CBSE Class 10 Physics HOTs Electricity Set 04

Question. Explain the role of fuse in series with any electrical appliance in an electric circuit. Why should a fuse with defined rating for an electric circuit not be replaced by one with a larger rating?
Answer: Fuse wire is a safety device connected in series with the live wire of circuit. It has high resistivity and low melting point. It melts when a sudden urge of large current passes through it and disconnects the entire circuit from the electrical supply. But, in case if we use a larger rating instead of a defined rating, then it will not protect the circuit as high current will easily pass through it and it will not melt.

Question. In what time is 400 joules of heat produced across a \( 16 \text{ } \Omega \) resistor at 80 V potential difference?
Answer: Given: \( H = 400 \text{ J}, t = ?, R = 16 \text{ } \Omega, V = 80 \text{ V} \)
Using, \( H = \frac{V^2t}{R} \)

\( \implies \) \( 400 = \frac{80^2 \times t}{16} \)

\( \implies \) \( t = \frac{400 \times 16}{80^2} \) or \( t = 1 \text{ s} \)

Question. State the factors on which the heat produced in a current carrying conductor depends. Give one practical application of this effect.
Answer: According to Joule’s law of heating effect, the heat (H) produced in a current carrying conductor depends upon
(i) square of current pass through it \( (H \propto I^2) \).
(ii) resistance of the conductor \( (H \propto R) \).
(iii) time for which current is passed in conductor \( (H \propto t) \).
Practical application of heating effect:
(i) Electric heater (ii) Fuse (Any one)

Question. A fuse wire melts at 5 A. If it is desired that the fuse wire of the same material melt at 10 A, then should the new fuse wire be of smaller or larger radius than the earlier one? Give reason for your answer.
Answer: Let the resistance of fuse wire that melts at 5 A be \( R_1 \). Then heat produced every second is,
\( H = I_1^2R_1 = 5^2R_1 = 25R_1 \dots(i) \)
Let the resistance of new fuse wire for current \( I_2 \) be \( R_2 \). For the same heat,
\( H = I_2^2R_2 = 10^2R_2 = 100R_2 \dots(ii) \)
From equation (i) and (ii), we have
\( 100R_2 = 25R_1 \)

\( \implies \) \( \frac{R_2}{R_1} = \frac{1}{4} \)
But \( R \propto \frac{1}{A} \) (A = Area of cross-section of wire)
\( \therefore \frac{R_2}{R_1} = \frac{A_1}{A_2} = \frac{1}{4} \implies A_2 = 4A_1 \)

\( \implies \pi r_2^2 = 4\pi r_1^2 \implies r_2 = 2r_1 \)
Therefore, the radius of new fuse wire would be larger and twice that of earlier one.

Question. (a) Why does the fuse wire not break when the allowed magnitude of current flows in the circuit?
(b) An electric heater of resistance \( 15 \text{ } \Omega \) draws 5 A current from the service mains in 1.5 hours. Calculate the rate at which heat is developed in the heater.
Answer: (a) When small magnitude of current flows through the fuse wire, small amount of heat is produced. This amount of heat is not enough to melt the fuse wire and is transmitted to the surroundings.
(b) The rate at which heat is developed \( \frac{H}{t} = \frac{I^2Rt}{t} = I^2R = (5)^2 \times 15 = 25 \times 15 = 375 \text{ Js}^{-1} \).

PRACTICE QUESTIONS

Question. Heat produced in a wire of resistance R due to current flowing at constant potential difference for a given time is proportional to
(a) \( R^2 \)
(b) \( R \)
(c) \( \frac{1}{R} \)
(d) \( \sqrt{R} \)
Answer: (c) \( \frac{1}{R} \)

Question. A coil develops heat at the rate of \( 800 \text{ Js}^{-1} \) when 20 volt is applied across its end. The resistance of the coil is
(a) \( 40 \text{ } \Omega \)
(b) \( 20 \text{ } \Omega \)
(c) \( 2 \text{ } \Omega \)
(d) \( 0.2 \text{ } \Omega \)
Answer: (c) \( 0.5 \text{ } \Omega \) [Note: \( R = \frac{V^2}{P} = \frac{20 \times 20}{800} = 0.5 \text{ } \Omega \). Option 'c' in OCR is 2. The correct value is 0.5.]

Question. Mention two practical disadvantages of heating effect of electric current.
Answer: (i) Loss of electrical energy in the form of heat in transmission lines. (ii) Reduces the life and efficiency of electronic components in devices like computers and mobile phones.

Question. An electric heater of resistance \( 6 \text{ } \Omega \) is operated for 10 minutes on a 220 V supply line. Calculate the amount of heat energy liberated in that time.
Answer: \( H = \frac{V^2t}{R} = \frac{220 \times 220 \times 600}{6} = 48,400 \times 100 = 4,840,000 \text{ J} = 4.84 \times 10^6 \text{ J} \).

Question. A current of 3 A passing through a conductor produce 90 J of heat in 10 seconds. What is the resistance of that conductor?
Answer: \( R = \frac{H}{I^2t} = \frac{90}{3^2 \times 10} = \frac{90}{9 \times 10} = 1 \text{ } \Omega \).

Question. (a) State the working principle of an electric fuse.
(b) Out of 5 A fuse and 15 A fuse which will you prefer to use for the lighting circuit?
(c) Given reason: (i) Electric bulbs are usually filled with chemically inactive gases. (ii) Fuse wire is placed in series with the device.
Answer: (a) Working principle: Heating effect of electric current (Joule's heating). It melts when current exceeds its rating.
(b) A 5 A fuse is preferred for lighting circuits.
(c) (i) To prevent oxidation of the filament and prolong its life. (ii) To ensure that if the fuse melts, the current to the device is immediately interrupted.

Electric Power and Energy

Question. If \( R_1 \) and \( R_2 \) be the resistance of the filament of 40 W and 60 W respectively operating 220 V, then
(a) \( R_1 < R_2 \)
(b) \( R_2 < R_1 \)
(c) \( R_1 = R_2 \)
(d) \( R_1 \geq R_2 \)
Answer: (b) \( R_2 < R_1 \)

Question. The resistance of hot filament of the bulb is about 10 times the cold resistance. What will be the resistance of 100 W-220 V lamp, when not in use?
(a) \( 48 \text{ } \Omega \)
(b) \( 400 \text{ } \Omega \)
(c) \( 484 \text{ } \Omega \)
(d) \( 48.4 \text{ } \Omega \)
Answer: (d) \( 48.4 \text{ } \Omega \)

Question. If P and V are the power and potential of device, the power consumed with a supply potential \( V_1 \) is
(a) \( \frac{V_1^2}{V^2}P \)
(b) \( \frac{V^2}{V_1^2}P \)
(c) \( \frac{V}{V_1}P \)
(d) \( \frac{V_1}{V}P \)
Answer: (a) \( \frac{V_1^2}{V^2}P \)

Question. For maximum power consumption, all given resistors should be connected in
(a) parallel
(b) series
(c) some in parallel and some in series
(d) none of the options
Answer: (a) parallel

Question. Which of the following does not represent electric power?
(a) \( I^2R \)
(b) \( IR^2 \)
(c) \( VI \)
(d) \( V^2/R \)
Answer: (b) \( IR^2 \)

Question. One kilowatt hour is equal to
(a) \( 36 \times 10^6 \text{ J} \)
(b) \( 3.6 \times 10^6 \text{ J} \)
(c) \( 0.36 \times 10^6 \text{ J} \)
(d) All of the options
Answer: (b) \( 3.6 \times 10^6 \text{ J} \)

Question. If \( R_1 \) and \( R_2 \) are respectively the filament resistances of a 200 W bulb and 100 W bulb design to operate on the same voltage, then
(a) \( R_1 = 4R_2 \)
(b) \( R_2 = 4R_1 \)
(c) \( R_2 = 2R_1 \)
(d) \( R_1 = 2R_2 \)
Answer: (c) \( R_2 = 2R_1 \)

Question. Two heater wires of same length and same material but of different thickness are connected in series across a power supply. The power dissipated will be
(a) more in thicker wire
(b) more in thinner wire
(c) same in both
(d) cannot say
Answer: (b) more in thinner wire

Question. The wattage of a bulb is 24 W when it is connected to a 12 V battery. Calculate its effective wattage if it operates on a 6 V battery (Neglect the change in resistance due to unequal heating of the filament in the two cases).
Answer: Given: \( P_1 = 24 \text{ W}, V_1 = 12 \text{ V}, P_2 = ?, V_2 = 6 \text{ V} \)
Using \( P = \frac{V^2}{R} \)
\( \frac{P_2}{P_1} = \frac{V_2^2}{V_1^2} \)

\( \implies \) \( P_2 = \left(\frac{V_2}{V_1}\right)^2 \times P_1 = \left(\frac{6}{12}\right)^2 \times 24 = \frac{1}{4} \times 24 = 6 \text{ W} \).

Question. For the same potential difference, out of the two, a room heater of 1000 W and an electric motor of 2 kW, which has a greater resistance?
Answer: Power of room heater \( P_1 = 1000 \text{ W} \). Power of electric motor \( P_2 = 2 \text{ kW} = 2000 \text{ W} \).
Using \( P = \frac{V^2}{R} \). For same potential \( P \propto \frac{1}{R} \).
\( \frac{P_1}{P_2} = \frac{R_2}{R_1} \implies \frac{1000}{2000} = \frac{R_2}{R_1} = \frac{1}{2} \).

\( \implies \) \( R_1 = 2R_2 \).
So room heater has greater resistance.

Question. In the circuit given below [25W, 40W, 60W bulbs in series]: (a) Would any bulb glow when plug key is in open position? (b) Write the order of brightness of the bulb when key is closed. Give reason.
Answer: (a) No bulb will glow when the plug key is in open position because no current will flow through the circuit.
(b) Power of bulb \( P = \frac{V^2}{R} \). Since \( V \) is same, \( R \propto \frac{1}{P} \). So, resistance order is \( R_{25} > R_{40} > R_{60} \). In series, current \( I \) is same. According to Joule's law, \( H \propto R \). Hence, \( H_{25} > H_{40} > H_{60} \). The brightness order is: 25W bulb > 40W bulb > 60W bulb.

Question. The electric power consumed by a device may be calculated by either of the two expression \( P = I^2R \) or \( P = \frac{V^2}{R} \). The first expression indicates that it is directly proportional to R whereas the second expression indicates inverse proportionality. How can the seemingly different dependence of P on R in these expression be explained?
Answer: (a) In series, the current in each resistor is same and constant. Therefore, \( P = I^2R \) is used for series connection.
(b) In parallel, voltage across each resistance is same and constant. Therefore \( P = \frac{V^2}{R} \) is used where resistors are connected in parallel combination.

Question. (a) It would cost a man Rs. 3.50 to buy 1.0 kWh of electrical energy from the Main Electricity Board. His generator has a maximum power of 2.0 kW. The generator produces energy at this maximum power for 3 hours. Calculate how much it would cost to buy the same amount of energy from the Main Electricity Board.
(b) A student boils water in an electric kettle for 20 minutes. Using the same mains supply he wants to reduce the boiling time of water. To do so should he increase or decrease the length of the heating element? Justify your answer.
Answer: (a) \( E = P \times T = 2.0 \text{ kW} \times 3 \text{ h} = 6 \text{ kWh} \).
Cost of buying electricity = \( 6 \times 3.50 = \text{Rs. } 21.0 \).
(b) To reduce the boiling time using the same mains supply, the rate of heat production (\( P \)) should be large. Since \( P = \frac{V^2}{R} \) and \( V \) is constant, \( R \) should be decreased. Since \( R \propto l \), the length of the heating element should be decreased.

Question. An electric motor rated 1100 W is connected to 220 V mains. Find: (i) The current drawn from the mains. (ii) Electric energy consumed if the motor is used for 5 hours daily for 6 days. (iii) Total cost of energy consumed if the rate of one unit is Rs. 5.
Answer: Given: \( P = 1100 \text{ W}, V = 220 \text{ V} \).
(i) Current drawn \( I = \frac{P}{V} = \frac{1100}{220} = 5 \text{ A} \).
(ii) Energy \( E = P \times t = 1100 \text{ W} \times 5 \text{ h} \times 6 \text{ days} = 33000 \text{ Wh} = 33 \text{ kWh} \).
(iii) Cost \( = 33 \text{ units} \times \text{Rs. } 5 = \text{Rs. } 165 \).

Question. (a) Write two point of difference between electric energy and electric power.
(b) Out of 60 W and 40 W lamps, which one has higher electrical resistance when in use.
(c) What is the commercial unit of electric energy? Convert it into joules.
Answer: (a) Difference between electric energy and electric power:
Electrical energy: (i) The work done or energy supplied by the source in maintaining the flow of electric current. \( H = VIt = \frac{V^2t}{R} = I^2Rt \). (ii) It is equal to the product of power and time \( E = P \times t \). (iii) SI unit is joule (J).
Electric power: (i) The time rate at which electric energy is consumed or dissipated by an electrical device. \( P = VI = \frac{V^2}{R} = I^2R \). (ii) It equals the rate of doing work by an energy source \( P = \frac{W}{t} \). (iii) SI unit is watt (W).
(b) For the same voltage, \( R \propto \frac{1}{P} \). Therefore, a 40 W lamp has higher electrical resistance.
(c) Commercial unit: Kilowatt hour (kWh). \( 1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J} \).

Question. (a) Define electric power. Express it in terms of potential difference V and resistance R.
(b) An electrical fuse is rated at 2A. What is meant by this statement?
(c) An electric iron of 1 kW is operated at 220 V. Find which of the following fuses that respectively rated at 1 A, 3 A and 5 A can be used in it.
Answer: (a) Electric power: It is the rate of doing work by an energy source or the rate at which electrical energy is consumed per unit time. \( P = \frac{V^2}{R} \).
(b) It means the maximum current that can safely flow through it is 2 A. The fuse wire will melt if the current exceeds 2 A.
(c) Given: \( P = 1000 \text{ W}, V = 220 \text{ V} \). Current drawn \( I = \frac{P}{V} = \frac{1000}{220} \approx 4.54 \text{ A} \). To run this electric iron, a rated fuse of 5 A should be used.

Question. The watt is a
(a) \( \text{Js}^{-1} \)
(b) V-A
(c) both (a) and (b)
(d) neither (a) nor (b)
Answer: (c) both (a) and (b)

Question. A 60 W electric lamp gives off energy in the form of light at a rate of 7.5 joule per second. What percentage of energy does the lamp transform into light energy?
Answer: Percentage efficiency \( = \frac{\text{Output Power}}{\text{Input Power}} \times 100 = \frac{7.5}{60} \times 100 = 12.5\% \).

Question. An electric motor takes 5 A from 220 V line. Determine the power of a motor and the energy consumed in 2 hour.
Answer: Power \( P = VI = 220 \times 5 = 1100 \text{ W} = 1.1 \text{ kW} \). Energy \( E = P \times t = 1.1 \text{ kW} \times 2 \text{ h} = 2.2 \text{ kWh} = 7.92 \times 10^6 \text{ J} \).

Question. (i) Define electric energy. In which form it appears? (ii) What is the difference between kilowatt and kilowatt hour?
Answer: (i) Electric energy is the work done by an electrical source to maintain current flow. It usually appears in the form of heat. (ii) Kilowatt (kW) is a unit of power, whereas kilowatt hour (kWh) is a unit of energy.

Question. (a) An electric kettle of 2 kW is used for 2 h. Calculate the energy consumed in (i) Kilowatt hour (ii) Joules.
(b) When a constant current is for a time of t seconds, how can you increase the heat produced to four times?
Answer: (a) (i) Energy \( E = 2 \text{ kW} \times 2 \text{ h} = 4 \text{ kWh} \). (ii) \( 4 \times 3.6 \times 10^6 \text{ J} = 1.44 \times 10^7 \text{ J} \). (b) Since \( H = I^2Rt \), heat can be increased four times by doubling the current (\( I \)) or making the resistance (\( R \)) four times.

Question. Calculate the total cost of running the following electrical devices in the month of September, if the rate of 1 unit of electricity is Rs. 6.00. (i) Electric heater of 1000 W for 5 hours daily. (ii) Electric refrigerator of 400 W for 10 hours daily.
Answer: (i) Heater energy \( = 1 \text{ kW} \times 5 \text{ h} \times 30 \text{ days} = 150 \text{ kWh} \). (ii) Fridge energy \( = 0.4 \text{ kW} \times 10 \text{ h} \times 30 \text{ days} = 120 \text{ kWh} \). Total units \( = 270 \text{ units} \). Total cost \( = 270 \times \text{Rs. } 6 = \text{Rs. } 1620 \).

Question. A bulb is rated 40 W; 220 V. Find the current drawn by it when it is connected to a 220 V supply. Also find its resistance. If the given bulb is replaced by a bulb of rating 25 W; 220 V, will there be any change in the value of current and resistance? Justify your answer and determine the change.
Answer: For 40 W bulb: \( I = \frac{40}{220} = 0.18 \text{ A} \); \( R = \frac{220^2}{40} = 1210 \text{ } \Omega \). For 25 W bulb: \( I = \frac{25}{220} = 0.11 \text{ A} \); \( R = \frac{220^2}{25} = 1936 \text{ } \Omega \). Yes, the current decreases and resistance increases because \( I \propto P \) and \( R \propto \frac{1}{P} \) for a constant voltage.

Question. A boy records that 4000 joule of work is required to transfer 10 coulomb of charge between two points of a resistor of \( 50 \text{ } \Omega \). The current passing through it is
(a) 2 A
(b) 4 A
(c) 8 A
(d) 16 A
Answer: (c) 8 A [Note: \( V = \frac{W}{Q} = 400 \text{ V} \). \( I = \frac{V}{R} = \frac{400}{50} = 8 \text{ A} \).]

Question. The least resistance obtained by using \( 2 \text{ } \Omega, 4 \text{ } \Omega, 1 \text{ } \Omega, \text{ and } 100 \text{ } \Omega \) is
(a) < \( 100 \text{ } \Omega \)
(b) < \( 4 \text{ } \Omega \)
(c) < \( 1 \text{ } \Omega \)
(d) > \( 2 \text{ } \Omega \)
Answer: (c) < \( 1 \text{ } \Omega \)

Question. Two wires of same length and area, made of two materials of resistivity \( \rho_1 \) and \( \rho_2 \) are connected in parallel to a source of potential V. The equivalent resistivity for the combination is
(a) \( \rho_1 + \rho_2 \)
(b) \( \frac{\rho_1\rho_2}{\rho_1 + \rho_2} \)
(c) \( \frac{\rho_1 + \rho_2}{\rho_1\rho_2} \)
(d) \( \frac{2\rho_1\rho_2}{\rho_1 + \rho_2} \)
Answer: (b) \( \frac{\rho_1\rho_2}{\rho_1 + \rho_2} \)