Chapter
3
Current Electricity Solutions SECTION - A School/Board Exam. Type Questions 1.
Define current.
Sol. Current is defined as rate of flow of charge through any area. 2.
State ohm’s law.
Sol. Ohm’s law says, at constant temperature current through a conductor is directly proportional to potential difference across it. 3.
Name the factors on which resistivity of the material depends.
Sol. Resistivity depends on material and temperature of material. 4.
Define current density.
Sol. It is defined as electric current per unit normal area of cross-section of conductor through which current passes. 5.
What do you mean by temperature coefficient of resistivity?
Sol. It is defined as change in resistivity per unit original resistivity per unit rise in temperature. 6.
Write the relation between electric current and drift speed.
Sol. I = neAvd 7.
How does the drift speed vary with relaxation time?
Sol. v d
eE m
vd 8.
Write Kirchhoff’s junction rule.
Sol. Sum of current meeting at a junction is equal to zero. 9.
Which conservation rule is valid in Kirchhoff’s loop rule?
Sol. Conservation of energy. 10. On which principle is meter bridge experiment based? Sol. Wheatstone bridge principle. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Current Electricity
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11. What is thermal speed of free electrons in a conductor? Estimate the order of thermal speed at 27°C. Sol. The rms speed at which free electrons move within the interatomic spaces of conductor is known as thermal speed. As they behave like gas molecules 3 kT M
3 1.38 10 23 300
105 m/s 1.6 10 19 12. Define emf of a cell. How is the emf achieved in daily life? V
Sol. emf of a cell is defined as the difference of potential across its two terminals, when there is no current through the cell. emf is achieved from storage cells, chemical cells, solar cells etc. 13. Obtain the relationship between current density and electric field inside a conductor. Sol. Current density : It is defined as current per unit normal area. It is denoted as I A SI unit of j is Am–2 j
...(1)
V is the potential difference across the conductor of length I, then electric field inside the conductor V l Using equation (1) & (2), Ohm’s law V = IR becomes E
...(2)
⎛I⎞ E l ⎜ ⎟ l ⎝ A⎠ E = j
Current density is directed along E Vectorially E j
or j E
where,
...(3)
1 is called conductivity of the material of the conductor.
14. How does the resistivity of conductor vary with temperature of conductor? Explain. Sol. Resistivity of a material is found to depend on its temperature. In conductors. Resistivity
m ne2
where
1 1 and . n When the temperature of a conductor increases, average speed of free electrons increases. As a result collision frequency increases or relaxation time decreases. In metals n is not dependent on temperature to any appreciable extent and increases with rise in temperature.
0.4
–8
T(10 m)
Different materials do not exhibit the same dependence on temperature. Resistivity of copper as a function of temperature T is shown below
0.2 0
50 100 Figure (A)
150
A
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Current Electricity
53
1.20
–8
T(10 m)
Resistivity of nichrome as a function of absolute temperature T
1.10 1.00
200 K 400 K 600 K 800 K
T (K)
Figure (B) Resistivity of a metallic conductor is approximately given by T = 0[1 + (T – T0)]
...(1)
where T = resistivity at T°C, 0 = resistivity at reference temperature T 0°C and = temperature coefficient of resistivity. For metal is positive as their resistivity increases with rise in temperature. Equation (1) implies that graph of T plotted against T would be a straight line as in figure (B). 15. Establish the relationship between emf (E) of a cell and potential difference (V) across the terminals of a cell. Sol. A current I flows from C to D. A steady current flows from P to N through the resistance R and flows N to P through the electrolyte. The electrolyte through which current passes has a finite resistance r known as internal resistance of the cell.
C
R
I P A
D N B
When the resistance R is infinite or P and N are in open circuit, there is no current through the electrolyte V ⎛ ⎞ ⎜I 0⎟ . ⎝ R ⎠ Potential difference between P and N = V = (VP – VA) + (VA – VB) + (VB – VN) = V+ + 0 + V–
Cells can be represented as P
= V + – V– =
r
N
emf of a cell is defined as the potential difference between the positive and negative electrodes in an open circuit. When R is finite, I 0, (Cell is discharging). Potential difference between P and N is Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Current Electricity
Solution of Assignment (Set-1)
V = VP – VA + VA – VB + VB – VA = V+ + (–Ir) + V– = V+ + V– – Ir = – Ir
V Ir We can represent it as P
...(1)
I
r
N
Internal resistance of the cells in the circuit may be neglected when the current I is such that >> Ir. Internal resistances of dry cells is much higher than common electrolytic cells. In case of storage cells, discharged cells can by charged by a driver cell E by supplying a steady current I through the cell from P to N as shown in figure.
I E P
N
A
B
In this case potential difference between P and N is V = VP – VA + VA – VB + VB – VN = V+ + Ir + V– = V+ + V– + Ir = + Ir V = + Ir It can be represented as P
I
r
N
Potential difference between terminals of cell may be equal to, greater than or less than the emf of cell. 16. Write the Wheatstone bridge principle. Sol. It is an application of Kirchhoff’s rules. The bridge is consisting of four resistances R1, R2, R3 and R4 as four sides of a square ABCD as shown in figure.
R2 A
I2
4
Ig
R4 C
G
I1 I
BI
R1 D
I3
R3
I
E
R2 R4 In Wheatstone bridge we consider the special case Ig = 0. R R 1 3 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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55
17. What is the source of emf in a chemical cell? Sol. An electrolytic cell consisting of two electrodes, called positive (P) and negative (N) immersed in an electrolytic solution as shown in figure.
C
D P
N A
B
Electrolyte Electrodes exchange charges with the electrolyte. Positive electrode P has a potential difference V+(V+ > 0) between itself and electrolyte solution A immediately adjacent to it. Negative electrode N has a potential difference –(V–)(V– > 0) relative to electrolyte B adjacent to it. When there is no current, the electrolyte is at same potential throughout so the potential difference between P and N is V+ – (–V–) = V+ + V–. This difference is known as electromotive force (emf) of the cell and denoted by . So, = V+ + V– > 0 18. What is the meaning of internal resistance of a cell? Write the factors affecting it. Sol. The opposition offered by the cell to flow of current through itself is known as internal resistance of the cell. It depdnds on 1. Concentration of electrolyte 2. Distance between electrodes 3. Area of electrodes 4. Temperature of electrolyte 19. How is the potential drop per unit length along the standard resistance wire of potentiometer changed? Sol. By changing 1. Current through the wire. 2. Length of the wire. 3. The value of resistance offered by Rheostat in series. 20. Write the advantage of potentiometer. Sol. It works in the zero current condition through the element across which potential is to be measured Its readings are more accurate than voltmeter and ammeter. Its sensitivity is high. 21. How can the percentage error in measurement of unknown resistance is minimised in meter bridge experiment? Sol. The percentage error in resistance R can be minimised by adjusting the balance point near the middle of the bridge (i.e., l1 is close to 50 cm) by making suitable choice of S. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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22. Drift speed of free electrons is of the order of few mm per second, but current establishes in a circuit with the speed of light. Explain. Sol. Electric field is built up in the conductor with the speed of light. 23. Discuss the use of transformer in power transmission and power distribution. Sol. Consider a device of resistance R to be operated at voltage V and current through it is I, then power of device P = VI. If resistance of connecting wires from power station to the device is RC then power dissipated in connecting wires is PC with PC I 2RC
P 2RC V2
Therefore, to drive a device of power P, the power wasted in the connecting wires PC
1 V2
RC As the distance of power station is very large, RC is considerable so to decrease PC, these wires carry current at enormous values of V and this is the reason for high voltage danger signs on transmission lines. These voltages are lowered to a value suitable for use by a device known as transformer. 24. What is the relation between mobility and drift velocity of charge carriers in a conductor? What are the carriers in electrolytes? Sol. It is defined as the magnitude of the drift velocity per unit electric field. vd E
...(1)
Its SI unit is m 2V–1 s –1 Its practical unit is cm 2V–1 s –1 We have v d
eE m
v d e E m
...(2)
Mobility of free electrons is independent of electric field and dimension of conductor. Carriers in electrolytes are anions and cations. 25. Why does the resistivity of semiconductors decrease with increase in temperature? Sol. When temperature of semiconductor increases, the formation of hole electron pairs increases. So carrier density increases. 26. What are ohmic and non-ohmic conductors? Draw potential versus current characteristic graph for ohmic and non-ohmic conductors. Give one example of each. Sol. Conductors which obey Ohm’s law strictly are known as ohmic conductors. Example : Metals
I
IV
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57
Conductors which do not obey Ohm’s law are known as non-ohmic conductors. Example : Semiconductor Ge, Si
I
V 27. What do you mean by resistance of a conductor? Write its SI unit and dimensional formula. Sol. The opposition offered by the conductor to the flow of current through it is known as resistance of conductor. SI unit – ohm Dimensional formula – [M1L2T–3A–2] 28. Draw the circuit diagram for Wheatstone bridge principle. Sol.
R1
R3 G
R2
E
R4
() K
29. Draw the circuit diagram used for comparing emf of two primary cells by potentiometer experiment. Sol. Comparison of emf of two cells of emf 1 and 2 : 1
1
2 B
3
2
A
G
N1
R
N2 C ( ) K1 30. Why does a conductor get heated when current passes through it? Sol. Due to collision of free electrons with lattice points, energy is transferred to lattice. This energy increases internal vibration in the lattice. The energy released during these vibration appear as heat. 31. Define drift velocity of free electrons in a conductor. Derive the relation between current and drift speed. Sol. When potential difference is applied across a conductor, average velocity of free electrons is independent of time, although electrons are accelerated. This phenomena is known as drift, and the velocity v i is known as drift velocity. Therefore, there will be net flow of charge across any area perpendicular to E .
A
B vd
–
E vdt Consider the area of cross-section A of conductor. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Distance travelled from left to right by free electrons in time t after crossing the area of cross-section = v d t . If n is number of free electrons per unit volume in the conductor, then number of free electrons passing the cross-section from left to right in time interval t is equal to number of free electrons in the volume nA v d t
Charge transported across the area of cross-section from left to right = neA v d t Total charge transported along E from right to left I t neA v d t
Current through the conductor I = neA v d
32. Find the total resistance when various resistors are connected in (i) Series (ii) Parallel Sol. (i) Series Consider two resistances R1 and R2 connected in series and potential difference V is applied across them.
V1 A
I
V2
R1
CI V
R2
B
I
Charge which leaves R1 must enter R2. Since current measures the rate of flow of charge, current I through both resistances are equal. Potential difference across R1, V1 = VA – VC = IR1 Potential difference across R2, V2 = VC – VB = IR2 Potential difference V across the combination V = VA – VB = (VA – VC) + (VC – VB)
V = V1 + V2 = I(R1 + R2) Equivalent resistance of the combination. RS =
V = R1 + R2 I
...(1)
If there are n resistances R1, R2, ..., Rn in series, then equivalent resistance RS = R1 + R2 + ... + Rn
...(2)
When resistances are used in series then their equivalent resistance is greater than greatest among them. (ii)
Parallel Consider two resistances R1 and R2 connected in parallel across AB.
I1 A
R1
I
I I2
B
R2
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59
The charge flowing in at A from the left flows out partly through R1 and partly through R2. So current I flowing in at A flows out as I1 and I2 through R1 and R2 respectively, hence I = I1 + I2
...(3)
Potential difference between A and B by applying Ohm law to R1 V = I1R1 Potential difference between A and B by applying Ohm’s law to R2 V = I2R2 V V 1 ⎞ ⎛ 1 V ⎜ ⎟ R1 R2 ⎝ R1 R2 ⎠ If the combination is replaced by equivalent resistance RP then by Ohm’s law
I = I1 + I 2 =
I
V RP
1 1 1 RP R1 R2
1 ⎞ ⎛ 1 RP ⎜ ⎟ ⎝ R1 R2 ⎠
1
...(4)
If n resistances R1, R2, ..., Rn are in parallel, then equivalent resistance 1 1 ⎞ ⎛ 1 ..... RP ⎜ Rn ⎟⎠ ⎝ R1 R2
1
When resistances are connected in parallel, then equivalent resistance is smaller than the smallest one among them. 33. Explain grouping of cells in series across a load resistance R and justify its use. Sol. Series A
I
1 r1
I B
2
I
r2
C
1 and 2 are emf of two cells used in series as shown in figure. Here r1 and r2 are their respective internal resistances. Current I is flowing from C to A. VA, VB and VC are potentials at points A, B and C respectively, then VAB = VA – VB = 1 – Ir1 and VBC = VB – VC = 2 – Ir2 Potential difference between terminals A and C is VAC = VA – VC = VA – VB + VB – VC = (1 + 2) – I(r1 + r2)
...(1)
We can replace the above combination by a single cell of emf eq and internal resistance req as shown below. eq I A C req We have VAC = eq – Ireq ...(2) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Comparing (1) and (2), we get eq = 1 + 2 and req = r1 + r2 In n cells are connected correctly in series, then eq = 1 + 2 + ..... + n and req = r1 + r2 + ..... + rn If cells are identical then eq = n and req = nr. If n identical cells each of emf E and internal resistance r are connected in series across a load resistance R, then current through the load resistance I I0 =
nE E . R nr R r
If R >> r, then I0 =
Eeq R req
nE . For a single cell across R, current R nr
nE E and I = = nI0. (Useful due to high current through load w.r.t. single cell). R R
34. Explain grouping of cells in parallel across a load resistance R and justify its use. Sol. Consider two cells of emf 1 and 2 connected in parallel across AB.
I1
I1
1 r1
A
I
2 I2
B
I2
r2
If I1 and I2 are the currents leaving positive electrodes of both cells, then total current from B to A I = I1 + I2
...(1)
VA and VB are potentials at A and B respectively, then potential difference between A and B, can be written as V = VA – VB = 1 – I1r1
...(2)
and V = VA – VB = 2 – I2r2
...(3)
Using equation (1), (2) & (3), we get I = I1 + I2
1 V 2 V ⎛ 1 2 ⎞ ⎛1 1⎞ ⎜ ⎟ V ⎜ ⎟ r1 r2 ⎝ r1 r2 ⎠ ⎝ r1 r2 ⎠
I
V
rr 1r2 2 r1 I 1 2 r1 r2 r1 r2
...(4)
We can replace the above combination by a single cell of emf eq and internal resistance req across AB as shown in the figure. A
eq req
I
B
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61
We have, VA = VA – VB = eq – Ireq
...(5)
Comparing (4) & (5), we have eq
1r2 2 r1 rr and req 1 2 r1 r2 r1 r2 ⎛ 1 1⎞ ⎜ ⎟ ⎝ r1 r2 ⎠
and
eq req
1
1 2 r1 r2
If n cells of emf 1, 2, ....., n and internal resistances r1, r2, ....., rn are in series, then 1 1 1 1 ..... req r1 r2 rn
...(6)
eq 1 2 n and r r r ..... r n eq 1 2
...(7)
For identical cells : r1 = r2 = ..... = rn = r and 1 = 2 = ..... = n =
req
r and eq = n
If n identical cells each of emf and internal resistance r are connected in parallel across a load resistance r then current through the resistance will be I load, I0 =
eq R req
. Rr
If r >> R, I0 =
R
r n
. For single cell across the
n = and I = = nI0 (Useful) Due to higher current that load w.r.t. single cell. r r
35. A cell of emf E and internal resistance r is connected across a load resistance R. Draw the graph between power P developed in the load and the load resistance R. Sol. Current through the load I =
Rr
Power delivered at the load P = I 2R =
R 2R (R r )2
Power delivered at load is maximum when Solving above equation R = r
2 r
2 Pmax (r r )2 4r P versus R graph will be
I
r
dP 0. dR
P 2 4r
r
R
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Solution of Assignment (Set-1)
36. State and explain Kirchhoff’s rules and deduce the Wheatstone bridge principle from these rules. Sol. Kirchhoff’s first rule : Junction rule At any junction, the sum of the currents entering the junction is equal to sum of currents leaving the junction. Explanation : When currents are steady, there is no accumulation of charges at any junction or at any point in a line, therefore, rate of flow of charge into the junction is equal to rate of flow of charge from the junction.
I2
I1 I5
O I3 I4
In the above figure applying junction rule I1 + I3 + I5 = I2 + I4 Kirchhoff’s second rule : Loop rule Algebraic sum of changes in potential around any closed loop involving resistors and cells in a loop is zero. Explanation : We know that electric potential depends on the location of point. Therefore, starting with any point if we come back to the same point the total change of potential must be zero. Wheatstone bridge It is an application of Kirchhoff’s rules. The bridge is consisting of four resistances R1, R2, R3 and R4 as four sides of a square ABCD as shown in figure.
R2 A
I2
4
Ig
R4 C
G
I1 I
BI
R1 D
I3
R3
I
E Across the diagonally opposite points between A and C, battery E is connected. This is called battery arm. To remaining two diagonally opposite points B and D, a galvanometer G is connected to detect current this line is known as galvanometer arm. Currents through all resistances and galvanometer are as shown in figure. In wheatstone bridge we consider the special case Ig = 0. Applying junction rule to junction B and D, we have I2 = I4 and I1 = I3 Applying loop rule to loop ABDA Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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I2R2 + 0 – I1R1 = 0
I1 R2 I2 R1
...(i)
Applying loop rule to loop BCDB I4R4 – I3R3 + 0 = 0
I2R4 – I1R3 = 0
I1 R4 I2 R3
(Using I4 = I2 and I3 = I1) ...(ii)
Using equation (i) & (ii), we have R2 R4 R1 R3
...(iii)
The equation (iii) relating the four resistor is called the balance condition for the galvanometer to give zero or null deflection. 37. Discuss the method of comparison of emf of cells by using potentiometer arrangement. Sol. Comparison of emf of two cells of emf 1 and 2 : 1
2
3
2
A
B
1
G
N1
R ( ) K1
C
N2
Circuit Diagram 1, 2, 3 represent the contact points of two way key. When 1 and 3 are connected the galvanometer G is connected to 1. At point N1 at distance l1 from A the galvanometer has no deflection. Applying Kirchhoff’s loop rule to A13GN1A 1 + 0 – l1 = 0
...(i)
Similarly, if another emf 2 is balanced against l2 (i.e., AN2) Apply Kirchhoff’s loop rule to A23GN2A, 2 + 0 – l2 = 0
...(ii)
From equation (i) & (ii), we get 1 l1 2 l2
If one of the cells is chosen as a standard known emf cell, then emf of other can be very easily calculated. Potentiometre has the advantage that it draws no current from the voltage source being measured. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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38. Discuss the method of determination of internal resistance of a cell by using potentiometer arrangement. Sol. Determination of internal resistance of a cell :
2
K2 ( )
R.B.
A
B
G
N2
R ( ) K1
C
N1
Circuit diagram The cell of emf whose internal resistance r is to be measured is connected to the circuit along with a resistance box (RB) as shown in figure. With K2 open balance is obtained at length l1 (i.e., AN1), then = l1
...(i)
With K2 closed, the cell sends current through the RB of resistance R. Now its terminal potential difference balances at length l2 (i.e., AN2). V = l2
...(ii)
R l 2 Rr
l Equation (i) R r : 1 Equation (ii) R l2
⇒
⎛l ⎞ r R ⎜ 1 1⎟ ⎝ l2 ⎠
...(iii)
39. Discuss the method of determination of unknown resistance by meter bridge experiment. Sol. It is the practical application of Wheatstone bridge. A standard wire AC of length one metre and of uniform crosssectional area is stretched and clamped between two thick metal strips bent at right angles as shown in figure. The end points, where the wire is clamped are connected to a cell through a key K1. The metal strip has two gaps across which resistors can be connected. One end of Galvanometer is connected to mid-point of the metal strip between the gaps. The other end of the galvanometer is connected to a jockey, which can slide over AC to make electrical connections by its knife edge. R is the unknown resistance to be determined. R S S is the standard known resistance from a resistance box. Let the jockey be in contact with point D. Length of portion AD of wire be l. Resistance of the portion AD is RAD = Rm l and resistance of the portion DC is RDC = Rm(100 – l1). Where Rm is the resistance per centimetre of the wire. Now, R, S, RAD and RDC represent four resistances of Wheatstone bridge. When the galvanometer shows zero deflection then length AD = l1. The balance condition gives
A
l1
G
C
D 100 – l1
Metre scale
K1
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Rcm l1 l1 R RAD . S RDC Rcm (100 l1 ) 100 l1
⎛ l1 ⎞ Unknown resistance R in terms of standard known resistance S is given by R S ⎜ . ⎝ 100 l1 ⎟⎠ 40. Explain the ohmic loss in a conductor carrying a current. Where does the power come from? Sol. Consider a conductor having terminals A and B. Current I is flowing from A to B. If potentials at A is VA and potentials at B is VB, then potential difference across AB is V = VA – VB > 0. I A
B
Amount of charge flown from A to B, in time interval t, Q = It. Potential energy of charge at A, UA = (Q)VA. Potential energy of charge at B, UB = (Q)VB Change in potential energy of charge in travelling from A to B, U = UB – UA = Q[VB – VA] = –QV = –IVt < 0 If a charges moves without collision, then kinetic energy of charges would also change. So applying conservation of total energy. Change in kinetic energy + change in potential energy = 0.
k = –U
k = IVt > 0
Actually charge carriers do not move with acceleration but with steady drift velocity due to collisions with ions and atoms during transit. The kinetic energy gain by charges is shared with the atoms during collisions. As a result, atoms vibrate more vigorously and this energy is dissipated as heat in the conductor during time interval t. i.e., W = IVt The energy dissipated per unit time is the power dissipated.
P
W VI t
Using Ohm’s law V = IR, we get
P = I 2R =
V2 (Power loss or ohmic loss) R
It is the power which heats up the coil of an electric bulb to incandescence, radiating out heat and light. Source of the power : We need an external source (like chemical cells) to keep a steady current through the conductor which is the source to supply this power. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solution of Assignment (Set-1)
41. R, and are temperature coefficient of resistance, temperature coefficient of resistivity and coefficient of linear expansion of a conductor. Derive the relation between them. Sol. If the reference temperature T0 = 0°C, then temperature coefficient of resistivity of metals Resistance of the conductor at t°C can be written as Rt = R0(1 + Rt).
T 0 d . 0 t dt
R = Temperature coefficient of resistance
Rt R0 dR R0 t Rdt
l Resistance of conductor R = . When temperature changes then R, , l and A all change. So for small A range of temperature
dR d dl dA R l A
1 dR 1 d 1 dl 1 dA R dt dt l dt A dt
R
and are coefficient of linear and superficial expansion of substance. 42. What are the limitation of ohm’s law? Explain them graphically. Sol. The proportionality of V and I does not hold for certain materials and devices used in electric circuits. Following are few types of deviations. 1. V ceases to be proportional to I for a good conductor.
V
I 2. Value of current is different for same potential difference on reversing the direction of V. I(mA)
–2
2
V
I(A)
Characteristic curve of diode 3. Value of potential is different for same current.
I 1 mA
V 5V 10 V Variation of current vs Voltage for GaAs Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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43. Define conductivity of a conductor and mobility of free electrons in it. Derive the relation between them. Sol. Mobility ( ) It is defined as the magnitude of the drift velocity per unit electric field. vd E We have v d
...(1)
eE m
v d e E m
...(2)
Mobility of free electrons is independent of electric field and dimension of conductor. Conductivity of a conductor is defined as reciprocal of resistivity.
1 ne 2 m
⎛ e ⎞ (ne ) ⎜ ⎟ ⎝ m⎠
ne
44. Discuss how the power wasted in connecting wires in transmission cables and also show that this power is inversely proportional to V 2. (V is the voltage at which equipment is operated) Sol. Power loss in transmission lines : Consider a device of resistance R to be operated at voltage V and current through it is I, then power of device P = VI. If resistance of connecting wires from power station to the device is RC then power dissipated in connecting wires is PC with PC I 2RC
P 2RC V2
Therefore, to drive a device of power P, the power wasted in the connecting wires PC
1 V2
RC As the distance of power station is very large, RC is considerable. So to decrease PC, these wires carry current at enormous values of V and this is the reason for high voltage danger signs on transmission lines. These voltages are lowered to a value suitable for use by a device known as transformer. 45. Discuss working of a cell through its emf and internal resistance. Sol. An electrolytic cell consisting of two electrodes, called positive (P) and negative (N) immersed in an electrolytic solution is as shown in figure.
P
N A
B
Electrolyte Electrodes exchange charges with the electrolyte. Positive electrode P has a potential difference V+(V+ > 0) between itself and electrolyte solution A immediately adjacent to it. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Current Electricity
Solution of Assignment (Set-1)
Negative electrode N has a potential difference –(V–)(V– > 0) relative to electrolyte B adjacent to it. When there is no current, the electrolyte is at same potential throughout. So the potential difference between P and N is V+ – (–V–) = V+ + V–. This difference is known as electromotive force (emf) of the cell and denoted by . So, = V+ + V– > 0 emf is a potential difference not a force. When a resistance R is connected across the cell as shown in figure A current I flows from C to D. A steady current flows from P to N through the resistance R and from N to P through the electrolyte. The electrolyte through which current passes has a finite resistance r known as internal resistance of the cell.
R
I
C
P A
D N B
When the resistance R is infinite or P and N are in open circuit, there is no current through the electrolyte V ⎛ ⎞ ⎜I 0⎟ . ⎝ R ⎠ Potential difference between P and N = V = (VP – VA) + (VA – VB) + (VB – VN) = V+ + 0 + V–
Cells can be represented as P
= V+ – V– = r
N
emf of a cell is defined as the potential difference between the positive and negative electrodes in an open circuit.
SECTION - B Model Test Paper 1.
Write the SI unit of current.
Sol. ampere (A). 2.
Write the dimensional formula of resistance.
Sol. M1L2T–3A–2 3.
Maximum resistance obtained by combining five resistance each of 10 is ________.
Sol. 50 . 4.
Emf of a electrolytic cell is independent of nature of electrolyte present. (True/False)
Sol. True. 5.
Maximum current from cell of emf 10 V and internal resistance 10 is ________.
Sol. 1 A. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-1)
6.
Current Electricity
69
When a potential difference is applied across a conductor, the nature of path of free electron between two consecutive collisions is generally ________.
Sol. Curved. 7.
Why manganin is used as wire bound resistances?
Sol. Low temperature coefficient of resistivity. 8.
What are the external conditions on which resistivity of a substance depends?
Sol. Temperature and pressure. 9.
What happens to the resistivity of semiconductors with the increase in temperature?
Sol. Decrease. 10. Which conservation law Kirchhoff’s loop rule is based on? Sol. Conservation of energy. 11. Define electric current. Sol. It is the rate of flow of charge through a cross-section of conductor. Average current Iav
Total charge passed Q t Time interval
Instantaneous current I =
dQ dt
12. What is mobility of free electrons? Sol. It is defined as drift velocity of free electrons per unit electric field inside the conductor
vd e E m
13. Write the expression for variation of resistivity with temperature. Sol. Resistivity of a metallic conductor is approximately given by T = 0[1 + (T – T0) where T = resistivity at T°C, 0 = resistivity at reference temperature T°C and = temperature coefficient of resistivity. 14. Define ohm’s law. What is the meaning of ohmic conductors? Sol. Ohm’s Law It was discovered by G.S. Ohm in 1828. If V be the potential difference between the ends of the conductor through which a current I is flowing, then Ohm’s law states that V I or V = RI where R is the proportionality constant known as Resistance of the conductor, SI unit of resistance are VA–1 or ohm (). Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
70
Current Electricity
Solution of Assignment (Set-1)
Resistance of the conductor depends on 1.
Dimensions of conductor and
2.
Material of conductor
Conductors obeying Ohm’s law are known as Ohmic conductors. For Ohmic conductor I vs V graph is a straight line through origin.
I
V
0
15. What is the meaning of resistance of conductor? Write the factors on which resistance depends. Sol. It is the opposition offered by the conductor to the flow of electric current through it. It depends on 1. Length of conductor 2. Area of cross-section of conductor 3. Material of conductor 4. Temperature of conductor 16. State Kirchhoff’s junction rule and loop rule. Sol. Kirchhoff’s first rule : Junction rule At any junction, the sum of the currents entering the junction is equal to sum of currents leaving the junction. Kirchhoff’s second rule : Loop rule Algebraic sum of changes in potential around any closed loop involving resistors and cells in a loop is zero. 17. State Joule’s law of heating effect in conductors. Sol. When current I flows through a conductor of resistance R for time interval t, then amount of heat (H) produced in the conductor is directly proportional to square of current, directly proportional to the resistance of conductor and directly proportional to the time interval for which current flow. H l 2 ⎫ ⎪ 2 R ⎬ ⇒ H I R t ⎪ ⎭ t 18. What is the source of heat produced in a conductor connected across a chemical cell? OR Why the balance point of meter bridge is preferred to be close to 50 cm? Sol. We need an external source like chemical cells to keep a steady current through the conductor, which is the source to supply the energy. OR Percentage error in the measurement of unknown resistance can be minimised by selecting the balance point of meter bridge close to 50 cm. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-1)
Current Electricity
71
19. Length of a conductor is increased by 1%. Calculate the percentage change in its resistance at constant volume. Sol. R
l A
R l A l 2 2% R l A l
20. Calculate resistance across AB of the network shown in figure.
10 5
10
A
10
B
10 Sol. Apply Wheatstone bridge principle. 21. Two resistances R1 and R2 are connected in parallel. Derive the effective resistance of their parallel connection. OR Two cells of emf E1, E2 and internal resistance r1, r2 respectively are connected in series. Find the equivalent emf of their series connection. Sol. Consider two resistances R1 and R2 are connected in parallel across AB.
I1 A
R1
I
I I2
B
R2
The charge flows in at A from the left and flows out partly through R1 and partly through R2. So current I flows in at A and flows out as I1 and I2 through R1 and R2 respectively, hence I = I1 + I2
...(1)
Potential difference between A and B by applying Ohm law to R1 V = I1R1 Potential difference between A and B by applying Ohm’s law to R2 V = I2R2
I = I1 + I2 =
V V 1 ⎞ ⎛ 1 V ⎜ ⎟ R1 R2 R R ⎝ 1 2 ⎠
If the combination is replaced by equivalent resistance RP then by Ohm’s law
I
V RP
1 1 1 RP R1 R2
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Current Electricity
1 ⎞ ⎛ 1 RP ⎜ ⎟ R R ⎝ 1 2 ⎠
Solution of Assignment (Set-1) 1
...(2) OR
Consider two cells of emf 1 and 2 connected in parallel across AC.
I1
I1
1 r1
A
I
2 I2
B
I2
r2
If I1 and I2 are the currents leaving positive electrodes of both cells, then total current from B to A I = I1 + I2
...(1)
VA and VB are potentials at A and B respectively, then potential difference between A and B, can be written as V = VA – VB = 1 – I1r1
...(2)
and V = VA – VB = 2 – I2r2
...(3)
Using equation (1), (2) & (3), we get I = I1 + I2
1 V 2 V ⎛ 1 2 ⎞ ⎛1 1⎞ ⎜ ⎟ V ⎜ ⎟ r1 r2 ⎝ r1 r2 ⎠ ⎝ r1 r2 ⎠
I
V
Effective emf =
rr 1r2 2 r1 I 1 2 Ir r1 r2 r1 r2 1r2 2 r1 r1 r2
22. Explain Wheat-stone bridge principle. Sol. Wheatstone bridge It is an application of Kirchhoff’s rules. The bridge is consisting of four resistances R1, R2, R3 and R4 as four sides of a square ABCD as shown in figure.
R2 A
I2
4
Ig
R4 C
G
I1 I
BI
R1 D
I3
R3
I
E Across the diagonally opposite points between A and C, battery E is connected. This is called battery arm. To remaining two diagonally opposite points B and D, a galvanometer G is connected to detect current this line is known as galvanometer arm. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-1)
Current Electricity
73
Currents through all resistances and galvanometer are as shown in figure. In wheatstone bridge we consider the special case Ig = 0. Applying junction rule to junction B and D, we have I2 = I4 and I1 = I3 Applying loop rule to loop ABDA I2R2 + 0 – I1R1 = 0
I1 R2 I2 R1
...(i)
Applying loop rule to loop BCDB I4R4 – I3R3 + 0 = 0
I2R4 – I1R3 = 0
I1 R4 I2 R3
(Using I4 = I2 and I3 = I1) ...(ii)
Using equation (i) & (ii), we have R2 R4 R1 R3
...(iii)
The equation (iii) relating the four resistor is called the balance condition for the galvanometer to give zero or null deflection. 23. Discuss the method of determination of unknown resistance by metre bridge method. Sol. It is the practical application of Wheatstone bridge. A standard wire AC of length one metre and of uniform cross-sectional area is stretched and clamped between two thick metal strips bent at right angles as shown in figure. The end points, where the wire is clamped are connected to a cell through a key K1. The metal strip has two gaps across which resistors can be connected. One end of Galvanometer is connected to mid-point of the metal strip between the gaps. The other end of the galvanometer is connected to a jockey, which can slide over AC to make electrical connections by its knife edge. R is the unknown resistance to be determined.
S
R
A
l1
G
D 100 – l1
C
Metre scale K1 S is the standard known resistance from a resistance box. Let the jockey be in contact with point D. Length of portion AD of wire be l1. Resistance of the portion AD is RAD = Rml1 and resistance of the portion DC is RDC = Rm(100 – l1), where Rm is the resistance per centimetre of the wire. Now R, S, RAD and RDC represent four resistances of Wheatstone bridge. When the galvanometer shows zero deflection then length AD = l 1 . The balance condition gives Rm l1 l1 R RAD . S RDC Rm (100 l1 ) 100 l1
⎛ l1 ⎞ Unknown resistance R in terms of standard known resistance S is given by R S ⎜ . ⎝ 100 l1 ⎟⎠ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
74
Current Electricity
Solution of Assignment (Set-1)
24. Find the current through the 20 load resistance and power dissipated in it.
2
10 V
3
15 V
20 Sol. I = 1 A, P = I 2R = 20 W 25. Find the effective resistance across AB. A
1
1
1
1 B
1
1
1
1
1
1
1
1
1 to 1
Sol. (1 3) 26. Drift speed of free electrons are of the order of 1 mm s–1 but current establishes in a circuit with the speed of light. Explain. Sol. The electric field sets up in the conductor at the speed of light. 27. Draw the circuit diagram for comparison of emf of two cells by using potentiometer and explain how to compare their emf’s. OR Two cells of emf E1 and E2 and internal resistances r1 and r2 are connected in parallel across a load resistance R as shown in figure. Derive the expression for current through R using Kirchhoff’s rules.
E1
r1
E2
r2 R
Sol. Comparison of emf of two cells of emf 1 and 2 :
1 2
N1
R ( ) K1
C
3
2
A
B
1
G
N2
Circuit Diagram Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-1)
Current Electricity
75
1, 2, 3 represent the contact points of two way key. When 1 and 3 are connected the galvanometer G is connected to 1. At point N1 at distance l1 from A the galvanometer has no deflection. Applying Kirchhoff’s loop rule to A13GN1A 1 + 0 – l1 = 0
...(i)
Similarly, if another emf 2 is balanced against l2 (AN2) Apply Kirchhoff’s loop rule to A23GN2A, 2 + 0 – l2 = 0
...(ii)
From equation (i) & (ii), we get 1 l1 2 l2
If one of the cells is chosen as a standard known emf cell, then emf of other can be very easily calculated. Potentiometre has the advantage that it draws no current from the voltage source being measured. OR
A
D E
I
E1
r1
E2
r2 R
I1
B
C I – I1 F
Currents are distributed in all branches by using Kirchhoff’s junction rule. Applying Kirchhoff’s loop rule to loop ABFEA E1 – I1r1 – IR = 0
...(i)
Applying Kirchhoff’s loop rule to loop DCFED E2 – (I – I1)r2 – IR = 0
...(ii)
Multiply equation (i) by r2 E1r2 – I1r1r2 – Ir2R = 0
...(iii)
Multiply equation (ii) by r1 E2r1 – Ir1r2 + I1r1r2 – Ir1R = 0
...(iv)
Equation (i) + equation (iv) E1r2 + E2r1 = I [(r1 + r2)R + r1r2] I
E1r2 E2 r1 R(r1 r2 ) r1 r2
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76
Current Electricity
Solution of Assignment (Set-1)
28. Explain drift of electrons and the origin of resistivity in conductors. OR Explain working of chemical cell. How emf is realised in a chemical cell and describe the relation between emf and internal resistance of the cell. Sol. In the absence of electric field a free electron suffer collision with heavy fixed ions and after collision it emerges with same speed but in random directions. If we consider N free electrons with velocity of i th electron ui (i = 1, 2, 3, ....., N) at a given cross-section, then average velocity of these free electrons through a 1 N ...(1) ∑u 0 N i 1 i When the electric field E is present inside the conductor, acceleration of each free electron is eE a ...(2) m
cross-section =
where –e = charge of electron and m = mass of electron. Velocity of the i th electron after time ti of its last collision ⎛ eE ⎞ v i ui ⎜ t ⎝ m ⎟⎠ i
...(3)
Now average velocity of these N free electrons at that cross-section 1 N 1 N eE N v d ∑ v i ∑ ui ∑t N i 1 N i 1 m i 1 i
eE N vd ∑t m i 1 i eE vd m
where
(Using equation 1)
...(4)
N
∑ ti
= the average time between successive collisions known as relaxation time.
i 1
From equation (4) it is concluded that average velocity of free electrons is independent of time, although electrons are accelerated. This phenomena is known as drift and the velocity v i is known as drift velocity, therefore there will be net flow of charge across any area perpendicular to E .
A
B vd
–
E vdt Consider the area of cross-section A of conductor. Distance travelled from left to right by free electrons in time t after crossing the area of cross-section = v d t . If n is number of free electrons per unit volume in the conductor, then number of free electrons pass the crosssection from left to right in time interval t is equal to number of free electrons in the volume
A v d t nA v d t Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-1)
Current Electricity
77
Charge transported across the area of cross-section from left to right = neA v d t
Total charge transported along E from right to left I t neA v d t
...(5)
Substituting the value of v d from equation (4) I
ne2 AE m
By definition of current I j A
...(6) (∵
j Current density)
ne2 E Equation (6) becomes j m
...(7)
ne2 E j is parallel to E so j m
...(8)
Equation (8) is as per the definition of Ohm’s law.
Conductivity of conductor =
ne 2 m
...(9)
n and are assumed to be constants for a given material at a given temperature and are independent of E . OR An electrolytic cell consisting of two electrodes, called positive (P) and negative (N) immersed in an electrolytic solution is as shown in figure.
P
N A
B
Electrolyte Electrodes exchange charges with the electrolyte. Positive electrode P has a potential difference V+(V+ > 0) between itself and electrolyte solution A immediately adjacent to it. Negative electrode N has a potential difference –(V–)(V– > 0) relative to electrolyte B adjacent to it. When there is no current, the electrolyte is at same potential throughout so the potential difference between P and N is V+ – (–V–) = V+ + V–. This difference is known as electromotive force (emf) of the cell and denoted by . So, = V+ + V– > 0 emf is a potential difference not a force. When a resistance R is connected across the cell as shown in figure A current I flows from C to D. A steady current flows from P to N through the resistance R and from N to P through the electrolyte. The electrolyte through which current passes has a finite resistance r known as internal resistance of the cell. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
78
Current Electricity
Solution of Assignment (Set-1)
C
R
I P
D N
A
B
When the resistance R is infinite or P and N are in open circuit, there is no current through the electrolyte V ⎛ ⎞ ⎜I 0⎟ . ⎝ R ⎠
Potential difference between P and N = V = (VP – VA) + (VA – VB) + (VB – VN) = V+ + 0 + V– = V+ – V– = Cells can be represented as P
r
N
emf of a cell is defined as the potential difference between the positive and negative electrodes in an open circuit.
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