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Karnataka State Syllabus Class 10 Science Chapter 12 Electricity
KSEEB Class 10 Science Electricity Intext Questions and Answers
Question 1.
What does an electric circuit mean?
Answer:
A closed conducting path through which electric current may flow is called an electric circuit.
Question 2.
Define the unit of current?
Answer:
One ampere is constituted by the flow of one column of charge per second that is
\(1 \mathrm{A}=\frac{1 \mathrm{C}}{1 \mathrm{S}}\)
Question 3.
Calculate the number of electrons constituting one coulomb of charge.
Answer:
Charge on an electron = 1.6 x 10-19 C.
If the value of charge is 1.6 x 10-19 C
then number of electron= 1
∴ If the value of charge is 1 C, then number of electrons
Question 4.
Name a device that help to maintain a potential difference across a conductor.
Answer:
A cell or a battery.
Question 5.
What is meant by saying that the potential difference between two points is 1V?
Answer:
The potential difference between two points in a current-carrying conductor is said to be 1 V when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.
\(1volt=\frac { 1\quad { joule } }{ 1{ coulomb } } \)
1 V = 1 JC-1
Question 6.
How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
Charge, Q = 1C
Potential difference, V = 6V
Work done W = VQ = 6V x 1C = 6J
The work done on each coulomb = 6J
Therefore, the energy, given to each coulomb of charge is also 6J.
Question 7.
On what factors does the resistance of a conductor depend?
Answer:
The resistance of a conductor depends:
- On its length
- On its area of cross-section
- On the nature of its material and
- On the temperature of the conductor.
Question 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?
Answer:
Current flows more easily through a thick wire. Because if the wire is thick resistance is less. Therefore thick wire is required.
Question 9.
Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half its former value. What change will occur in the current through it?
Answer:
We know that, \(I=\frac{V}{R}\)
Since the resistance remains constant, so the current is directly proportional to a potential difference. If the potential difference is halved, the current also gets halved.
Question 10.
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer:
Coils of electric toasters and electric irons are made of an alloy rather than a pure metal because
- The resistivity of an alloy is much higher than that of pure metal and
- An alloy does not undergo oxidation easily even at high temperatures.
Question 11.
Use the data in table 12.2 to answer the following
(i) Which among iron and mercury is a better conductor?
(ii) Which material is the best conductor?
Answer:
(i) Among iron and mercury, iron is better conductor of electricity because resistivity of iron 10 x 10-8 Q m) is less than that of mercury (98.4 x 10-8 Q m).
(ii) We know that a good conductor of electricity should have a low resistivity and a poor conductor of electricity will have a high resistivity. Silver has the lowest resistivity of 1.60 x 10-8 Q m, which means that silver offers the. the least resistance to the flow of current through it, so silver is the best conductor of electricity.
Question 12.
Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 v each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor and a plug key, all connected in series.
Answer:
The potential difference of each cell = 2 V
The total potential difference (or voltage ) of 3 cells = 3 x 2V = 6V
Question 13.
Redraw the circuit of Q.1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12Ω resistor. What would be the readings in the ammeter and the voltmeter?
Answer:
It is noted that the ammeter has been. put in series with the circuit and the voltmeter has been put in parallel with the 12 Ω resistor.
(i) Calculation of current reading in the ammeter:
Here, R1 = 5Q, R2 = 8Q and R3 = 12Ω
These three resistors are connected in series
∴ total three resistance
R = R1 + R2 + R3= 5 + 8 + 12 = 25Ω
Potential difference, V = 6v
Current; I = ?
Apply ohms law,
V = IR
\(I=\frac{V}{R}=\frac{6}{25}=0.24 \mathrm{A}\)
Therefore, ammeter will show a reading of 0.24A.
(ii) Calculation of potential difference reading across 12 Ω resistor.
Current, I = 0.24A. Resistance, R = 12Ω, potential difference, V = ? Applying ohms law
V = IR = 0.24 x 12 = 2.88V
Therefore, the voltmeter reading is 2.88 V.
Question 14.
Judge the equivalent resistance when the following are connected in parallel
(i) 1Ω and 106Ω
Answer:
(i) When a number of resistances are connected in parallel, then their combined resistance is less than the smallest individual resistance. Therefore, equivalent resistance will be less than 1Ω
(ii) In this case, also the equivalent resistance will be less than 1 Ω.
Question 15.
An electric lamp of 100 Ω, a toaster of resistance 50Q 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?
Answer:
Here, R1= 100Ω, R2 = 50Ω, R3 = 500Ω
Equivalent resistance = R Resistors are connected in parallel,
Since the electric iron connected to the same source that takes as much current as all the three appliances so Resistance of the electric iron
\(=\frac{125}{4} \Omega=31.25 \Omega\)
Current through the electric iron = 7.04 A
Question 16.
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer:
A parallel circuit divides the current through the electrical gadgets. The total resistance in a parallel circuit is decreased. This is helpful particularly when each gadget has different resistance and requires different current to operate properly.
Question 17.
How can three resistors of resistance 2Ω, 3Ω and 6Ω be connected to give a total resistance of
(i) 4Ω.
(ii) 1Ω?
Answer:
(i) As the total resistance (equivalent resistance) is 4Ω, the 6Ω resistor cannot be in series, so, it must be in parallel with some other resistors. In a parallel connection, the equivalent resistance (4Ω) has to be less than all, the resistances. So, the resistors of 2Ω and 3Ω cannot be in parallel at one time with 6Ω.
So, the resistors have to be in a mixed combination, Let us consider the combination shown in the figure. The equivalent resistance between B and C (which are in parallel)
The resistance between A and D = 2Ω + 2Ω = 4Ω
So, the combination shown in the figure is true.
(ii) Here, R, = 2Ω, R2 = 3Ω, R3 = 6Ω, and R = 6Ω
Since the equivalent resistance of the combination is of lesser value than any of the resistors of the combination, it is clear that the resistors should be connected in parallel, It can be further confirmed by using the formula.
Therefore, resistors should be connected in parallel.
Question 18.
What is
(i) the highest,
(ii) the lowest total resistance that can be secured by combinations of four coils of resistance 4Ω, 8Ω, 12Ω, 24Ω?
Answer:
(i) The highest can be secured by series combination and is equal to R = 4Ω + 8Ω + 12Ω + 24Ω = 48Ω
(ii) The lowest total resistance can be, secured by parallel combination, which is given by
Question 19.
Why does the cord of an electric heater not glow while the heating element does?
Answer:
The cord of the electric heater is made of copper, it does not glow because negligible heat is produced in it by the passing current due to its extremely low resistance. The heating element of an electric heater is made of nichrome. It glows because large amount of heat is produced in it by the passing electric current due to its high resistance.
Question 20.
Compute the heat generated while transforming 90,000 coulombs of charge in hour through a potential difference of 50V.
Answer:
Here; Q = 96,000c,
t = 1 hour = 60 x 60 sec and V = 50V
Question 22.
An electric iron of resistance 20Ω takes a current of 5A. Calculate the heat developed in 30s.
Answer:
Here, R = 20Ω, I = 5 A, t = 30s
Heat developed = l2 Rt
5 x 5 x 20 x 30
= 15,000 J = 15kJ
Question 23.
What determines the rate at which energy is delivered by a current?
Answer:
Electric power.
Question 24.
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 hour.
Answer:
Power, P = VI
= 220 x 5 = 1,100 w= 1.1 KW
Energy Consumed, E = P x t
= 1.1KW x 2h = 2.2 kwh
KSEEB Class 10 science Electricity Textbook Exercise Questions and Answers
Question 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 ratio R/R’ is
(a) 1/25
(b) 1/5
(c) 5
(d)25
Answer:
Resistance of this wire,
The equivalent resistance of the 5 wires in parallel is R’ Then
Hence, the correct answer is (d)
Question 2.
Which of the following terms does not represent electrical power in a circuit?
(a) I2R
(b) IR2
(c) VI
(d) V2/R
Answer:
(b) IR2
Question 3.
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
(a)100W
(b) 75 W
(c) 50 W
(d) 25 W
Answer:
Hence, the correct answer is (d)
Question 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 combinations would be
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
Answer:
Suppose the resistance of each one of, the two wires is R.
The equivalent resistance of the series combination,
Rs = R + R = 2R
The heat produced in time t,
Thus, the correct answer is (c)
Question 5.
How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer:
A voltmeter is always connected parallel in the circuit to measure the potential difference between two points.
Question 6.
A copper wire has diameter 0.5 mm and resistivity of 1.6 x 10-8 ft m. What will be the length of this Wire to make its resistance 10 ft? How much does the resistance change if the diameter is doubled
Answer:
= 0.1964 x 10-6m2
Resistance, R = 10Q, Resistivity,
ρ= 1.6 x 10-8 Qm, length, l = ?
Thus, on doubling the diameter, the area of cross-section becomes 4 times and the resistance becomes one – fourth.
Question 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
Plot a graph between V and I and calculate the resistance of that resistor.
Answer:
Question 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.
Answer:
The potential difference, V= 12V
Current, I = 2.5 mA = 2.5 x 10-3 A
We know that V = IR
\(\text { or, } \mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{12}{2.5 \times 10^{-3}}\)
= 4.8 x 10-3ft = 4.8 k
Question 9.
A battery of 9 V is connected in series with resistors of 0.2Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 ft resistor?
Answer:
Resistors are connected in series so, equivalent resistance
R = 0.2Ω + 0.3Ω + 0.4Ω + 0.5Ω + 12Ω
= 13.4Ω
Potential difference V = 9v
Current, through the circuit
\(I=\frac{V}{R}=\frac{9}{13.4}=0.67 \mathrm{A}\)
Question 10.
How many 176 ft resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer:
Current, I = 5A
Potential difference, V = 220V
\(R=\frac{V}{I}=\frac{220}{5}=44 \Omega\)
Resistance of parallel circuit,
Number of required resistors = 4.
Question 11.
Show how you would connect three resistors, each of resistance 6Ω, so that the combination has a resistance of (i) 9 ft, (ii) 4Ω.
Answer:
(i) In order to get a resistance of 9Ω, We connect the given resistors (each of resistance of 6Ω) in the following way.
Resistance of the combination = 6Ω + 3Ω = 9Ω
(ii) In order to get a resistance of 4Ω, we connect two resistors in series and third in parallel as shown in figure. Resistors AB and BC are in series, therefore, Rs = 6Ω + 6Ω = 12Ω
Now, Rs is parallel with the third (6Ω)
∴ Equivalent resistance of combination (Rp ) is given by
Question 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?
Answer:
Potential difference, V = 220V
Power of each bulb, P = 10 W
Resistance of each bulb
Let n be the number of bulbs to be connected in parallel to obtain resistance R’
Question 13.
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 ft resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Answer:
Potential difference, V = 220V
Resistance of each coil, R = 24ft
Case 1:
When coils A and B are used separately, current through each coil,
\(I=\frac{V}{R}=\frac{220}{24}=9.2 \mathrm{A}\)
Case II:
When coils A and B are connected in series, the equivalent resistance in the circuit.
Rs = 24 + 24 = 48Ω
Current \(I=\frac{V}{R_{s}}=\frac{220}{48}=4.6 \mathrm{A}\)
Case III:
When coils A and B are connected in parallel, the equivalent resistance (Rp) is given by
Question 14.
Compare the power used in the 2 ft resistor in each of the following circuits:
(i) a 6 V battery in series with 1 ft and 2 ft resistors, and
(ii) a 4 V battery in parallel with 12 ft and 2 ft resistors.
Answer:
(i) Equivalent resistance of 1Ω and 2Ω in series, R = 1Ω+ 2Ω = 3Ω
Potential difference, V = 6v
Current, \(I=\frac{V}{R}=\frac{6}{3}=2 A\)
Current in series circuit is same
Current in 2ft resistor = 2A
Power in 2Q resistor, P = I2 R
= 22 x 2 = 8W
(ii) Potential difference across 2Q
resistor = 4v
\(\text { Power, } P=\frac{V^{2}}{R}=\frac{4^{2}}{2}=8 \mathrm{W}\)
P : P’ = 8W : 8W = 1:1
Question 15.
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?
Answer:
Since, two lamps are connected in parallel, so its equivalent resistance is given by
Question 16.
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer:
Energy used by TV set
E1 = P x t
= 250 x W x 1h = 250Wh
Energy used by toaster,
E2 = P x t
\(=1200 \times \frac{10}{60}=200 \mathrm{Wh}\)
Thus, TV set in one hour uses more energy than the toaster uses in 10 minutes.
Question 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.
Answer:
Here, R = 8Ω, I = 15 A
t= 2h = 2 x 60 x 60s
Heat developed,
H = I2 Rt = (15)2 x 8 x 2 x 60 x 60 J
The rate at which heat is developed is power,
\(P=\frac{W}{t}\)
Question 18.
Explain the following.
(i) Why is the tungsten used almost exclusively for filament of electric lamps?
(ii) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(iii) Why is the series arrangement not used for domestic circuits?
(iv) How does the resistance of a wire vary with its area of crosssection?
(v) Why are copper and aluminum wires usually employed for electricity transmission?
Answer:
(i) Pure tungsten has a high resistivity and a high melting point (nearly 3000°c). When an electric current is passed through the filament, the electric energy is converted to heat and light energy due to the heating of the filament to a very high temperature. Due to the high melting point of tungsten, the filament does not melt.
(ii) The resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise (bum) readily at higher temperatures. Therefore, conductors of eletric heating devices, such as toasters and electric irons, are made of an alloy rather than pure metal.
(iii) The series arrangement is not used for domestic circuits because:
(a) If connected in series total resistances will increase therefore, current flowing through the circuit will be low.
(b) if one appliance is switched off or gets damaged than all other appliances will also stop working because their electricity supply will be cut off.
(iv) The resistance of a wire is inversely proportional to its cross-sectional area. Thus, a thick wire has less resistance, and a thin wire has more resistance
(v) copper and aluminium wires are usually employed for electricity transmission because copper and aluminium have very low resistivities.
Multiple choice questions:
Question 1.
The unit of charge is
(a) ampere
(b) coulomb
(c) farad
(d) Volt
Answer:
(b) coulomb
Question 2.
The SI unit of electric current is
(a) Ohm
(b) Volt
(c) Ampere
Answer:
(c) Ampere
Question 3.
Volt is the SI unit of
(a) Potential difference
(b) current
c) resistance
(d) Charge.
Answer:
(d) Charge.
Question 4.
The resistance R1 and R2 are connected in parallel. The equivalent resistance of the combination is ….
(a) R1 + R2
(b) R1 – R2
(c) \(\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}\)
(d) \(\frac{\mathrm{R}_{1}+\mathrm{R}_{2}}{\mathrm{R}_{1} \mathrm{R}_{2}}\)
Answer:
(c) \(\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}\)
Question 5.
Three resistances of 4Ω, 5Ω, and 20Ω are connected in parallel. Their combined resistant
(a) 2Ω
(b) 4Ω
(c) 5Ω
(d) 20Ω
Answer:
(a) 2Ω
Question 6.
Three resistances of 1Ω each are connected to form a triangle. The resistance between any two terminals is
(a) 3Ω
(b) \(\frac{1}{2} \Omega\)
(c) \(\frac{2}{3} \Omega\)
(d) \(\frac{3}{2} \Omega\)
Answer:
(c) \(\frac{2}{3} \Omega\)
Question 7.
The device used for measuring current is
(a) galvanometer
(b) ammeter
(c) voltameter
(d) potentiometer
Answer:
(b) ammeter
Question 8.
Electrical power is given by
(a) \(P=\frac{V}{I}\)
(b) \(P=\frac{I}{V}\)
(c) \(P=\frac{I}{V I}\)
(d) P = VI
Answer:
(d) P = VI
Question 9.
Kilowatt-hour is the unit of
(a) electric power
(b) electric resistance
(c) electric potential
(d) electric energy.
Answer:
(d) electric energy.
Question 10.
The electrical appliances in the houses are connected with each other in
(a) Parallel
(b) Series
(c) a combination of series & parallel circuits
(d) none of the above
Answer:
(a) Parallel
KSEEB Class 10 Science Electricity Additional Questions and Answers
Question 1.
Tap water conducts electricity whereas distilled water does not, why?
Answer:
Tap water contains dissolved salts and minerals which ionise in water. Tap water conducts electricity due to the presence of these ions. Distilled water is a covalent compound containing very few ions and almost does not conduct, electricity.
Question 2.
What constitutes an electric current?
Answer:
The flow of electric charges across a cross-section of a conductor constitutes an electric current. For example, a stream of electrons moving through a conducting wire constituent an electric current.
Question 3.
How much is the charge on an electron? Can a charge less than this value exist?
Answer:
Charge on an electron, e = 1.6 x 10-19C.
A charge less than this value cannot exist.
Question 4.
State the relationship between 1 ampere and 1 coulomb?
Answer:
Ampere: If one coulomb of charge flows through any section of a conductor in one second, then current through it is said to be one ampere.
\(1ampere=\frac { 1{ coulomb } }{ 1{ second } } or\quad 1{ A\quad }\)
\(=\frac{1 \mathrm{C}}{1 \mathrm{S}}=1 \mathrm{CS}^{-1}\)
The smaller units of current are milliampere and microampere
1 milliampere = 1 mA = 10-3 A
1 microampere = 1 pA = 10-6 A
Question 5.
Whatis an electric circuit? Distinguish between an open and a closed circuit.
Answer:
Electric circuit: A closed and continuous path along which an electric current flows is called an electric circuit.
In atorch, a switch provides a conducting link between a battery ( a number cells placed in proper order) and a bulb. When the switch is turned ON, an electric current flows through the bulb and it gives light. Such a continuous and closed path of an electric current is called an electric circuit. When the circuit is broken any where ( or the switch is turned OFF); the current stops flowing and the bulb does not glow. An electric circuit through which no current flows is called an open circuit. An electric circuit through which current flows continuously is called a closed circuit.
Question 6.
What is an ammeter? How is it connected in a circuit?
Answer:
Ammeter: An ammeter (ampere + meter) is a device used to measure electric current in a circuit.
Question 7.
Draw a circuit diagram consisting of a cell, an electric bulb, an ammeter and a plug key.
Answer:
Figure 12.3 Shows the schematics diagram of a typical electric circuit comprising a cell, an electric bulb, an ammeter and a plug key, clearly, the electric current flows in the circuit from the positive terminal of the cell to the negative terminal of the cell through the bulb and the ammeter.
Question 8.
An electric bulb draws a current of 0.2 A when the voltage 220 volts. Calculate the amount of electric charge flowing through it in one hour.
Answer:
Hence, I = 0.2A, t = 1 h-= 3600 S
Charge q = I x t = 0.2 x 3600 = 720 C.
Question 9.
Show that one ampere is equivalent to a flow of 6.25 x 1018 elementary charges.
Answer:
Here I = 1 A, t = 1 S
Total charge q = I + t= 1 x 1 = 1c .
Charge on one electrons = 1.6 x 10-19 C
= \(Numberofelectrons=\frac { { Total\quad Change } }{ { Charge\quad on\quad one\quad electron } } \)
= \(=\frac { 1{ C } }{ 1.6\times 10^{ -19 }{ C } } \)
= 6.25 x 1018
Question 10.
What is SI unit of electric potential?
Answer:
It is volt and has the symbol V.
Question 11.
What is meant by the potential difference between two points? What is the SI unit of potential difference?
Answer:
Potential difference: The potential difference between two points in an electric field is the amount of work done in bringing a unit positive charge from one point to the other.
In fig. 12.4 A and B are two points located in the electric field of a charge Q placed at origin O. If W work is done in bringing a charge Q0 from B to A, then potential difference between A and B is
\(\mathrm{v}=\frac{\mathrm{W}}{\mathrm{Q}_{0}}\) or potencial difference = \(=\frac { { Work\quad done } }{ { Charge } } \)
The SI unit of potential difference is joule / coulomb or volt.
Question 12.
Define the SI units of electric potential difference.
Answer:
Volt: The potential difference between two points is 1 volt if one joule of work is done in bringing a positive charge of one coulomb from one point to the other point.
Question 13.
Express work done in an electric field in terms of charge and potential difference.
Answer:
Work done = Charge x Potential difference.
W = Q x V
Question 14.
Find the amount of work done in moving a charge of 5mc across two points showing a potential difference of 100V.
Answer:
W = QV = 5 x 10-3 x 100J = 0.5J.
Question 15.
List in a tabular form two difference between a voltmeter and an ammeter.
Answer:
| Voltmeter | Ammeter |
| 1. It measures potential difference between two points in a circuit | 1. It measures current passing through a circuit. |
| 2. It is connected in parallel with the component across which p.d is measured | 2. It is connected in series in a circuit. |
Question 16.
What is meant by resistance of a conductor?
Answer:
Resistance: The resistance of a conductor is its property by virtue of which it opposes the flow of current through it. It is equal to the ratio of the potential difference applied across the conductor to the current flowing. through it.
\(Resistance=\frac { { Potential\quad difference } }{ { Current } } or\quad R=\frac { V }{ I } \)
Question 17.
Name and define the SI unit of resistance?
Answer:
SI unit of resistance is ohm: The resistance of a conductor is said to be 1 ohm if a current of 1 ampere flows through it on applying a potential difference of 1 volt across its ends.
\({ 1\quad ohm }=\frac { 1{ volt } }{ 1{ ampere } } { or }1\Omega =\frac { 1{ V } }{ 1{ A } } \)
Question 18.
Name the physical quantity whose unit is volt/ampere.
Answer:
resistance.
Question 19.
What is the function of a rheostat in an electric circuit?
Answer:
It changes the current in a circuit due to the change in its resistance.
Question 20.
What would be the resistance of a conductor if the current flowing through it is 0.35A when the potential difference across it is 1.4 V?
Answer:
Here V = 1.4V, I = 0.35A
\(\text { Resistance, } \mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{1.4}{0.35}=4.0 \Omega\)
Question 21.
A graph between potential difference (v) and current (I) is given in figure
(i) What is the relation between V and I.
(ii) Find the resistance of the conductor using the graph.
Answer:
(i) As the graph between V and I is linear, so V α I.
(ii) \({ R }=\frac { { V }_{ 2 }-{ V }_{ 1 } }{ { I }_{ 2 }-{ I }_{ 1 } } =\frac { 12.0-8.0 }{ 3.0-2.0 } =\frac { 4 }{ 1 } =4\Omega \)
Question 22.
List the factors on which the resistance of a conductor depends. Write the formula showing relation of resistance with these factors.
Answer:
Factor affecting the resistance: At a constant temperature, the resistance of a conductor depends on the following factors:
(i) Length: Resistance R of a conductor is directly proportional to its length L, i.e.,
R α L
(ii) Area of cross-section: Resistance R of conductor is inversely proportional to its area of cross-section A, i.e., \(\mathrm{R} \alpha \frac{1}{\mathrm{A}}\)
(iii) Nature of the material: Resistance also depends on the nature of the material of which the conductor is made. The resistance of a copper wire is much less than that of a nichrome wire of same length and area of cross-section combining the above factors, we get.
\(\mathrm{R} \alpha \frac{\mathrm{L}}{\mathrm{A}} \Rightarrow \mathrm{R}=\rho \frac{\mathrm{L}}{\mathrm{A}}\)
The proportionality constant p is called resistivity of specific resistance which depends on the nature of the material.
Question 23.
A wire of resistivity p is stretched to three times its length. What will be its new resistivity?
Answer:
The resistivity remains unchanged as it does not depend on length, it depends on the nature of the material of the wire.
Question 24.
Two wires of equal length one of copper and the other of nichrome have the same resistance. Which wire will be thicker?
Answer:
Nichrome wire is thicker than copper wire.
Question 25.
What happens to the resistance of a conductor when its area of cross-section is increased?
Answer:
\(\mathrm{R} \alpha \frac{l}{\mathrm{A}}\) So the resistance decreases when A area of cross-section is increased.
Question 26.
A given length of a wire is doubled on itself and this process is repeated once again. By what factor does the resistance of the wire change?
Answer:
Question 27.
Aluminium wire has radius 0.25 mm and length 75m. If the resistance of the wire is 10Ω; Calculates the resistivity of aluminium.
Answer:
Here r = 0.25mm = 0.25 x 10-3m, L = 75m, R= 10Ω
Question 28.
In the circuit given below, find (i) total resistance (ii) current shown by ammeter
Answer:
(i) The resistance of 3Ω and 2Ω are in series . Their equivalent resistance – 3Ω + 2Ω = 5Ω
This combination is in parallel with the resistance of 5Ω
So net resistance R is given by
Question 29.
Consider the following circuit diagram. If R1 = R2 = R3 = R4 = Rs = 3Ω, Find the equivalent resistance of the circuit.
Answer:
The equivalent resistance of the series combination of R2 and R3.
Equivalent resistance of the parallel combination of R4 and R’,
\(\mathrm{R}^{\prime \prime}=\frac{\mathrm{R}_{4} \mathrm{R}^{\prime}}{\mathrm{R}_{4}+\mathrm{R}^{\prime}}=\frac{3 \times 6}{3+6}=2 \Omega\)
Now R1, R” and R2 are inseries. There fore, the equivalent resistance of the entire circuit.
R = R, +R” + R5 = 3 + 2 + 3 = 8Ω
Question 30.
The figure given below shows a part of an electric circuit, the reading of the ammeter is 3.0 A. Find the currents through 10Ω and 20Ω resistors.
Answer:
Current through 10Ω resistor,
Question 31.
A lamp rated 100 Watt at 200 V is connected to the mains electric supply.
(i) What amount of current is drawn from the supply line if the voltage is 220 V?
(ii) What is its resistance?
Answer:
Given P = 100W, V = 220V
Question 32.
Two identical resistors, each of resistance 10 ohms, are connected (i) in series (ii) in parallel, in turns to a battery of 6 volts. Calculate the ratio of power consumed.
Answer:
Here, R1 = 10Ω; R2 = 10Ω
Total resistance of the series combination.
(ii) Here R1 = 10ft, R2 = 10ft Total resistance RP of the parallel combination is given by
Ratio of the power consumed in the two combination,
Question 33.
The potential difference between the terminals of an electric heater is 110V when it draws a current of 5 A from the source. What current will the heater draw and what will be its wattage if the potential difference is increased to 220 V? Consider that the resistance of the heater element does not change with temperature?
Answer:
The resistance of heater element,
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