12. ELECTRICITY
Questions:
1. The SI unit of electric current is: [SSLC July, 2018-19] (a) Ampere (b) Ohm (c) Volt (d) Watt
Ans: (a) Ampere
2. Define the SI unit of potential difference.
Ans: If amount of work done in bringing one coulomb charge from one point to the other is 1 joule then potential difference between two points is said to be 1 volt.
3. State the relationship between 1 ampere and 1 coulomb.
Ans: 1 ampere = 1 coulomb
1 second
4. 12.400 J of heat is produced in 4 s in a 4Ω resistor. Find potential difference across the resistor.
Ans: V2 R x t = 400
Or V2 = 400
Or V = 20 volt
5. State the factors on which at a given temperature the resistance of a cylindrical conductor depends. State the SI unit of resistivity. [SSLC July, 2018-19]
Ans: Factors on which the resistance of a cylindrical conductor depends:
(i) Area of cross-section of the conductor.
(ii) Nature of the material. SI unit of resistivity is ohm−m
6. State the physical quantity which is equal to the ratio of potential difference and current. Define its SI unit. V
Ans: Electrical Resistance, R = V
I
The resistance of a conductor is said to be one ohm if the potential difference applied across its ends is 1 volt and a current of 1 A flows through it. Its SI unit is ohm (Ω)
1 ohm = 1 volt
1 ampere
7. Calculate the work done in moving a charge of 2 coulombs across two points having a potential difference of 12 V.
Ans: q = 2C V
= 12V W
= Vxq
= 12V x 2C
= 24J.
8. List in a tabular form two differences between a voltmeter and an ammeter.
Ans: Difference between voltmeter and Ammeter: Voltmeter Ammeter
9. Draw a schematic circuit diagram for a circuit in which three resistors R1, R2 and R a plug key under closed condition, an ammeter are joined in series with a 5V battery. Also a voltmeter is connected to measure the potential difference across the resistor R1.
Ans: Circuit diagram to show R1, R2 and R3 connected in series with a battery, ammeter and a key. Voltmeter V is connected parallel to R1. Direction of current.
10. Name and define S.I. unit of resistance. Calculate the resistance of a resistor if the current flowing through it is 200 mA, when the applied potential difference is 0.8 V.
Ans: The SI unit of resistance is ohm. One ohm is the resistance offered, by a conductor when the current passing through it is one ampere and the potential difference across its ends is volt. By Ohm’s law, V = IR
R = V
I
= 0.8 V
0.2 A
= 4 ohm
11. In an electric field the work done in bringing a 2 coulomb charge from infinity to a point A is 10 joules and in bringing the same charge to some another point B is 20 joules. Find the potential difference between two points A and B. What would be the work done if the same charge is brought directly from A to B?
Ans: Work done = 10 Joule
Charge = 2 Coulomb 20 J
Potential of point A = VA = 2C
Given, work done = 20 Joule Charge = 2 coulomb
= 5 V 20 J Potential of point B = VB = = 10 V 2C
Potential difference between two pints A and B is
V = VB − VA
= 10 V − 5V = 5V
Work sone directly from A to B Work done
= Potential difference x charge
= 5V x 2C
= 10 Joule
12. What does an electric circuit mean? Name a device that helps to maintain a potential difference across a conductor in a circuit. When do we say that the potential difference across a conductor is 1 volt? Calculate the amount of work done in shifting a charge of 2 coulombs from a point A to B having potentials 110 V and 25 V respectively.
Ans: Electric circuit: The closed path along which an electric current flows is called an ‘electric circuit’. The device that helps to maintain a potential difference across a conductor in a circuit are – Electric cell, electric battery, electric generator
1 Volt: The potential difference between two points in an electric field is said to be one volt if one joule of work has to be done in bringing a positive charge of one coulomb from one point to another. 1 joule 1 J
1 volt = or 1 volt = 1 coulomb 1
C
Work done = V (Potential difference) x Q (Coloumb)
W = V x Q
= 85 x 2
= 170 Joule
13. (a) Two identical resistors each of resistance 10 ohm are connected in:
(i) series, (ii) parallel. in turn to a battery of 6V. Calculate the ratio of power consumed by the combination of resistor in the two cases.
(b) List two factors on which the resistance of a conductor depends.
(c) Write a difference between an ammeter and voltmeter.
Ans:
b) Resistance of a conductor depends on:
(i) Length of the conductor (ii) Area of cross section
14. Three resistances of 4Ω, 5Ω and 20Ω are connected in parallel. Their combined resistance is: (a) 2 Ω (b) 4 Ω (c) 5 Ω (d) 6 Ω
Ans: (a) 2Ω
15. On what principle is an electric bulb based? [NCERT Exemplar]
Ans: Heating effect of current.
16. Find the minimum resistance that can be made using five resistors, each of 5Ω. [NCERT Exemplar]
Ans: By connecting resistors in parallel, resistance 1Ω is obtained R 5Ω Req = = = 1Ω n 5
17. A potential difference of 220 V is applied across a resistance of 440 Q in an electrical appliance. Calculate the current drawn and heat energy produced in 20 seconds. [SSLC July, 2018-19]
Ans:
18. Explain two disadvantages of series arrangement for household circuit.
Ans: (i) Current is constant in series combination, so it impractical to connect a bulb and an electric heater in series.
(ii) When one component fails, the circuit is broken and none of the components work.
19. Out of the two wires X and Y shown below, which one has greater resistance & Justify your answer.
Ans: Wire Y, because R ∝ I Resistance of a conductor is directly proportional o the length of the conductor, whose area of cross- section is the same.
20. Explain the following: [SSLC 2018-19, July]
(i) The elements of electric heating devices such as bread-toasters and electric iron are made of an alloy rather than of a pure metal.
(ii) Series arrangement is not used for domestic circuits.
(iii) Copper and aluminium wires are usually employed for electricity transmission.
Ans: (i) Resistivity of an alloy is generally higher and it does not oxidize easily. (ii) In series arrangement, same current will flow through all the appliances which is not required and the equivalent resistance becomes higher, hence the current drawn becomes less. (iii) They are extremely good conductors having a low value of resistivity.
21. State Ohm’s Law. Draw a circuit diagram to verify this law indicating the positive and negative terminals of the battery and the meters. Also show the direction of current in the circuit.
Ans: Ohm’s Law: It states that “Physical conditions remaining same, the current flowing through a conductor is directly proportional to the potential difference across its two ends”.
i.e V α I V = IR
where the constant of proportionality R is called the electrical resistance or resistance of the conductor. Diagram to Verify Ohm’s Law:
22. Two devices of ratings 44W; 200V and 11W; 220V are connected in series. The combination is connected across a 440V mains. The fuse of which of the two devices is likely to burn when switch is on 7 Justify your answer.
Ans: The fuse of device of rating 11W; 220V will burn.
R = V2
W
∴ Resistance of 11W device will be four times than that of the device of 44W.
∴ Voltage across 11W = 352V
∴ Voltage across 44W = 88V 352V across the device of 11W; 220V
rating is sufficient to burn the fuse of the device.
23. Find the current drawn from the battery by the network of four resistors shown in the figure.
Ans: Two combinations of two parallel resistors of 10t each connected in series.
24. Find out the following in the electric circuit given in the figure.
(i) Effective resistance of two 8Ω resistors in the combination
(ii) Current flowing through 4Ω resistor
(iii) Potential difference across 4Ω resistor
(iv) Power dissipated in 4Ω resistor
(v) Difference in reading of ammeter A1 and A2 (if any).
Ans: (i) Effective resistance of two 8Ω resistors 1 R
= 1 8 + 1 8
= 2 8 R = 4Ω
(ii) Current flowing through the circuit = current flowing through 4Ω Equivalent resistance of the circuit = 4 + 4 = 8Ω
Current flowing in the circuit = V R = 8V 8Ω = 1A
(iii) Potential difference across 4 = V1 = IR1 = 1A x 4Ω = 4V
(iv) Power dissipated in 4Ω = P = VI = 4V x 1A
= 4 watt (v) Both A1 and A2 show same reading because current remains same in series combination.
25. Two wires X and Y are of equal length and have equal resistances. If the resistivity of X is more than that of Y, which wire is thicker and why? For the electric circuit given below calculate:
(i) Current in each resistor (ii) Total current drawn from the battery and (iii) Equivalent resistance of circuit Ans. V = JR wire A is thicker
Ans: V = IR wire A is thicker 6 1
(i) For 30 Ω, I = = Ampere or 0.2 A 30 5 6 For 10 Ω, I = = 0.6 Amphere 10 6 For 5 Ω, I = = 1.2 Ampere 5
(ii) 1 R
= 1 30 + 10 + 15 = 1 3
⟹ R = 3 V 6 V = IR ⟹ I = = = 2 Ampere R 3
(iii) Equivalent Resistance = 3
Textbook Exercises:
1. 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 R′ ratio is: R 1 (a) 25
Ans: (d) 25 (b) 1 5 (c) 5 (d) 25
Ans: (d) 25
2. Which of the following terms does not represent electrical power in a circuit?
(a) I2R (b) IR2 (c) VI (d) V2 R
Ans: IR2
3. An electric bulb is rated 220V and 100 W When it is operated on 110 V, the power consumed will be— (a) 100W (b) 75W (c) 50W (d) 25W
Ans: (d) 25 W
4. 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 (a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (4) 4 : 1
Ans: (c) 1 : 4
5. How is a voltmeter connected in the circuit to ‘measure the potential difference between two points?
Ans: Voltmeter is connected in parallel to the resistor.
6. A copper wire has diameter 0.5 mm and resistivity of 1.6 x 10− 8Ω m. What will be the length of this wire to make its resistance 10Ω ? How much does the resistance change if the diameter is doubled?
7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below— I (amperes) 0.5 1.0 2.0 3.0 4.0 V (volts) 1.6 3.4 6.7 10.2 13.2 Plot a graph between V and I and calculate the resistance of the resistor.
8. 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.
9. A battery of 9 V is connected in series with of 0.2Ω, 0.3Ω, O.4Ω, 0.5Ω, and 12Ω, respectively. How much current would flow through the 12Ω resistor?
10. How many 176Ω resistors (in parallel) are required to carry 5A on a 220 V line?
Ans:
11. Show how you would connect three resistors, each of resistors 6Ω, so that the combination has a resistance of (i) 9Ω, (ii) 4Ω.
Ans:
12. 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:
13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Ans:
14. Compare the power used in the 2Ω resistor in each of the following circuits:
(i) a 6 V battery in series with 1Ω and 2Ω resistors, and
(ii) a 4 V battery in parallel with 12Ω and 2Ω resistors.
Ans:
15. Two lamps, one rated 100W 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:
16. Which uses more energy, a 250 W TV set in 1 hr or a 1200 W toaster in 10 minutes?
Ans: For TV set E = P x t = 0.25 x 1 = 0.25 kwh
For Toaster E = 1.2 x 1 6 =
0.2kwh
∴ TV set consumes more energy than a toaster.
17. An electric heater of resistance 8Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
Ans: R = 8Ω, I = 15A, t = 2h = 2 x 60 x 60 s
H = I2Rt = (15)2 x 8 x 2 x 60 x 60
Rate at which heat produced: = H t
= (15)2 x 8 x 2 x 60 x 60
2 x 60 x 60
= 1800 W
18. 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-toaster 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 a wire vary with its area of cross-section? (e) Why are copper and aluminium wires usually employed for electricity transmission?
Ans: (a) Because tungsten is. a metal of very high melting point and it does not oxidise even at high temperature.
(b) As alloy has high melting point so it can be heated to a large amount till getting red hot without melting.
(c) In series combination
(i) Equivalent resistance is maximum, hence, less current flows in the circuit.
(ii) If one appliance is switched off then all other appliances will also stop working.
(iii) The voltage of external power supply is divided among the various appliances.
(iv) Same current flows in all appliances. Hence, high rating appliances can’t draw more current from the circuit.
(d) Resistance is inversely proportional with the cross- sectional area of wire, R
(e) Because they have low resistivity.
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