NCERT Solution Class 11 Physics Chapter 3: Current Electricity
NCERT Solution for Class 11 Physics Chapter 3 is based on Current Electricity. These principles are very important to understand if you are a student or have aspirations of studying the behavior of electric circuits and its constituents in the future. These are the solutions that explain Ohm's law and show the current, voltage, and resistance as interlinked features that affect resistance in clear and simple ways. Such foundational knowledge is critical for current electricity at the Class 12 level. Current Electricity In the class 11 Physics chapter 3, which is all about the sources of current and the motion of charges in the conductor. NCERT Solutions for Class 11 Physics Chapter 3: Current Electricity, apart from covering the concepts, also introduces students to a detailed discussion on the types of circuits including the series and parallel circuits. Although about the analysis and solution of circuit problems through Kirchhoff's laws and other methods. Since a number of examples and practice problems are there associated with the Class 12 physics chapter 3, thus students can work on to secure a proper understanding of the chapter on current electricity.
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The NCERT Solution Class 11 Physics Chapter 3: Current Electricity are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 2023-2024. These solutions provide thorough explanations with a step-by-step approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Access Answers to NCERT Solution Class 11 Physics Chapter 3: Current Electricity
Students can access the NCERT Solution Class 11 Physics Chapter 3: Current Electricity. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Physics much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Current Electricity
Aheating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds toa steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10−4 °C −1.
Given Supply Voltage = 230 Volt
And it draws current I1 = 3.2 Ampere
Resistance is given by R1 = V/I
R1 = 230V/3.2A
R1 = 71.87 Ω
Steady value of current I2 = 2.8Ampere
Resistance R α is given by R2 = 230V/2.8A
R2 = 82.14 Ω
Temperature coefficient of nichrome wire α = 1.70 × 10-4 per °C
Initial temperature of nichrome T1 = 27 ° C
Steady Temperature attained by nichrome = T2
The formula is: R2 = R1[1 + α(T2-T1)]
⇒ R2 – R1 = R1α(T2 – T1)
⇒ R2 – R1 = R1α (T2 – T1)
The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery?
Given that Emf of the battery, E = 12 V
Battery has an Internal resistance of R = 0.4 Ω
The Maximum current drawn from the battery is given by = I
According to Ohm’s law,
E = IR
I=E / R I=12/0.4 = 30 A
The maximum current drawn from the given battery is 30 A.
A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Electromotive force (E) of battery = 10 volt.
Internal Resistance r of the battery = 3 Ω
Current in the Circuit (I) = 0.5 A
Let the resistance of the required resistor is R Ω
Let the Terminal voltage of the register be V.
As per Ω ’s law,
(a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.
The diagram is given as:
A) When three resistors are connected in series then the total resistance is equal to sum of individual resistance.
Total resistance is equal to = 1 + 2 + 3 = 6 Ω
(b) Let the current flowing in the circuit be I.
Electromotive force EMF, (E) = 12V
Total resistance in the circuit R = 6 Ω
According to Ohm ’s law, V = IR
(a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.
(b)
At room temperature (27.0 °C) the resistance of a heating element is 100Ω. What is the temperature of the element if the resistance is found to be 117Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10–4 °C–1.
Given: Temperature coefficient of filament,
α = 1.70 × 10-4 per °C
Let T1 be the temperature of element, R1 = 100Ω (Given: T1 = 27°C)
Let T2 be the temperature of element, R2 = 117Ω
To find T2 = ?
The formula is: R2 = R1[1 + α(T2-T1)]
⇒ R2 – R1 = R1α(T2 – T1)
⇒ R2 – R1 = R1α (T2 – T1)
A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10−7 m2, and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?
Given, the length of wire l = 15 metre
Area of cross section of wire a = 6.0 × 10-7 metre square
Resistance of material of wire,R = 5 Ω
Resistivity of material of wire = ρ
A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.
Resistance of silver wire at 27.5 °C R1 = 2.1 Ω
Resistance of silver wire at 100 ° C R2 = 2.7 Ω
Let the temperature coefficient of silver wire is given by α.
The formula for α is: R2 = R1[1 + α(T2-T1)]
⇒ R2 – R1 = R1α(T2 – T1)
⇒ R2 – R1 = R1α (T2 – T1)
Determine the current in each branch of the network shown in fig 3.30:
The current through the labelled diagram shows the current flowing the respective branches:
For the closed circuit ABDA, potential is zero i.e.,
10I2 + 5I4 − 5I3 = 0
2I2 + I4 −I3 = 0
I3 = 2I2 + I4 … (1)
For the closed circuit BCDB, potential is zero i.e.,
5(I2 − I4) − 10(I3 + I4) − 5I4 = 0
5I2 + 5I4 − 10I3 − 10I4 − 5I4 = 0
5I2 − 10I3 − 20I4 = 0
I2 = 2I3 + 4I4 … (2)
For the closed circuit ABCFEA, potential is zero i.e.,
−10 + 10 (I1) + 10(I2) + 5(I2 − I4) = 0
10 = 15I2 + 10I1 − 5I4
3I2 + 2I1 − I4 = 2 … (3)
From equations (1) and (2), we obtain
I3 = 2(2I3 + 4I4) + I4
I3 = 4I3 + 8I4 + I4
− 3I3 = 9I4
− 3I4 = + I3 … (4)
Putting equation (4) in equation (1), we obtain
I3 = 2I2 + I4
− 4I4 = 2I2
I2 = − 2I4 … (5)
It is evident from the given figure that,
I1 = I3 + I2 … (6)
Putting equation (6) in equation (1), we obtain
3I2 + 2(I3 + I2) − I4 = 2
5I2 + 2I3 − I4 = 2 … (7)
Putting equations (4) and (5) in equation (7), we obtain
5(−2 I4) + 2(− 3 I4) − I4 = 2
− 10I4 − 6I4 − I4 = 2
17I4 = − 2
I4 = -2/17Ampere
Equation (4) reduces to
I3 = − 3(I4)
I3 = -3 × -2/17
I3 = 6/17Ampere
I2 = -2(I4)
I2 = -2 × -2/17
I2 = 4/17Ampere
I2 – I4 = 6/17Ampere
I3 + I4 = 6/17Ampere
I1 = I2 + I3 = 4/17 + 6/17 = 10/17 Ampere
Therefore, current in branch AB = 4/17Ampere
In branch BC = 6/17Ampere
In branch CD = -4/17Ampere
In branch AD = 6/17Ampere
In branch BD = -2/17Ampere
Total current = 4/17 + 6/17 + -4/17 + 6/17 + -2/17 = 10/17Ampere.
(a) In a metre bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?
(b) Determine the balance point of the bridge above if X and Y are interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?
The diagram is shown below:
(b) If X and Y are interchanged, then l1 and 100−l1 get interchanged.
The balance point of the bridge will be 100−l1 from A.
100−l1 = 100 − 39.5 = 60.5 cm
Therefore, the balance point is 60.5 cm from A.
(c) When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflection. Hence, no current would flow through the galvanometer.
A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
Electromotive force EMF of battery = 8Volt
Internal resistance of battery r = 0.5 Ω
Supply Voltage V = 120Volt
Resistance of resistor R = 15.5 Ω
Effective voltage in circuit = V1
Since Resistance R is connected in series Hence We can write V1 = V-E
V1 = 120V-8V = 112 V
Current flowing in the circuit:
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?
Given EMF of the cell E1 = 1.25Volt
Balance point of the potentiometer l1 is given 35cm
Since the cell is replaced by another cell of EMF E2
New balance point of potentiometer l2 is 63cm
.
The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 1028 m−3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10−6 m2 and it is carrying a current of 3.0 A.
Number density of free electrons in a copper conductor, n = 8.5 × 10^28 per meter cube
Length of Copper Wire l = 3 meter
Area of cross section of wire A = 2 × 10-6 meter square
Current drawn by wire is given 3 Ampere
Since I = nAeVd
Here Vd is the drift velocity and “e” is the electron charge 1.6 × 10^-19Coulomb.
The earth’s surface has a negative surface charge density of 10−9 C m−2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 106 m.)
Surface Charge density of earth d = 10-9 Coulomb per meter square
Current over the entire globe = 1800 Ampere
Radius of earth r = 6.37 × 106 meter
Surface area of earth A = 4×π×radius×radius
A = 4×π×6.37×106×6.37×106
A = 5.09 × 1014 m2
Charge on the earth surface q = d × A
q = 10-9 × 5.09 ×1014 m2
q = 5.09 × 105 Coulomb
Let the time taken to neutralize earth surface = t
(a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?
(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?
A) Number of secondary cells n = 6
Emf of each secondary cell E = 2 Volt
Internal resistance of each cell r = 0.015 Ω
Resistance of resistor R = 8.5 Ω
Let current drawn from supply = I
Hence maximum current drawn from cell is 0.005Ampere and since a large current is required to start the motor of a car and hence cell cannot be used to start the motor.
Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. (ρAl = 2.63 × 10−8 Ω m, ρCu = 1.72 × 10−8 Ω m, Relative density of Al = 2.7, of Cu = 8.9.)
Resistivity of aluminium, p1 = 2.63 × 10−8 Ω m
Relative density of aluminium, d1 = 2.7
Let l1 be the length of aluminum wire and m1 be its mass.
Resistance of the aluminum wire = R1
Area of cross-section of the aluminum wire = A1
Resistivity of copper, p2 = 1.72 × 10−8 Ω metre
Relative density of copper, d2 = 8.9
Let l2 be the length of copper wire and m2 be its mass.
Resistance of the copper wire = R2
Area of cross-section of the copper wire = A2
We can write,
R1 = (p1 × l1)/A1
R2 = (p2 × l2)/A2
Since R1 = R2
(p1 × l1)/A1 = (p2 × l2)/A2
And l1 = l2
Therefore p1/A1 = p2/A2
A1/A2 = (2.63 × 10-8)/(1.72 × 10-8)
⇒ A1/A2 = 2.63/1.72
Mass of the aluminium wire,
m1 = Volume × Density
= A1 × l1 × d1 = A1 l1 d1 …..1
Mass of the copper wire, m2 = Volume × Density
= A2 × l2 × d2 = A2 l2 d2 …….2
Dividing equation (1) by equation (2), we obtain
m1/m2 = (A1 l1 d1)/(A2 l2 d2)
Since l1 = l2
m1/m2 = A1 d1/A2 d2
Since A1/A2 = 2.63/1.72 calculated above
⇒ m1/m2 = (2.63 × 2.7)/(1.72 × 8.9)
⇒ m1/m2 = 0.46
Since m1 is smaller than m2 Hence, aluminium is lighter than copper. Since aluminium is lighter, it is preferred for overhead power cables over copper.
What conclusion can you draw from the following observations on a resistor made of alloy manganin?
|
Current A |
Voltage V |
Current A |
Voltage V |
|
0.2 |
3.94 |
3.0 |
59.2 |
|
0.4 |
7.87 |
4.0 |
78.8 |
|
0.6 |
11.8 |
5.0 |
98.6 |
|
0.8 |
15.7 |
6.0 |
118.5 |
|
1.0 |
19.7 |
7.0 |
138.2 |
|
2.0 |
39.4 |
8.0 |
158.0 |
It can be inferred from the given table that the ratio of voltage with current is a constant, which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys Ohm’s law. According to Ohm’s law, the ratio of voltage with current is the resistance of the conductor. Hence, the resistance of manganin is 19.7 Ω.
Answer the following questions:
(a) A steady current flows in a metallic conductor of non-uniform cross- section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed?
(b) Is Ohm’s law universally applicable for all conducting elements?
If not, give examples of elements which do not obey Ohm’s law.
(c) A low voltage supply from which one needs high currents must have very low internal resistance. Why?
(d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?
(a) When a steady current flows in a metallic conductor of non-uniform cross-section, the current flowing through the conductor is constant. Current density, electric field, and drift speed are inversely proportional to the area of cross-section. Therefore, they are not constant.
(b) No, Ohm’s law is not universally applicable for all conducting elements. Vacuum diode semi-conductor is a non-ohmic conductor. Ohm’s law is not valid for it.
(c) According to Ω ’s law V = I×R
If Voltage V is low, then Resistance R must be very low so that high current can be drawn from source.
(d) In order to prohibit the current from exceeding the safety limit, a high tension supply must have a very large internal resistance. If the internal resistance is not large, then the current drawn can exceed the safety limits in case of a short circuit.
Choose the correct alternative:
(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (1022/103).
(a) Alloys of metals usually have greater resistivity than that of their constituent metals.
(b) Alloys usually have lower temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy, manganin, is nearly independent of increase of temperature.
(d) The resistivity of a typical insulator is greater than that of a metal by a factor of the order of 1022.
(a) Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?
(b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?
(c) Determine the equivalent resistance of networks shown in Fig. 3.31.
(a) Total number of resistors = n
Resistance of each resistor = R
Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32. Each resistor has 1 Ω resistance.
The resistance of each resistor connected in the given circuit, R = 1 Ω
Equivalent resistance of the given circuit = R’
The network is infinite. Hence, equivalent resistance is given by the relation,
R’ = 2 + (R’/R’ + 1)
or R’2 -2R’ -2 = 0
On solving R’ = (2+√4+8)/2
(Negative value of R’ cannot be accepted)
R’ = 1 + √3 Ω
R’ = 1 + 1.73 = 2.73 Ω
Internal resistance of the circuit, r = 0.5 Ω
Hence, total resistance of the given circuit = 2.73 + 0.5 = 3.23 Ω
Supply voltage, V = 12 V
According to Ω ’s Law, current drawn from the source is given by the ratio, = 12/3.23 = 3.72 A
Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
(a) What is the value ε ?
(b) What purpose does the high resistance of 600 kΩ have?
(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell?
(e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?
(f ) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?
a) Constant emf of the given standard cell, E1 = 1.02 V
Balance point on the wire, l1 = 67.3 cm
A cell of unknown emf, ε, replaced the standard cell. Therefore, new balance point on the wire,
l = 82.3 cm
The relation connecting emf and balance point is,
E1/l1 = E/l
E = l × E1/l1
E = 82.3 cm × 1.02 V/67.3 cm
E = 1.247 Volt.
The value of unknown emf is 1.247 V.
(b) Explanation:
The purpose of using the high resistance of 600 kΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.
(c) The balance point is not affected by the presence of high resistance.
(d) The point is not affected by the internal resistance of the driver cell.
(e) The method would not work if the driver cell of the potentiometer has an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then we cannot obtain balance point on the wire.
(f) The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a high percentage of error. The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.
Figure 34 shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε?
Resistance of the standard resistor, R = 10.0 Ω
Balance point for this resistance, l1 = 58.3 cm
Current in the potentiometer wire = i
Hence, potential drop across R, E1 = iR
Resistance of the unknown resistor = X
Balance point for this resistor, l2 = 68.5 cm
Hence, potential drop across X, E2 = iX
The relation connecting emf and balance point is,
E1/E2 = l1/l2
iR/iX = l1/l2
X = l1 × R/l2
X = 68.5 × 10/58.3
X = 11.749 Ω
Therefore, the value of the unknown resistance, X, is 11.75 Ω .
Explanation:
If we fail to find a balance point with the given cell of emf, E, then the potential drop across R and X must be reduced by putting a resistance in series with it. Only if the potential drop across R or X is lesser than the potential drop across the potentiometer wire AB, a balance point is obtained.
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