CBSE Class 10 Science Electricity VBQs Set 05

Case Based Questions

Read the passage given below and answer the following questions :
The rate of flow of charge is called electric current. The SI unit of electric current is Ampere (A). The direction of flow of current is always opposite to the direction of flow of electrons in the current.
The electric potential is defined as the amount of work done in bringing a unit positive test charge from infinity to a point in the electric field. The amount of work done in bringing a unit positive test charge from one point to another point in an electric field is defined as potential difference.
\( V_{AB} = V_B - V_A = \frac{W_{BA}}{q} \)
The SI unit of potential and potential difference is volt.

Question. The 2 C of charge is flowing through a conductor in 100 ms, the current in the circuit is
(a) 20 A
(b) 2 A
(c) 0.2 A
(d) 0.02 A
Answer: (a) 20 A

Question. Which of the following is true?
(a) Current flows from positive terminal of the cell to the negative terminal of the cell outside the cell.
(b) The negative charge moves from lower potential to higher potential.
(c) The direction of flow of current in same as the direction of flow of positive charge.
(d) All of the options
Answer: (d) All of the options

Question. The potential difference between the two terminals of a battery, if 100 joules of work is required to transfer 20 coulombs of charge from one terminal of the battery to other is
(a) 50 V
(b) –5 V
(c) 0.5 V
(d) 500 V
Answer: (a) 50 V

Question. The number of electrons flowing per second in a conductor if 1 A current is passing through it
(a) \( 6.25 \times 10^{20} \)
(b) \( 6.25 \times 10^{19} \)
(c) \( 6.25 \times 10^{18} \)
(d) \( 6.25 \times 10^{-19} \)
Answer: (c) \( 6.25 \times 10^{18} \)

Question. The voltage can be written as
(a) Work done \( \times \) charge \( \times \) time
(b) \( \frac{\text{Work done}}{\text{Current } \times \text{ time}} \)
(c) \( \frac{\text{Work done } \times \text{ time}}{\text{Current}} \)
(d) Work done \( \times \) charge
Answer: (b) \( \frac{\text{Work done}}{\text{Current } \times \text{ time}} \)

Read the passage given below and answer the following questions :
The heating effect of current is obtained by transformation of electrical energy in heat energy. Just as mechanical energy used to overcome friction is covered into heat, in the same way, electrical energy is converted into heat energy when an electric current flows through a resistance wire. The heat produced in a conductor, when a current flows through it is found to depend directly on (a) strength of current (b) resistance of the conductor (c) time for which the current flows.
The mathematical expression is given by \( H = I^2Rt \).
The electrical fuse, electrical heater, electric iron, electric geyser etc. all are based on the heating effect of current.

Question. What are the properties of heating element?
Answer: A heating element should have high resistivity and a high melting point so that it can produce a large amount of heat without melting.

Question. What are the properties of electric fuse?
Answer: An electric fuse wire should have a low melting point and relatively high resistance compared to the connecting wires of the circuit.

Question. Initially 16 J of heat is dissipated in a heating device, now when the current is doubled and time is halved, then how much heat is dissipated?
Answer: Heat \( H = I^2Rt \).
Given \( H_1 = 16 \text{ J} \).
New current \( I' = 2I \) and new time \( t' = t/2 \).
New heat \( H_2 = (2I)^2 \times R \times (t/2) \)

\( \implies H_2 = 4I^2 \times R \times \frac{t}{2} \)

\( \implies H_2 = 2 \times I^2Rt \)

\( \implies H_2 = 2 \times 16 = 32 \text{ J} \).

Question. A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. Find the new radius of the wire.
Answer: For a fuse wire, the current \( I \) at which it melts is related to its radius \( r \) as \( I^2 \propto r^3 \).
\( \frac{I_2^2}{I_1^2} = \frac{r_2^3}{r_1^3} \)

\( \implies \left(\frac{10}{5}\right)^2 = \left(\frac{r_2}{r_1}\right)^3 \)

\( \implies 4 = \left(\frac{r_2}{r_1}\right)^3 \)

\( \implies r_2 = r_1 \times \sqrt[3]{4} \approx 1.587 r_1 \).
The new radius should be approximately 1.587 times the original radius.

A & R Questions

In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.

Question. Assertion : Fuse wire must have high resistance and low melting point.
Reason : Fuse is used for small current flow only.
Answer: (c) Assertion is correct statement but reason is wrong statement.

Question. Assertion : Electron has a negative charge.
Reason : Electrons move always from a region of higher potential to a region of lower potential.
Answer: (c) Assertion is correct statement but reason is wrong statement.

Question. Assertion : When a wire is stretched to three times of its length, its resistance becomes 9 times.
Reason : Resistance is directly proportional to length of wire.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).

Multiple Choice Questions

Question. When the lightning strikes, 100 C of charge flow in 0.02 s. What is the current?
(a) 0.2 A
(b) 200 A
(c) 0.05 A
(d) 5000 A
Answer: (d) 5000 A

Question. When a charge of 5 C flows through a bulb, 10 J of energy in the form of light and heat is dissipated. What is the potential difference across the bulb?
(a) 50 V
(b) 2 V
(c) 10 V
(d) 0.5 V
Answer: (b) 2 V

Question. A battery of 3 V is connected to a lamp. When 4 C of charge flows through the lamp, the energy dissipated from the lamp is 10 J. What is the work done by the battery to drive the charge of 4 C around the circuit?
(a) 3 J
(b) 4 J
(c) 10 J
(d) 12 J
Answer: (d) 12 J

Question. Which device maintains a potential difference across a conductor?
(a) Cell or battery eliminator
(b) Ammeter
(c) Voltmeter
(d) None of the options
Answer: (a) Cell or battery eliminator

Question. Wire A and wire B has the following ratios \( \frac{\text{length } L_A}{\text{length } L_B} = \frac{5}{18} \); \( \frac{\text{diameter } D_A}{\text{diameter } D_B} = \frac{2}{3} \); \( \frac{\text{resistivity } \rho_A}{\text{resistivity } \rho_B} = \frac{4}{9} \). What is the ratio of the resistance of wire A to the resistance of wire B?
(a) 5 : 18
(b) 10 : 54
(c) 18 : 5
(d) 54 : 10
Answer: (a) 5 : 18

VSA Type Questions

Question. Three resistors of 3 \( \Omega \), 6 \( \Omega \) and 4 \( \Omega \) are connected in series. Calculate the total resistance of the combination.
Answer: Total resistance \( R_s = R_1 + R_2 + R_3 = 3 + 6 + 4 = 13\text{ }\Omega \).

Question. An electric bulb of resistance 80 \( \Omega \) draws a current of 0.75 A. Find the line voltage.
Answer: Using Ohm's Law: \( V = IR = 0.75 \times 80 = 60 \text{ V} \).

SA I Type Questions

Question. What do you understand by the term “Potential difference between two points is 1 volt”? Name the device that helps to measure the potential difference across a resistor?
Answer: Potential difference between two points is 1 volt if 1 joule of work is done in moving an electric charge of 1 coulomb from one point to the other. The device used to measure potential difference is a Voltmeter.

Question. Why are filaments of incandescent lamps made of thin tungsten wire?
Answer: Tungsten is used because it has a very high melting point (\( 3380^\circ\text{C} \)) and does not oxidize readily at high temperatures. It is made thin to increase its resistance (\( R \propto 1/A \)), which produces more heat and light when current flows through it.

Question. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many bulbs can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Answer: Maximum total power \( P_{total} = V \times I = 220 \times 5 = 1100 \text{ W} \).
Number of bulbs \( n = \frac{P_{total}}{P_{bulb}} = \frac{1100}{10} = 110 \).
Therefore, 110 bulbs can be connected in parallel.

Question. What should be the characteristics of a heating element?
Answer: A heating element should have high resistivity and a high melting point.

SA II Type Questions

Question. The equivalent resistance of series combination of four equal resistors is S. If they are joined in parallel, the total resistance is P. The relation between S and P is given by S = nP, then what is the minimum possible value of n?
Answer: Let the resistance of each resistor be \( R \).
Equivalent resistance in series, \( S = 4R \).
Equivalent resistance in parallel, \( P = R/4 \).
Given \( S = nP \)

\( \implies 4R = n(R/4) \)

\( \implies n = 16 \).

Question. For a heater rated at 4 kW and 220 V, calculate (a) the current, (b) the resistance of the heater, (c) the energy consumed in 2 hours, (d) the cost, if 1 kW h is priced at 50 paise.
Answer: (a) Current \( I = \frac{P}{V} = \frac{4000 \text{ W}}{220 \text{ V}} \approx 18.18 \text{ A} \).
(b) Resistance \( R = \frac{V}{I} = \frac{220}{18.18} \approx 12.1 \text{ }\Omega \).
(c) Energy \( E = P \times t = 4 \text{ kW} \times 2 \text{ h} = 8 \text{ kWh} \).
(d) Cost \( = 8 \text{ kWh} \times 0.50 \text{ Rs./kWh} = \text{Rs. } 4.00 \).

Question. The specific resistance of the material of a wire is \( 44 \times 10^{-8}\text{ }\Omega\text{ m} \). If the resistance of the wire is 14 \( \Omega \) and its diameter is 1 mm, calculate the length of the wire.
Answer: Given \( \rho = 44 \times 10^{-8}\text{ }\Omega\text{ m} \), \( R = 14\text{ }\Omega \), diameter \( D = 1\text{ mm} = 10^{-3}\text{ m} \).
Radius \( r = 0.5 \times 10^{-3}\text{ m} \).
Area \( A = \pi r^2 = \frac{22}{7} \times (0.5 \times 10^{-3})^2 \approx 0.785 \times 10^{-6}\text{ m}^2 \).
Using \( R = \rho \frac{l}{A} \)

\( \implies l = \frac{RA}{\rho} = \frac{14 \times 0.785 \times 10^{-6}}{44 \times 10^{-8}} \approx 25 \text{ m} \).

Question. When two unknown resistances are connected in series and in parallel, their equivalent resistances are 9 \( \Omega \) and 2 \( \Omega \) respectively. What are the unknown resistances?
Answer: Let the resistances be \( R_1 \) and \( R_2 \).
In series: \( R_1 + R_2 = 9 \) ... (i)
In parallel: \( \frac{R_1R_2}{R_1 + R_2} = 2 \implies R_1R_2 = 2 \times 9 = 18 \) ... (ii)
Using \( (R_1 - R_2)^2 = (R_1 + R_2)^2 - 4R_1R_2 = 9^2 - 4(18) = 81 - 72 = 9 \).
\( R_1 - R_2 = 3 \) ... (iii)
Solving (i) and (iii): \( 2R_1 = 12 \implies R_1 = 6\text{ }\Omega \) and \( R_2 = 3\text{ }\Omega \).