Q.1. The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0. 4 Ώ, What is the maximum current that can be drawn from the battery ?
Q.2. A battery of emf 10 V and internal resistance 3 Ώ 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?
Q.3. (a) Three resistors 1 Ώ , 2Ώ, and 3Ώ are combined in serves. 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.
(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.
Sol. (a) Here, R1 = 1 Ώ : R2 = 2Ώ : R3 = 3 Ώ : V = 12 V
In series, total resistance Rs = R1 + R2+ R3 = 1+2+3= 6 Ώ
(b) Current through he circuit , I = V/ Rs = 12/6 = 2 A
∴ Potential drop across R1 = I R1 = 2×1 = 2 V
Potential drop across R2 = I R2 = 2×2 =4V
Potential drop across R3 = I R3 = 2×3=6V
In series, total resistance Rs = R1 + R2+ R3 = 1+2+3= 6 Ώ
(b) Current through he circuit , I = V/ Rs = 12/6 = 2 A
∴ Potential drop across R1 = I R1 = 2×1 = 2 V
Potential drop across R2 = I R2 = 2×2 =4V
Potential drop across R3 = I R3 = 2×3=6V
Q.4. (a) Three resistors 2 Ώ, 4Ώ, and 5 Ώ 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) 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.
Sol. (a) Here, R1 = 2 Ώ ; R2 = 4 Ώ ; R3 = 3Ώ ; V = 20 V
In parallel combination, total resistance Rp is given by
(b) Current through R1 = V/ R1 = 20/2 = 10 A ; Current through R2= 20/4 = 5 A
Current through R3 = 20/5 = 4 A
In parallel combination, total resistance Rp is given by
(b) Current through R1 = V/ R1 = 20/2 = 10 A ; Current through R2= 20/4 = 5 A
Current through R3 = 20/5 = 4 A
Q.5. At room temperature (27.00 C) the resistance of a heating element is 100 Ώ What is the temperature of the element if the resistance is found to be 117 Ώ, given that the temperature co- efficient of the material of the resistor is 1.70 10-4 0C-1.
Q.6. A negligibly small current is passed through a wire of length 15 m an2d uniform cross section 6.0×〖10〗^(-7) m2, and its resistance is measured to be 5.0 Ώ. What is the resistivity of the material at the temperature of the experiment ?
Q.7. A silver wire has a resistance of 2.1 Ώ at 27.5°C, and a resistance of 2.7 Ώ at 1000C, Determine the temperature coefficient of resistivity of silver.
Q.8. A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.00C ? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 10-4 0 C-1.
Q.9. Determine the current in each branch of the following network .
Sol. The currents the various arms of the circuit have been shown in
According to Kirchhoff’s second law;
In a closed circuit EABCE ;
1 10 = 10(i1-i2)+ 10 i1 + 5 (i1 – i3)
According to Kirchhoff’s second law;
In a closed circuit EABCE ;
1 10 = 10(i1-i2)+ 10 i1 + 5 (i1 – i3)
Or 10 = 25 i1 +10 i2 – 5 i3
Or 2 = 5 i1 + 2 i2 – i3
In a closed circuit BCDA ;
10 i1 + 5 i3 – 5 i2 = 0
Or 2 i1 + i3 – i2 = 0
Or i2 = 2 i1 + i3
In a closed circuit BCDB ;
5 (i1 – i3) – 10 (i1 + i3 ) – 5 i3 = 0
Or 5 i1 – 10 i2 – 20 i3 = 0 or i1 = 2i2 + 4 i3 ….(iii)
Form (ii) and (iii) ;
I1 = 2(2 i1 + i3) + 4 i3 = 4 i1 + 6 i3 or 3 i1 = – 6 i3 or i1 = – 2 i3 …..(iv)
Putting this value in (ii) ; i2 = 2 (-2 i3) + i3 = -3 i3
Putting values in (i) ; 2 = 5(-2 i3) + 2 (-3 i3) – i3 or 2 = -17 i3 or i3 = -2/17 A
From (iv) ; i1 = -2 (-2/17)= 4/17 A ;
From (v) ; i2 = – 3 (- 2/17) = (6/17) A
i1 + i2 = (4/17) + (6/17) = (10/17) A
I1 – i3 = 4/17 – (-2/17) = (6/17) A
I2 + i3 = 6/17 + (- 2 /17) = 4/17 A
Q – 10. (a) In a meter bridge, the balance point is found to be at 39.5 cm from the left and A, if an unknown resistor X is in the left gap and a known resistor Y of resistance 12.5 Ω is an the right gap. Determine made of thick copper strips ?
(b) Determine the balance point of the above bridge 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 ?
(b) Determine the balance point of the above bridge 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 ?
Thick copper strips are used to minimise resistance of the connections which being too small, can be neglected in the formula.
(b) As X and Y are interchanged, therefore, l1 and l2 (i.e., lengths) are also interchanged.
Hence, I = 100 – 39.5 = 60.5 cm
(c) The position of balance point remains unchanged. The galvanometer will show no current.
(b) As X and Y are interchanged, therefore, l1 and l2 (i.e., lengths) are also interchanged.
Hence, I = 100 – 39.5 = 60.5 cm
(c) The position of balance point remains unchanged. The galvanometer will show no current.
Q.11. A storage battery of emf 8.0 V and internal resistance of 0.5 Ω is being charged by a 120 V d. c. 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 ?
Sol. here, e. m. f. of the battery = 8.0 V ;
Voltage of d. c. supply = 120 V
Internal resistance battery, r = 0.5 Ω ;
External resistance, R = 15.5 Ω
Since a storage battery of e. m. f. 8 V is charged with a d. c. supply 120 V, the effective e. m. f. in the circuit is given by
E = 120 – 8 = 112 V
Total resistance of the circuit = R + r = 15.5 + 0.5 = 16.0 Ω
∴ Current in the circuit during charging is given by,
Voltage of d. c. supply = 120 V
Internal resistance battery, r = 0.5 Ω ;
External resistance, R = 15.5 Ω
Since a storage battery of e. m. f. 8 V is charged with a d. c. supply 120 V, the effective e. m. f. in the circuit is given by
E = 120 – 8 = 112 V
Total resistance of the circuit = R + r = 15.5 + 0.5 = 16.0 Ω
∴ Current in the circuit during charging is given by,
V = 120 – 108.5 = 11.5 V
The series resistor limits the current drawn from the external source of d. c. supply. In its absence the current will be dangerously high.
Q.12. in a potentiometer arrangement, a cell of emf 1.25 V given 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 ?
Q.13.. The number density of free electrons in a copper conductor is estimated at 8.5 n1028 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 wore is 2.010-6 m2 and it is carrying a current of 3.0 A.
Q.14.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 low conductivity of the lower atmosphere) in a current of only 1800 A over the entire glove. If there is a 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 thunder storms and lightning in different parts of the globe). Radius of the earth = 6.37 106 m.
Q.15. (a) six lead- acid type of secondary cells each of emf 2.0 V and internal resistance 0.0. Ω are joined in series to provide a supply to a resistance of 8.5Ω. 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 ?
(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 ?
This amount of current cannot start a car because to start the motor, the current required is 100 A for few seconds.
Q.16.Two wires of equal length, one of aluminium and the outer of copper have the same resistance. Which of the two wires is lighter ? Hence explain why aluminium wires are preferred for overhead power cables. Given, For Al , p1 = 2.63 m, For Cu, p2 = 1.7210-8 Ω m. Relative density of Al = 2.7, of Cu = 8.9.
Sol.. Given for aluminum wire; R1 = R ; l1 = l ;
Relative density d1 = 2.7
Let A1, A2 be the area of cross section for aluminum wire and copper wire.
It shows that copper wire is 2.16 time =s heavier than aluminium wire. Since the same value of length and resistance, aluminium were has lesser mass then copper wire, therefore aluminium wire is preferred of overhead power cables. A heavy cable may sag down owing to its own weight.
Q.17. 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
0.4 0.6 0.8 1.0 2.0 |
3.94
7.87 11.8 15.7 19.7 39.4 |
3.0
4.0 5.0 6.0 7.0 8.0 |
59.2
78.8 98.6 118.5 138.2 158.0 |
Sol. Since the ratio of voltage and current for different readings is same so Ohm’s law is valid to high accuracy. The resistivity of the alloy manganin in nearly independent of temperature.
Q.18. Answer the following question : (a) A steady current flows in metallic conductor of non-uniform cross-section. Explain which of these quantities is constant along the conductor : current, current density, electric field and 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 low internal resistance. Why?
(d) A high tension (HT) supply of say 6 kV must have a very large internal resistance. Why?
(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 low internal resistance. Why?
(d) A high tension (HT) supply of say 6 kV must have a very large internal resistance. Why?
Sol. Only current through the conductor of non- uniform area of cross- section is constant as the remaining quantities vary inversely with the areas of cross- section of the conductor.
(b) Ohm’s law is not applicable for non ohmic elements. For example ; vacuum tubes, semi- conducting diode, liquid electrolyte etc.
(c) As. L max =e. m. f . /internal resistance, so for maximum current, internal resistance should be least.
(d) A high tension supply must have a large internal resistance otherwise, if accidently the circuit is shorted, the current drawn will exceed safety limit and will cause damage to circuit.
(b) Ohm’s law is not applicable for non ohmic elements. For example ; vacuum tubes, semi- conducting diode, liquid electrolyte etc.
(c) As. L max =e. m. f . /internal resistance, so for maximum current, internal resistance should be least.
(d) A high tension supply must have a large internal resistance otherwise, if accidently the circuit is shorted, the current drawn will exceed safety limit and will cause damage to circuit.
Q.19. Choose the current alternatives :
(a) Alloys of metals usually have (greater/less) resistivity than that constituent metals.
(b) Alloys usually have much (lower/ higher) temperature coefficients of resistance then pure metals.
(c) The resistivity of the alloy manganin in (nearly independent of/ increases rapidly) with increase of temperature.
(d) The resistivity of a typical insulator (e.g. amber) is greater then that of a metal by a factor of the order of (1022 or 103).
(a) Alloys of metals usually have (greater/less) resistivity than that constituent metals.
(b) Alloys usually have much (lower/ higher) temperature coefficients of resistance then pure metals.
(c) The resistivity of the alloy manganin in (nearly independent of/ increases rapidly) with increase of temperature.
(d) The resistivity of a typical insulator (e.g. amber) is greater then that of a metal by a factor of the order of (1022 or 103).
Sol. (a) greater (b) lower (c) nearly independent of (d) 1022.
Q.20. (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 you 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 and (b).
(b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will you 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 and (b).
Sol.(a) For maximum effective resistance, the n resistors must be connected in series.Maximum effective resistance, in Rs = nR.
For minimum effective resistance, the n resistors must be connected in parallel.
Maximum effective resistance, Rp = R/n
(b) It is to be noted that (a) the effective resistance of parallel combination of resistors is less then the individual resistance and (b) the effective resistance of series combination of resistors is more than individual resistance.
Case (i) Parallel combination of 1 Ω and 2 Ω is connected in series with 3 Ω.
Effective resistance of 1 Ω and 2 Ω in parallel will be given by Rp 
Case (iii) All the resistances are to be connected in series. Now
∴Equuivalent resistance=1+2+3=6Ω
Case (iv) All the resistance are to be connected in parallel
(c) For the given net work is a series combination of 4 equal units. Each unit has 4 resistances in which, two resistance (1 Ω each in series) are in parallel with two other resistances (2 Ω each in series).
For the five resistances each of value R, are connected in series.
Their effective resistance = 5 R
Q.21. Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the following infinite net work ; each resistor has 1 Ω resistance.
Sol. Let x the equivalent resistance of infinite net work. Since the net work is infinite, therefore, the addition of one more unit of three resistances each of value of 1 Ω across the terminals will not alter the total resistance of net work, i. e., it should remain x .
Therefore, the network would appear as shown in and its total resistance should remain x. Here the parallel combination of x and 1 Ω is in series with the two resistors of 1 Ω each. The resistance of parallel combination is
The vale of resistance can be negative, therefore, the resistance of net work
Therefore, the network would appear as shown in and its total resistance should remain x. Here the parallel combination of x and 1 Ω is in series with the two resistors of 1 Ω each. The resistance of parallel combination is
The vale of resistance can be negative, therefore, the resistance of net work
Q.22. shows a potentiometer with a cell of 2.0 V and internal resistance 0.4 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 amperes) gives a balance point at 67.3cm. length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 K Ω 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 out to be at 82.3 cm length of the wire.
(a) What is the value of E ?
(b) What purpose does the high resistance of 600 k Ω have ?
(c) Is the balance point affected by the high resistance ?
(d) Is the balance point affected by the internal resistance of the driver cell ?
(e) Would the method work in the absolute 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 extremely small emf, say of the order of a few mV (such as the typical emf of a thermocouple) ? If not, how will you modify the circuit ?
(a) What is the value of E ?
(b) What purpose does the high resistance of 600 k Ω have ?
(c) Is the balance point affected by the high resistance ?
(d) Is the balance point affected by the internal resistance of the driver cell ?
(e) Would the method work in the absolute 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 extremely small emf, say of the order of a few mV (such as the typical emf of a thermocouple) ? If not, how will you modify the circuit ?
Sol .(a) Here, E1 = 1.02 V ; l1 = 67.3 cm ; E2 = E = ? ; l2 = 82.3cm
(b) The purpose of using high resistance of 600 k Ω is to allow very small current through the galvanometer when the movable contact is far from the balance point.
(c) No, the balance point is not affected by the presence of this resistance.
(d) No, the balance point is not affected by the internal resistance of the driver cell.
(e) No, the method will not work as the balance point will not be obtained on the potentiometer wire if the e. m. f. of the driver cell is less than the e. m. f. of the other cell.
(f) The circuit will not work for measuring extremely small e. m. f. because in that case, the balance point will be just close to the end A. To modify the circuit, we have to use a suitable high resistance in series with the cell of 2.0 V. This would decrease the current in the potentiometer were. Therefore, potential difference/cm of wire will decrease. Then only extremely small e. m. f. can be measured.
Q.23. 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 e. m. f. E?
Sol.Here, l1 = 58.3 cm ; l2 = 68.5 cm ; R = 10 Ω ; X = ?
Let / be the current in the potentiometer wire and ε_1 and ε_2 be the potential drops across R and X respectively . When connected in circuit by closing respective key. Then
If there is no balance point with given cell of e. m. f. it means potential drop across R or X is greater than the potential drop across the potentiometer wire AB. In order to obtain the balance point, the potential drops across R and X are to be reduced, which is possible by reducing the current in R and X. For that, either a suitable resistance should be put in series with R and X or a cell of smaller e. m. f. E should be used. Another possible way is to increase the potential drop across the potentiometer wire by increasing the voltage of driver cell.
Let / be the current in the potentiometer wire and ε_1 and ε_2 be the potential drops across R and X respectively . When connected in circuit by closing respective key. Then
If there is no balance point with given cell of e. m. f. it means potential drop across R or X is greater than the potential drop across the potentiometer wire AB. In order to obtain the balance point, the potential drops across R and X are to be reduced, which is possible by reducing the current in R and X. For that, either a suitable resistance should be put in series with R and X or a cell of smaller e. m. f. E should be used. Another possible way is to increase the potential drop across the potentiometer wire by increasing the voltage of driver cell.
Q.24. The shows a 2.o V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm. length of the potentiometer. Determine the internal resistance of the cell. External
