Answer The Following Question.
1. What does an electric circuit mean?
Ans: An electric circuit is a closed and continuous path through which electric current flows. It typically consists of electrical components like a power source, wires, and a load (e.g., a bulb).
2. Define the unit of current.
Ans: The SI unit of electric current is the ampere (A). One ampere is defined as the current when 1 coulomb of charge flows through a cross-section of a conductor in 1 second.
3. Calculate the number of electrons constituting one coulomb of charge.
Ans : Charge on one electron = 1.6 x 10-19 coulomb.
No of electron in one coulomb of charge = 1/1.6 x 10-19
= 6.25 x 1018
Your solution is correct! Here’s a cleaner version:
Calculate the number of electrons constituting one coulomb of charge.
Charge on one electron = 1.6 x 10-19 columb.
Number of electrons in one coulomb of charge == 1/1.6 x 10-19
= 6.25 x 1018 electrons
Thus, 6.25 x 1018 electrons constitute one coulomb of charge.
4. Name a device that helps to maintain a potential difference between across a conductor.
Ans : Battery
5. What is meant by saying that the potential difference between two points is 1 v?
Ans: The potential difference between two points is said to be 1 volt if 1 joule of work is required to move a charge of 1 coulomb from one point to another.
6. How much energy is given to each coulomb of charge passing through a 6-volt battery?
Ans : Potential difference (V) = 6 V
Charge (Q) = 1 C
Energy = total work done (W) = Q x V = 1×6 = 6 joule.
7. On what factor does the resistance of a conductor depend?
Ans : The resistance of a conductor depends upon the following factors:
(a) Length of the conductor
(b) Cross-sectional area of the conductor
(c) Material of the conductor
(d) Temperature of the conductor
8. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Ans: The current flows more easily through a thick wire than a thin wire of the same material, when connected to the same source. This is because resistance decreases with increased thickness. A thicker wire offers less resistance to the flow of current compared to a thinner wire.
9. Let the resistance of an electric component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Ans: What change will occur in the current when the potential difference decreases to half?
According to Ohm’s Law I=V/R , if the potential difference is halved while the resistance remains constant, the current will also decrease to half of its original value.
10. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Ans: Coils are made of alloys because:
Alloys have higher resistivity than pure metals, which helps in generating more heat.
Alloys are less likely to oxidize at high temperatures, giving the coil a longer lifespan.
11. Use the data in Table 12.2 to answer the following:
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?
Ans : (a) Resistivity of iron = 10.0×10−8Ωm
Resistivity of mercury =94.0×10−8Ωm
Resistivity of mercury is more than that of iron. This implies that iron is a better conductor than mercury.
(b) It can be observed from Table 12.2 that the resistivity of silver is the lowest among the listed materials. Hence, it is the best conductor.
12. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V, each, a 5 Ώ resistor, 8 Ώ resistors and a 12 Ώ and a plug key, all connected in series.
Ans : The schematic diagram of circuit is as follows:
13. Redraw the circuit of question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ώ resistors. What would be the reading in the ammeter and voltmeter?
Ans : Here ammeter A has been joined in series of circuit and voltmeter V is joined in parallel to 12 ohms’ resistor.
Total voltage of battery V = 3×2 = 6 V.
Total resistance R = R1+ R2+ R3 = 5 Ω +8 Ω +12 Ω = 25 Ω
Ammeter reading (current) = I = V/R = 6/25 = 0.24 A.
Voltmeter reading = IR = 0.24 x 12 = 2.88 V.
14. Judge the equivalent resistance when the following are connected in parallel:
(a) 1 Ω and 106 Ω
(b) 1 Ω, 103 Ω and 106 Ω
Ans : When the resistances are joined in parallel, the resultant resistance in parallel arrangement is given by:
1/R = 1/R1 + 1/R2 + 1/R3
(a) 1/R = 1/1+ 1/106 = 1+ 10-6
R = 1 Ω
(b) 1/R = 1/1 + 1/103 + 1/106 = 1 + 10-3 + 10-6
R= 1 Ω
15. An electric lamp of 100 Ω, a toaster of resistance 50 Ω and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Ans : Here, voltage (V) = 220 V
R1 = 100 Ω, R2 = 50 Ω and R3 = 500 Ω
1/R = 1/R1 + 1/R2 + 1/R3 —–
1/R = 1/100+1/50 +1/500 = 16/500
R = 500/16 = 31.25 Ω
The resistance of electric iron, which draws as much current as all three appliances take together = R = 31.25 Ω.
Current passing through electric iron (I) = V/R = 220/31.25 = 7.04 A.
16. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Ans: Your solution is correct! Here’s a refined and more concise version:
Advantages of connecting electrical devices in parallel with the battery:
Same voltage across each device: Each device receives the same voltage, and the current through each device depends on its resistance.
Independent control: Separate on/off switches can be applied to each device.
Reduced total resistance: The total resistance decreases, allowing more current to be drawn from the battery.
Device independence: If one device is damaged, the others continue to work properly.
17. How can three resistors of resistance 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω (b) 1 Ω?
Ans : (a) If we connect resistance of 3 Ω and 6 Ω in parallel and then resistance of 2 Ω is connected in series of the combination, then total resistance of combination is 4 Ω.
(b) If all the three resistance are joined in parallel the resultant resistance will be 1 Ω.
18. What is (a) the highest, (b) the lowest total resistance that can be secured by combination of four resistances of 4 Ω, 8 Ω, 12 Ω and 24 Ω?
Ans : (a) To obtain highest resistance, all the four resistances must be connected in series arrangement. In that case resultant R = R1 + R2 + R3
= 4+8+12 48 Ω
(b) To obtain lowest resistance, all the four resistance must be connected in parallel arrangement. 1/R = 1/R1 + 1/R2 + 1/R3
= 1/4 +1/8 +1/12 + 1/24 = 12/24 Ω
= 24/12 = 2 Ω
19. Why does the cord of an electric heater not glow while the heating element does?
Ans: The cord of an electric heater does not glow while the heating element does because of the difference in resistance between the two.
Heating element: The heating element is made of a material with a high resistance (like nichrome), which generates a lot of heat when current passes through it. The high resistance causes the element to heat up, reaching a temperature where it starts to glow.
Cord: The cord, on the other hand, is made of a material with low resistance (such as copper or aluminum), so it does not generate significant heat. As a result, while the cord carries the same current, the heat produced in the cord is much less, and it does not reach a temperature high enough to cause it to glow.
Thus, while both the cord and the heating element carry the same current (since they are in series), the difference in resistance results in the heating element glowing but the cord remaining cool.
20. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Ans : Charge transferred (Q) = 96000 C, time = 1 hour = 60 x 60 = 3600 s and potential difference (V) = 50 V.
Heat generated (H) = VIt = V.Q = 50 x 96000 = 4800000 j = 4.8 x 106 j.
21. An electric iron of resistance 20 takes a current of 5 A. Calculate the heat developed in 30 s.
Ans : Resistance of electric iron (R) = 20 Ω, current (I) = 5 A and time = 30 s.
Heat generated (H) = I2Rt = 52 x 20 x 30 = 15000 j.
22. What determines the rate at which energy is delivered by a current?
Ans : Electric power determines the rate at which energy is delivered by a current.
23. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and energy consumed in 2 h.
Ans : It is given that current drawn by electric motor (I) = 5 A. the line voltage V = 220 V time (t) = 2 h.
Power of motor (P) = P = VI = 220 x 5 = 1100 W and the energy consumed (E) = Pt
1100 W x 2 h = 2200 Wh or 2.2 kWh.
24. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is:
(a) 1/25
(b) 1/5
(c) 5
(d) 25
Ans : (d) 25
25. Which of the following terms does not represent electrical power in a circuit?
(a) I2R
(b) IR2
(c) VI
(d) V2/R
Ans : (b) IR2
26. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be:
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Ans : (d) 25 W
27. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combination would be:
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
Ans : (c) 1:4
28. How is a voltmeter connected in the circuit to measure the potential difference between two points?
Ans : A voltmeter is always connected in parallel to resistance across the point between which the potential difference is to be measured.
29. A copper wire has diameter 0.5 mm and resistivity of 1.6 x 10-8m. what will be the length of this wire to make its resistance 10? How much does the resistance change if the diameter is doubled?
Ans : Diameter of wire (d) = 0.5 mm, resistivity (ρ) 1.6 x 10-8 Ωm, resistance (R) = 10 Ω.
R = ρL/A
L= πD2R/4ρ
= 22 x (5 x 10-4)2/ 7 x 4 x 1.6 x 10-8 = 122.5 m
If the diameter is doubled for given length of given material resistance is inversely proportional to the cross-section area of wire.
30. The value of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:
Plot a graph between V and I and calculate the resistance of that resistor.
Ans : From the given data the I-V graph is a straight line as shown below:
The slope of the line gives the value of Resistance(R) as,
Slope = 1/R = BC/AC = 2/6.8
R = 6.8/2 = 3.4 Ω
Therefore,the resistance of register is 3.4Ω.
31. When a 12 v battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Ans : Voltage of battery = V = 12 V, Current (I) = 2.5 mA = 2.5 x 10-3 A
Resistance (R) = V/I = 12V/ 2.5 x 10-3 A = 4800 Ω.
32. A battery of 9 V is connected in series with resistance of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω respectively. How much current would flow through the 12 resistor?
Ans : Potential difference (V) = 9 V.
Total resistance (R) = R1+ R2+ R3+R4 +R5
= 0.2 +0.3 + 0.5 + 0.5 + 12 = 13.4 Ω
Current in the circuit (I) = V/R = 9 V / 13. 4 Ω = 0.67 A.
In series circuit same current flows through all the resistance, hence current of 0.67 A will flow through 12 Ω resistor.
33. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Ans : Let a resistor of 176 Ω are joined in parallel. Then their combined resistance (R)
1/R = 1/176 + 1/176 …… times = n/176 or R = 176/n Ω
It is given that V= 220 V and I = 5 A
R = V/I or 176/n = 220/5 = 44 Ω
n = 176/44 = 4, 4 resistors should be joined in parallel.
34. Show how you would connect three resistors, each of resistance 6 Ω so that the combination has resistance of (i) 9 Ω (ii) 4 Ω.
Ans : It is given here that R1 = R2 = 6 Ω.
(i) To get net resistance of 9 Ω we should join three resistors as below:
(ii) To get 4 Ω net resistance we should join three resistors as below:
35. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Ans : Each bulb is rated as 10 W, 220 V, It draws a current (I) = P/V = 10 W/220 V
= 1/22 A.
As the maximum allowable current is 5 A and all lamps are connected in parallel, hence maximum number of bulbs joined in parallel with each other = 5 x 22 = 110.
36. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B. Each of 24 Ω resistances, which may be used separately, in series or in parallel. What are the currents in the three cases?
Ans : It is given that potential difference (V) = 220 V.
Resistance of coil (A) = Resistance of coil (B) = 24 Ω
(i) When either coil is used separately, the circuit (I) = V/R = 220 V/ 24 Ω
= 9.2 A.
(ii) When two coils are used in series total resistance (R)
= R1 + R2 = 24 +24 = 48 Ω
Current flowing (I) = V/ R = 220 V/ 48 Ω = 4.6 A.
(iii) When two coils are joined in parallel. Total resistance (R) = 1/24 + 1/24
= 2/24, R = 12 Ω.
Current (I) = V/R = 220 V / 12 Ω = 18.3 A.
37. Compare the power used in the 2 Ω resistor in each of the following circuits:
(i) a 6-volt battery in series with 1 Ω and 2 Ω resistors and,
(ii) a 4 V battery in parallel with 12 Ω and Ω resistors.
Ans : (i) When a 2 Ω resistor is joined t a 6 V battery in series with 1 Ω and 2 Ω resistors. Total resistance (R) = 2 + 1 + 2 = 5 Ω.
Current (I) = 6 V/5 Ω = 1.2 A
Power used in 2 A resistor = I2R = 2.88 W
(ii) When 2 Ω resistor is joined to a 4 V battery in parallel with 12 Ω resistor and 2 Ω resistors, the current flowing in 2 Ω = 4 V/ 2 Ω = 2 A/.
Power used in 2 Ω resistor = I2R = 8 W
Ratio = 2.88/8 = 0.36: 1.
38. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Ans : Current drawn by 1st lamp rated 100 W at 220 V = P/V = 100/ 220 = 5/11 A.
Current drawn by 2nd lamp rated 60 W at 220 V = 60/220 = 3/11 A.
In parallel arrangement the total current = I1 +I2 = 3/11+ 5/11 = 8/11 = 0.73 A.
39.Which uses more energy, a 250 W TV set in 1 hour, or a 1200 W toaster in 10 minutes?
Ans : Energy used by a TV set of power 250 W in 1 hour = P x t = 250 Wh.
Energy used by toaster of power 1200 W in 10 minute (10/60 h)
= P x t = 1200 W x 10/60 h = 200 Wh.
40. An electric heater of resistance 8 draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Ans : Resistance of electric heater (R) = 8 Ω, current (I) = 15 A.
Rate at which heat developed in the heater = I2Rt/t = 15 x 15 x 8 = 1800 W.
41. Explain the following:
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electric transmission?
Ans: Here are the explanations in a more concise and clear format:
(a) Why is tungsten used almost exclusively for the filament of electric lamps?
Tungsten is used for the filament of electric lamps because it has a very high melting point and can withstand the high temperatures generated when the filament glows. This allows the filament to glow without melting.
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
The conductors of heating devices are made of alloys because alloys have higher resistivity and higher melting points compared to pure metals. This allows them to generate more heat without getting damaged or oxidizing at high temperatures.
(c) Why is the series arrangement not used for domestic circuits?
In a series circuit, the same current flows through all appliances. This is not practical in domestic circuits because different appliances require different amounts of current, and if one appliance fails, the entire circuit is broken, causing all appliances to stop working.
(d) How does the resistance of a wire vary with its area of cross-section?
The resistance of a wire is inversely proportional to its cross-sectional area. As the area increases, the resistance decreases because there is more space for the electrons to flow through.
(e) Why are copper and aluminium wires usually employed for electric transmission?
Copper and aluminium wires are used for electric transmission because they are good conductors of electricity with low resistivity. They are also ductile, meaning they can be drawn into thin wires, which makes them ideal for transmission over long distances.