NCERT Solutions for Class 10 Science Chapter 12 Electricity

These NCERT Solutions for Class 10 Science Chapter 12 Electricity Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Electricity NCERT Solutions for Class 10 Science Chapter 12

Class 10 Science Chapter 12 Electricity InText Questions and Answers

In-text Questions (Page 200)

Question 1.
What does an electric circuit mean?
Answer:
A continuous and closed path of an electricity current is called an electric circuit.

Question 2.
Define the unit of current.
Answer:
Ampere is the unit of current. When one coulomb of charge flow per second in the circuit, it is known as one ampere.
1A = \(\frac{1 \mathrm{C}}{1 \mathrm{~S}}\)

Question 3.
Calculate the number of electrons constituting one coulomb of charge.
Answer:
We know that, charge on one electron = 0.6 × 10^-19 C
Let 1 C contain ‘n’ electrons
so, 1 C = n × 0.6 × 10^-19 C
or

In-text Questions (Page 202)

Question 1.
Name a device that helps to maintain a potential difference across a conductor.
Answer:
Battery or Cell.

Question 2.
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 one volt when one joule of work is done to move a charge of one coulomb from one point to the other.

Question 3.
How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
We know that, the amount of energy or work is given by the formula.
W = VQ
= 6 V × 1 C
= 6 V – C = 6 J

In-text Questions (Page 209)

Question 1.
On what factors does the resistance of a conductor depend?
Answer:
The resistance of a conductor depends upon two factors.
(i) Length of the conductor ; Resistance of a conductor is directly proportional to its length i.e.,
R ∝ 1
(ii) Area of cross section of the conductor :
The resistance of conductor is directly proportional to its area of cross section.

Question 2.
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 will flow more easily through a thick wire because the resistance of a wire is inversely proportional to its area of cross section. Thick wire has a large surface area than a thin wire.

Question 3.
Let the resistance of an electrical 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?
Answer:
Let ‘R’ be the resistance, and ‘V’ be the potential difference than ‘I’ current flow through the wire, so from Ohm’s Law
V = iR ……….(i)
When potential difference is half of its initial value than, from Ohm’s Law
\(\frac{\mathrm{V}}{2}\) = i₁ R …….(ii)
Now from equation (i) and (ii) we get

The current become half of its initial value (i).

Question 4.
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer:
The resistivity of an alloy is generally higher than that of its constituent metals but alloys do not oxidise readily at high temperature so they are commonly used in elecricity toasters and electric irons.

Question 5.
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?
Answer:
(a) Iron because its resistivity is lower than mercury.
(b) Silver is the best conductor because it has lowest resistivity.
Table 12.2 s Electrical resistivity of some substances at 20°C

In-text Questions (Page 213)

Question 1.
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:

Question 2.
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 Ω resistor. What would be the readings in the ammeter and the voltmeter?
Answer:

According to Ohm’s Law
V = IR
since R = R₁ + R₂ + R₃
= 5Ω + 8Ω + 12Ω
R= 25 W
And V = 6 volt
6 volt = 1 × 25 Ohm
I = \(\frac{6 \text { volt }}{25 \mathrm{Ohm}}\) = 0.24 Ampere
V¹ = -MR
V¹ = 0.24 \(\frac{\text { volt }}{\mathrm{Ohm}}\) × 12 ohm
V¹ = 2.88 volt
So the reading in the ammeter = 0.24 Ampere
And reading of the volmeter = 2.88 volt.

In-text Questions (Page 216)

Question 1.
Judge the equivalent resistance when the following are connected in parallel.
(A) 1 Ω and 10⁶ Ω
(b) 1 Ω and 10³ Ω, and 10⁶ Ω.
Answer:
(a) Let R is the equivalent resistance

(b) Let ‘R’ is the equivalent resistance.

Question 2.
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?
Answer:

Question 3.
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer:
In series the current is constant throughout the electric circuit Thus, it is obviously impracticable to connect an electric bulb and and an electric heater in series because they require currents of widely different values to operate properly. Another major disadvantages of a series circuit is that when one component fails the circuit is broken and none of the components works. On the other hand, 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 require different current to operate properly.

Question 4.
How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of
(a) 4 Ω
(b) 1 Ω?
Answer:
If 3 Ω and 6 Ω are in series than resultant resistant resistant say (R) can be calculated as

And if R and 2 Ω in series then resultant resistor (R¹) can be calculated as
R¹ = R + 2Ω
= 2Ω + 2Ω
R¹ = 4 Ω

(b) If all the resistance are in parallel than resultant resistance (R) can be calculated as follows:

Question 5.
What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω ?
Answer:
(a) When the four resistances are in parallel, the resultant resistance will be lowest.
Let the resultant resistance be ‘R’ so

(b) When the four resistance are in series, the resultant resistance will be highest.
Let the resultant resistant be ‘R’ so.
R₁ = 4Ω + 8Ω + 12Ω + 24Ω
R₁ = 48 Ω

In-text Questions (Page 2018)

Question 1.
Why does the cord of an electric heater not glow while the heating element does?
Answer:
The cord of an electric heater not glow because it is less resistive while the heating element glow because it is purely resistive.

Question 2.
Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Answer:
We know that the heat generated in time is
H = V I t
Now I = \(\frac{\mathrm{Q}}{\mathrm{t}}=\frac{96000 \mathrm{C}}{60 \times 60 \mathrm{Sec}}\)
= 26.67 Amp.
now, H = V I t
= 50 volt × 26.67 Amp. × 3600 Sec
= 4800600 J

Question 3.
An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 sec.
Answer:
Given I = 5 A
R = 20 W
t = 30 S
H = ?
H = I² R t
= (5 A)² × 20 Ω × 30 Sec
= 25 A² × 20 Ω × 30Sec
= 15000 J

In-text Questions (Page 220)

Question 1.
What determines the rate at energy is delivered by a current ?
Answer:
It is electric power (p) given by the formula,
P = V I = I²R = \(\frac{\mathrm{V}^{2}}{\mathrm{R}}\)

Question 2.
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer:
Given I = 5 A
V = 220 V
since P = V × I
= 220 V × 5 A
= 1100 V – A
P = 1100 W
Energy consumed in ‘2’ hours
= V × I × t
= 220 volt × 5 A × 2 × 60 × 60 sec
= 7920000 J

Class 10 Science Chapter 12 Electricity Textbook Questions and Answers

Page no. 221

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 ration R/R’ is
(a) \(\frac { 1 }{ 25 }\)
(b) \(\frac { 1 }{ 5 }\)
(c) 5
(d) 25
Answer:
(d) 25

Question 2.
Which of the following terms does not represent electric power in a circuit ?
(a) I²R
(b) IR²
(c) V I
(d) \(\frac{\mathrm{V}^{2}}{\mathrm{R}}\)
Answer:
(b) IR²

Question 3.
An electric bulb is rated 230 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
Answer:
(d) 25 W

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 an electric circuit. 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:
(c) 1 : 4

Question 5.
How is a voltmeter connected in the circuit to measure the potential difference between two points ?
Answer:
A voltmeter is connected in the circuit in parallel to measure the potential difference between two points.

Question 6.
A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10^-8 Ω m. What will be the length of this wire to make its resistance 10 W ? How much does the resistance change if the diameter is doubled ?
Answer:
Given
Diameter = 0.5 mm
so radius (r) = \(\frac {0.5}{2}\) mm
= \(\frac{0.5}{2 \times 10^{3}}\) metre
Resistivity (ρ) = 1.6 × 10 10^-8 Ωm
R = 10 W
A = πr²
l = ?
We know that


Second Case

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 the resistor.
Answer:
Resistance (R) = tan Q = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)

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:
By Ohm’s Law

Question 9.
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.5 Ω, and 12 Ω respectively. How much current would flow through the 12 Ω resistor ?
Answer:
Let the resultant resistor be R.
So, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω
R = 13.4 Ω
By Ohm’s Law
V = I × R
I = \(\frac{V}{R}=\frac{9 V}{13.4} \mathrm{Ohm}\)
= 0.671 A.
In series current remain same in all the resistors. So current in 12 Ω resistor will be 0.671 Amp.

Question 10.
How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line ?
Answer:
By Ohm’s Law
V= I × R
R = \(\frac{\mathrm{V}}{\mathrm{I}}=\frac{220 \mathrm{Volt}}{5 \text { Ampere }}\)
= 44 Ohm
Let ‘n’ resistors are connected in parallel.

So the no. of resistors will be ‘4’.

Question 11.
Show how you would connect three resistors each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (b) 4 Ω.
Answer:
(i)

Two resistance in parallel and their resultant should be in series with the third resistance.

Two resistance in series and their resultant should be parallel with the third resistance.

Question 12.
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 Ω. 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 ?
Answer:
The resistance of an electric bulb can be calculated as follows :
P = \(\frac{\mathrm{V}_{2}}{\mathrm{R}}\)
R = \(\frac{220 \times 220}{10}\) = 22 × 220
= 4840 Ohm.
If the allowable current is 5 A and voltage is 220 V than resistance
R¹ = \(\frac{V}{I}=\frac{220 \mathrm{~V}}{5 \mathrm{Amp}}\)
= 44 Ohm.
Let the number of lamps are ‘n’, so

So the number of lamps will be 110.

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 Ω resistance, which may be used separately, in series or in parallel. What are the currents in the three cases ?
Answer:
Case I when the coils are in series,
Let resultant resistance be R, so
R = 24 Ω + 24 Ω = 48 Ω
Now, I = \(\frac{\mathrm{V}}{\mathrm{R}}=\frac{220 \mathrm{~V}}{48 \Omega}\) = 4.58 A
Case when the coil ‘A’ and ‘B’ are in parallel, let the resultant resistance be ‘R’

Question 14.
Compare the power used in the 2 Ω resistor in each of the following circuit (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer:
(i) The resultant resistance (R)
= 1 Ω + 2 Ω = 3 Ω
= (R₁)(R₂)

(ii) When a battery in parallel with 12 Ω and 2 Ω resistors, then voltage remain same in both the resistance. Current will be different.
So power (P) = \(\frac{V^{2}}{R}=\frac{(4)^{2}}{2}\)
= \(\frac {16}{2}\) = 8 J
∴ \(\frac{P}{P^{1}}=\frac{8 J}{8 J}\)
P : P¹ = 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:
Resistance of bulb rated 100 W at 220 say R1

for the bulb rated 60 W at 220 V, say resistance is R₂
So, R₂ = \(\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{220 \times 220}{60}\)
= 806.6 Ohm
When R₁ and R₂ are in series, the resultant resistance say R, so

Question 16.
Which uses more energy, a 250 W.T.V. set in 1 hr, or a 1200 W toaster in 10 minutes ?
Answer:
Energy used by T.V. in one second = 250 J
So energy used by T.V. in 1 hr i.e., 3600 sec
= 250 × 3600 J = 900000 J
Energy used by a toaster in one second = 1200 J
So energy used by a toaster in 10 minutes i.e., 600 sec
= 1200 × 600 J = 720000 J
Hence T.V. will used more energy than a toaster.

Question 17.
An electric heater of resistance 8 W draws 15 A from the service mains 2 hours. Calculate the rate of which heat is developed in the heater.
Answer:
The rate at which heat is developed is called power.
So P = V × I
= \(\frac{\mathrm{V}^{2}}{\mathrm{R}}\)
Since V = I × R
So, = I² × R
= (15A)² × 8 Ohm
= 225 A² × 8 Ohm
= 1800 J/Sec

Question 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-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 a wire vary with its area on cross-section ?
(e) Why are copper and aluminium wires usually employed for electricity transmission ?
Answer:
(a) The resistivity of tungsten does not change with high temperature. It has very high melting point. It can operate upto 2700° C. So tungsten is almost used exclusively for filament of electricity lamps.
(b) Because alloys do not oxidise on heating at high temperature.
(c) Because when light are in series if one bulb fuses or one tube light stops working, the circuit breaks and all light go off.
(d) The resistance of a given wire is inversely proportional to area of cross-section of the wire.
(e) Because copper and aluminium have low resistivity that is why they are good conductors of electricity so they are usually employed for electricity transmission.

Class 10 Science Chapter 12 Electricity Textbook Activities

Activity 12.1 (Page 203)

• Set up a circuit as shown in Fig,, consisting of a nichrome wire XY of length, say 0.5 m, an ammeter, a voltmeter and four cells of 1.5 V each. (Nichrome is an alloy of nickel, chromium, manganese, and iron metals.)

• First use only one cell as the source in the circuit. Note the reading in the ammeter 1, for the current and reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Tabulate them in the Table given.
• Next connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the nichrome wire and potential difference across the nichrome wire.
• Repeat the above steps using three cells and then four cells in the circuit separately.

Question 1.
Calculate the ratio of V to I for each pair of potential difference V and current J.
Answer:

Question 2.
Plot a graph between V and I, and observe the nature of the graph.
Answer:
In this Activity, you will find that approximately the same value for V/I is obtained in each case. Thus the V-I graph is a straight line that passes through the origin of the graph, as shown in Fig. Thus, V/I is a constant ratio.

Activity 12.1 (Page 205)

• Take a nichrome wire, a torch bulb, a 10 W bulb and an ammeter (0 – 5 A range), a plug key and some connecting wires.
• Set up the circuit by connecting four dry cells of 1.5 V each in series with the ammeter leaving a gap XY in the circuit, as shown in Fig.

• Complete the circuit by connecting the nichrome wire in the gap XY. Plug the key. Note down the ammeter reading. Take out the key from the plug. [Note: Always take out the key from the plug after measuring the current through the circuit. ]
• Replace the nichrome wire with the torch bulb in the circuit and find the current through it by measuring the reading of the ammeter.
• Now repeat the above step with the 10 W bulb in the gap XY.
• Are the ammeter readings differ for different components connected in the gap XY? What do the above observations indicate?
• You may repeat this Activity by keeping any material component in the gap. Observe the ammeter readings in each case. Analyse the observations.

Question 1.
In this Activity we observe that the current is different for different components. Why do they differ?
Answer:
Certain components offer an easy path for the flow of electric current while the others resist the flow. We know that motion of electrons in an electric circuit constitutes an electric current. The electrons, however, are not completely free to move within a conductor. They are restrained by the attraction of the atoms among which they move. Thus, motion of electrons through a conductor is retarded by its resistance. A component of a given size that offers a low resistance is a good conductor. A conductor having some appreciable resistance is called a resistor. A component of identical size that offers a higher resistance is a poor conductor. An insulator of the same size offers even higher resistance.

Activity 12.3 (Page 206)

• Complete an electric circuit consisting of a cell, an ammeter, a nichrome wire of length 1 [say, marked (1)] and a plug key as shown in Fig.
• Now, plug the key. Note the current in the ammeter.
• Replace the nichrome wire by another nichrome wire of same thickness but twice the length, that is 21 [marked (2) in the Fig.].
• Note the ammeter reading.
• Now replace the wire by a thicker nichrome wire, of the same length 1 [marked (3)]. A thicker wire has a larger cross-sectional area. Again note down the current through the circuit.

Fig. : Electric circuit to study the factors on which the resistance of conducting wires depends

• Instead of taking a nichrome wire, connect a copper wire [marked (4) in Fig.] in the circuit. Let the wire be of the same length and same area of cross-section as that of the first nichrome wire [marked (1)] Note the value of the current.
• Notice the difference in the current in all cases.
• Does the current depend on the length of the conductor?
• Does the current depend on the area of cross-section of the wire used?

It is observed that the ammeter reading decreases to one-half when the length of the wire is doubled. The ammeter reading is increased when a thicker wire of the same material and of the same length is used in the circuit. A change in ammeter reading is observed when a wire of different material of the same length and the same area of cross-section is used. On applying Ohm’s law, we observe that the resistance of the conductor depends (i) on its length, (ii) on its area of cross-section, and (iii) on the nature of its material. Precise measurements have shown that resistance of a uniform metallic conductor is directly proportional to its length (l) and inversely proportional to the area of cross-section (A).

Activity 12.4 (Page 210)

• Join three resistors of different values in series. Connect them with a battery, an ammeter and a plug key, as shown in Fig. You may use the resistors of values like 1 Ω, 2 Ω, 3 Ω etc., and a battery of 6 V for performing this Activity.
• Plug the key. Note the ammeter reading.
• Change the position of ammeter to anywhere in between the resistors. Note the ammeter reading each time.

Question 1.
Do you find any change in the value of current through the ammeter?
Answer:
Observation : We observe that the value of the current in the ammeter is the same, independent of its position in the electric circuit. It means that in a series combination of resistors the current is the same in every part of the circuit or the same current through each resistor.

Activity 12.5 (Page 211)

• In Activity 12.4, insert a voltmeter across the ends X and Y of the series combination of three resistors, as shown in Fig. 12.8 of text-book.

• Plug the key in the circuit and note the voltmeter reading. It gives the potential difference across the series combination of resistors. Let it be V. Now measure the potential difference across the two terminals of the battery. Compare the two values.
• Take out the plug key and disconnect the voltmeter. Now insert the voltmeter across the ends X and P of the first resistor, as shown in Fig.
• Plug the key and measure the potential difference across the first resistor. Let it be V₁.
• Similarly, measure the potential difference across the other two resistors, separately. Let these values be V₂ and V₃ respectively.
• Deduce a relationship between V, V₁, V₂ and V₃.

We observe that the potential difference V is equal to the sum of potential differences V₁, V₂, and V₃. That is the total potential difference across a combination of resistors in series is equal to the sum of potential difference across the individual resistors. That is,
V = V₁ + V₂ + V₃

Activity 12.6 (Page 213)

• Make a parallel combination, X’. of three resistors having resistances R₁, R₂, and R₃, respectively. Connect it with a battery, a plug key and an ammeter, as shown in Fig. Also connect a voltmeter in parallel with the combination of resistors.
• Plug the key and note the ammeter reading. Let the current be I. Also take the voltmeter reading. It gives the potential difference V, across the combination. The potential difference across each resistor is also V. This can be checked by connecting the voltmeter across each individual resistor (see Fig.).

• Take out the plug from the key: Remove the ammeter and voltmeter from the circuit. Insert the ammeter in series with the resistor R₁, as shown in Fig. Note the ammeter reading, I₁.

• Similarly, measure the currents through R₂ and R₃. Let these be I₂ and I₃, respectively.

Question 1.
What is the relationship between I, I₁, I₂ and I₃?
Answer:
Observation : It is observed that the total current I, is equal to the sum of the separate currents through each branch of the combination.
I = I₁ +I₂ + I₃

Class 10 Science Chapter 12 Electricity Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
Why resistance becomes more in series combination ?
Answer:
Because resistance is directly proportional to the length in series.

Question 2.
Why resistance becomes less in parallel combination ?
Answer:
Because resistance is inversely proportional to the length in parallel.

Question 3.
What is the difference between kilowatt and kilo hour ?
Answer:
Kilowatt is the unit of electric power whereas kilowatt hour is the unit of electric power.

Question 4.
What is the main property of charge ?
Answer:
Unlike charges attract each other while like charges repel each other.

Question 5.
What are the special property of a heating wire ?
Answer:
High resistance and high melting point.

Short Answer Type Questions

Question 1.
What is the resistance of an electric arc lamp, if the lamp uses 20 A when connected to a 220 Volt line ?
Answer:
Here, V = 220 V
I = 20 A
Since R = \(\frac{\mathrm{V}}{\mathrm{I}}\)
Or R = \(\frac{220}{22}\) = 11 Ohm

Question 2.
If you connect three resistors having values 2Ω, 3Q and 5 Q in parallel, then will the value of the total resistance be less than 2 Ω, or greater than 5 Ω, or lie between 2 Ω and 5 Ω ? Explain.
Answer:
Since the resistors are connected in parallel.

This show that the resistance is less than 2.

Long Answer Type Questions

Question 1.
How much work is done during flow of current for a given time in a wire ?
Answer:
As the conductor offers resistance to the flow of current, work is done continuously to move the electrons or maintain the flow of current.

The work done in carrying charge 0 through potential difference is given by
W = Q × V
Q = I × t
W = I × t × V
According to the Ohm’s law,
V = 1R
On putting the value of V in equation (i), we get
W = I²Rt.
Since, energy is the capacity or ability to do work, therefore we can also say that H [electrical energy] = I²Rt.

Multiple Choice Questions

Question 1.
The element of an electric heater is made by:
(a) Nichrome
(b) Aluminium
(c) Tungsten
(d) Platinium
Answer:
(a) Nichrome

Question 2.
In V-I, graph the slope of the line gives:
(a) Current
(b) Potential difference
(c) Resistance
(d) Reciprocal of potential difference
Answer:
(c) Resistance

Question 3.
The resistance of a straight conductor does not depend on its:
(a) Resistivity
(b) Materials
(c) Length
(d) Shape of cross section
Answer:
(d) Shape of cross section

Question 4.
The unit of electric charge is :
(a) Coulomb
(b) Volt
(c) Ohm
(d) Ampere
Answer:
(a) Coulomb

Question 5.
The unit of resistance of:
(a) Ohm
(b) Volt
(c) Ampere
(d) Coulomb
Answer:
(a) Ohm