at 200 V; 100 W. What is its resistance?
(b) Calculate the energy consumed by 3 such bulbs if they glow continuously for 10 hours for complete month of November.
(c) Calculate the total cost if the rate is Rs. 6.50 per unit.

Answer: (a) \( R = \frac{V^2}{P} = \frac{200 \times 200}{100} = 400 \text{ }\Omega \). (b) Energy per day = \( 3 \times 100 \text{ W} \times 10 \text{ h} = 3000 \text{ Wh} = 3 \text{ kWh} \).
\( \implies \) For November (30 days), Total Energy = \( 3 \times 30 = 90 \text{ kWh (units)} \). (c) Cost = \( 90 \times 6.50 = \text{Rs. } 585 \).

Question. (a) What is meant by the statement, “The resistance of a conductor is one ohm”?
(b) Define electric power. Write an expression relating electric power, potential difference and resistance.
(c) How many 132 \( \Omega \) resistors in parallel are required to carry 5 A on a 220 V line?

Answer: (a) It means when a potential difference of 1 V is applied across the conductor, 1 A of current flows through it. (b) Electric power is the rate of electrical energy consumption. \( P = \frac{V^2}{R} \). (c) Required total resistance \( R_{eq} = \frac{V}{I} = \frac{220}{5} = 44 \text{ }\Omega \).
\( \implies \) In parallel, \( \frac{1}{R_{eq}} = \frac{n}{R} \)
\( \implies \) \( \frac{1}{44} = \frac{n}{132} \)
\( \implies \) \( n = \frac{132}{44} = 3 \). So, 3 resistors are required.