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CONTACT1FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
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1. (a) Estimate the average drift speed of conduction electrons in a copper wire of crosssectional area, 1.0 × 10-7 m2 carrying a current of 1.5 A. Assume that each copper atom
contributes roughly one conduction electron. The density of copper is 9.0 × 103 kg/m3 and
its atomic mass is 63.5 u.
(b) Compare the drift speed obtained above with,
(i) thermal speeds of copper atoms at ordinary temperatures,
(ii) speed of propagation of electric field along the conductor which causes the drift motion.
•
Solution:
(a) The direction of drift velocity of conduction electrons is opposite to the electric field direction, i.e.,
electrons drift in the direction of increasing potential. The drift speed vd is given by vd = (I/neA).
Now, e = 1.6 × 10-19 C, A = 1.0 × 10-7 m2, I = 1.5 A. The density of conduction electrons, n is equal
to the number of atoms per cubic meter. A cubic meter of copper has a mass of 9.0 × 103 kg. Since
6.0 × 1023 copper atoms have a mass of 63.5 g.
which gives,
(b) (i) At a temperature T, the thermal speed of a copper atom of mass M is obtained from
[<(1/2)Mv2> = (3/2) kBT ] and is thus typically of the order of
, where kB is the
Boltzmann constant. For copper at 300 K, this is about 2 × 102 m/s. This indicates the random
vibrational speeds of copper atoms in a conductor. The drift speed of electrons is much smaller,
about 10-5 times the typical thermal speed at ordinary temperatures.
(ii) An electric field traveling along the conductor has a speed of an electromagnetic wave, i.e., 3.0 ×
108 m/s. The drift speed of electrons is much smaller by the order of 10-11.
2. (a) In the previous example, the electron drift speed is estimated to be only a few
mm/s for currents in the range of a few amperes? How then is current established almost
the instant a circuit is closed?
(b) The electron drift arises due to the force experienced by electrons in the electric field
inside the conductor. But force should cause acceleration. Why then do the electrons
acquire a steady average drift speed?
(c) If the electron drift speed is so small, and the electron's charge is small, how can we
still obtain large amounts of current in a conductor?
1
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(d) When electrons drift in a metal from lower to higher potential, does it mean that all the
'free' electrons of the metal are moving in the same direction?
(e) Are the paths of electrons straight lines between successive collisions in the (i) absence
of electric field (ii) presence of electric field?
Solution:
(a) Electric field is established throughout the circuit, almost instantly causing at every point a local
electron drift. It takes a little time for the current to reach the steady value.
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•
(b) Each 'free' electron does accelerate, increasing its drift speed until it collides with a positive ion of
the metal. It loses its drift speed after collision but starts to accelerate and increases its drift speed
again only to suffer a collision again and so on. On the average, the electrons acquire only a drift
speed.
(c) Simple, because the electron number density is enormous ~ 1029 /m3.
(d) By no means. The drift velocity is superposed over the large random velocities of electrons.
(e) In the absence of electric field, the paths are straight lines; in the presence of electric field, the
paths are curved.
3. An electric toaster uses nichrome for its heating element. When a negligibly small
current passes through it, its resistance at room temperature (27.0о C) is found to be 75.3
Ω. When the toaster is connected to a 230 V supply, the current settles, after a few seconds,
to a steady value of 2.68 A. What is the steady temperature of the nichrome element? The
temperature coefficient of resistance of nichrome averaged over the temperature range
involved, is 1.70 × 10-4 оC -1.
•
Solution:
When the current is small, the temperature T1 of the element is the same as the room temperature.
When the toaster is connected to the supply, its initial current will be slightly higher than the steady
current of 2.68 A. This will cause an increase in resistance and a slight decrease in current. The
resistance R2 at the steady temperature T2 is
R2 = 230 V / 2.68 A = 85.8 Ω
Using the relation
R2 = R1 [1 +
With
(T2 – T1)]
= 1.70
10-4 оC–1, we get
in
T2 – T1 =
T2 = (820 + 27.0)
= 847
Thus, the steady temperature of the heating element is 847
.
4. The resistance of the platinum wire of a platinum resistance thermometer at the ice
point is 5 Ω and at steam point is 5.23 Ω. When the thermometer is inserted in a hot bath,
the resistance of the platinum wire is 5.975 Ω. Calculate the temperature of the bath.
2
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•
Solution:
R0 = 5 Ω, R100 = 5.23 Ω and Rt = 5.975 Ω
Now, t =
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=
.
5. A network of resistors is connected to a 16 V battery with internal resistance of 1 Ω as
shown in the figure.
(a) Compute the equivalent resistance of the network
(b) Obtain the current in each resistor
(c) Obtain the voltage drops VAB, VBC and VCD.
•
Solution:
(a) The network is a simple series and parallel combination of resistors. First the two 4 Ω resistors in
parallel are equivalent to a resistor = [(12 6)/(12+6)] Ω = 4 Ω.
The equivalent resistance R of the network is obtained by combining these resistors (2 Ω and 4 Ω)
with 1 Ω in series, i.e., R = 2 Ω + 4 Ω + 1 Ω = 7 Ω.
(b) The total current I in the circuit is
Consider the resistors between A and B. If I1 is the current in one of the two arms. If I1 is the
current in one of the 4 Ω resistors and I2 the current in the other,
I1
4 = I2
4
That is I1 = I2, which is otherwise obvious from the symmetry of the two arms. But I1 + I2 = I = 2
A. Thus I1 = I2 = 1 A.
That is current in each 4 Ω resistors is 1 A. Current in 1 Ω resistor between B and C would be 2 A.
Now, consider the resistances between C and D. If I3 is the current in the 12 Ω resistor, and I4 in the
6 Ω resistor.
I3
12 = I4
6, i.e., I4 = 2 I3
But, I3 + I4 = I = 2 A
3
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MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
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Thus, I3 = (2/3) A, I4 = (4/3) A
That is, the current in the 12 Ω resistor is (2/3) A, while the current in the 6 Ω resistor is (4/3) A.
(c) The voltage drop across AB is VAB = I1
4=1A
4 Ω = 4 V,
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This can also be obtained by multiplying the total current between A and B by the equivalent
resistance between A and B, i.e.,
VAB = 2 A
2 Ω = 4 V.
The voltage drop across CD is
VCD = 12 Ω
I3 = 8 V.
This can alternately be obtained by multiplying total current between C and D by the equivalent
resistance between C and D.
VCD = 2 A
4Ω=8V
The total voltage drop across AD is 4 V + 2 V + 8 V = 14 V. The terminal voltage of the battery is 14
V, while its emf is 16 V. The loss of the voltage is accounted for by the internal resistance 1 Ω of the
battery.
6. A battery of 10 V and negligible internal resistance is connected across the diagonally
opposite corners of a cubical network consisting of 12 resistors each of resistance 1 Ω.
Determine the equivalent resistance of the network and the current along each edge of the
cube.
•
Solution:
The network is not reducible to a simple series and parallel combinations of resistors. There is,
however, a clear symmetry in the problem which we can exploit to obtain the equivalent resistance
of the network.The paths AA’, AD and AB are obviously symmetrically placed in the network. Thus,
the current in each must be the same, say, I. Further, at the corners A’, B and D, the incoming
current I must split equally into the two outgoing branches. In this manner, the current in all the 12
4
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MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
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edges of the cube are easily written down in terms of I, using Kirchoff’s first rule and the symmetry
in the problem.
Next, take a closed loop, say ABCC’EA, and apply Kirchoff’s second law:
-IR - (1/2)IR – IR + = 0.
Where R is the resistance of each edge and
the emf of battery. Thus,
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= (5/2) IR
The equivalent resistance Req of the network is
Req =
For R = 1 Ω, Req = (5/6) Ω and for
= 10 V, the total current (= 3I) in the network is3 I = 10
V/(5/6) Ω = 12 A, i.e., I = 4 A
The current flowing in each edge can now be read off from the figure.
7. Determine the current in each branch of the network shown in the figure.
•
Solution:
Each branch of the network is assigned an unknown current to be determined by the application of
Kirchoff’s rules. To reduce the number of unknowns at the outset, the first rule of Kirchoff is used at
every junction to assign the unknown current in each branch. We then have three unknowns I1,
I2 and I3, which can be found by applying the second rule of Kirchoff to three different closed loops.
Kirchoff’s second rule for the closed loop ADCA gives,
10 - 4(I1 - I2) + 2(I2 + I3 - I1) - I1 = 0
that is, 7I1 - 6I2 - 2I3 = 10
For the closed loop ABCA, we get
10 - 4I2 - 2(I2 + I3) - I1 = 0
that is, I1 + 6I2 + 2I3 = 10
For the closed loop BCDEB, we get
1) = 0
5
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that is, 2I1 – 4I2 – 4I3 = -5
Equations are three simultaneous equations in three unknowns. These can be solved by the usual
method to give
I1 = 2.5 A, I2 = (5/8) A, I3 = (15/8) A
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The currents in the various branches of the network are
AB: (5/8) A, CA: (5/2) A, DEB: (15/8) A
AD: (15/8) A, CD: 0 A, BC: (5/2) A
It is easily verified that Kirchoff’s second rule applied to the remaining closed loops does not provide
any additional independent equation, that is, the above values of currents satisfy the second rule for
every closed loop of the network. For example, the total voltage drop over the closed loop BADEB
5V +
equal to zero, as required by Kirchoff’s second rule.
8. In a meter bridge experiment , the null point is found at a distance of 33.7 cm from A.
If now a resistance of 12 Ω is connected in parallel with S, the null point occurs at 51.9 cm.
Determine the values of R and S.
•
Solution:
From the first balance point we get
R/S = 33.7 / 66.3
After S is connected in parallel with a resistance of 12 Ω, the resistance across the gap changes from
S to Seq, where
Seq =
And the new balance condition gives
Substituting the value of R/S ,we get
which gives S = 13.5 Ω. Using the value of R/S above, we get R = 6.86 Ω.
6
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9. A resistance of R Ω draws current from a potentiometer. The potentiometer has a total
resistance R0 Ω as in the figure. A voltage V is supplied to the potentiometer. Derive an
expression for the voltage across R when the sliding contact is in the middle of the
potentiometer.
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•
Solution:
While the slide is in the middle of the potentiometer only half of its resistance (R0/2) will be between
A and B, say R1, will be given by the following expression:
The total resistance between A and C will be sum of resistance between A and B and B and C, i.e.,
R1 + R0/2
The current flowing through the potentiometer will be
The voltage V1 taken from the potentiometer will be the product of current I and resistance R1,
V1 = IR1 =
Substituting for R1, we have
.
10. The storage battery of a car has an emf of 12V. If the internal resistance of the
battery is 0.4W, what is the maximum current that can be drawn from the battery?
7
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•
Solution:
The voltage of the car battery V = 12V
The internal resistance R = 0.4W
... The maximum current I =
= 30A.
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11. A battery of emf 10V and internal resistance 3W is connected to a resistor. If the
current in the circuit is 0.5A, what is the resistance of the resistor? What is the terminal
voltage of the battery when the circuit is closed?
•
Solution:
The emf of the battery E = 10V
Internal resistance r = 3W
The current in the circuit I = 0.5A
... The resistance of the resistor R =
- r = ( 10/0.5 ) - (3) = 17 W
... The terminal voltage VT = IR = 17 × 0.5 = 8.5 V.
12. Three resistors 1W, 2W and 3W are combined in series.
(a)What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 12V and negligible internal
resistance, obtain the potential drop across each resistor.
•
Solution:
(a) The value of resistor R1 = 1W
The value of resistor R2 = 2W
The value of resistor R3 = 3W
... The total resistance R = R1 + R2 + R3 = 1 + 2 + 3 = 6W.
(b) The emf of the battery E = 12V
The current in the circuit I =
= 2A
... Potential drop across R1 = I × R1 = 2 × 1 = 2V
Potential drop across R2 = I × R2 = 2 × 2 = 4V
Potential drop across R3 = I × R3 = 2 × 3 = 6V.
13. Three resistors 2W, 4W and 5W are combined in parallel.
(a)What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 20V and negligible internal
resistance,determine the current through each resistor, and the total current drawn from
the battery.
•
Solution:
a) The value of resistor R1 = 2W
The value of resistor R2 = 4W
The value of resistor R3 = 5W
8
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... The total resistance =
R = (20/19) W
R = 1.05
=
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b) The emf of battery E = 20 V
The resistance of battery R = (20/19) W
... The total current I =
The current through resistor R1 =
= 10A
The current through resistor R2 =
= 5A
The current through resistor R3 =
= 4A
... The total current drawn from the battery = R1 + R2 + R3 = 10 + 5 + 4 = 19 A
The total current drawn from the battery = 19 A.
14. At room temperature (27.0°C) the resistance of a heating element is 100W. What is
the temperature of the element if the resistance is found to be 117W, given that the
temperature coefficient of the material of the resistor is 1.70 × 10-4 ºC-1.
•
Solution:
The resistance at 27º C, R0 = 100W
The resistance at tº C, Rt = 117W
The temperature coefficient of the given material a = 1.70 × 10-4 ºC-1
The temperature t => Rt = R0 [1 + a (t - 27)]
117
= 100[1 + 1.7 × 10-4 (t - 27)]
or t - 27
... t
=
= 1000°C
= 1000 + 27
= 1027°C.
15. A negligibly small current is passed through a wire of length 15m and uniform cross
section 6 × 10-7m2, and its resistance is measured to be 5.0 W. What is the resistivity of the
material at the temperature of the experiment?
•
Solution:
The resistance through the wire R = 5.0W
The length of the wire l
= 15 m
9
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= 6 × 10-7m2
The cross section of it A
... The resistivity of the material r =
= 2 ×10-7 Wm.
16. A silver wire has a resistance of 2.1 W at 27.5 °C, and a resistance of 2.7W at 100°C.
Determine the temperature coefficient of resistivity of silver.
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•
Solution:
The resistance of silver wire at 27.5 °C 'R27.5' = 2.1W
The resistance of silver wire at 100 °C 'R100' = 2.7W Let the temperature coefficient be a
... Rt = R0(1 + at)
R100 = 2.7 = R0(1 + 100a)
R27.5 = 2.1 = R0(1 + 27.5a)
=>
2.7(1 + 27.5a) = 2.1(1 + 100a)
... a = 0. 0039°C-1.
17. A heating element using nichrome connected to a 230V supply draws an initial
current of 3.2A, which settles after a few seconds to a steady value of 2.8A. What is the
steady temperature of the heating element if the room temperature is 27.0°C ? The
temperature co-efficient of nichrome averaged over the temperature range involved is 1.7 ´
10-4 °C-1.
•
Solution:
The supply voltage E
The initial current I1
... The resistance R1
= 230V
= 3.2A
=
= 71.875 W
The resistance R2
=
= 82.143 W
The temperature coefficient a = 1.7 × 10-4 °C-1
The temperature t for resistance R2 = R1[ 1 + a (t - 27)]
82.143 = 71.875[1 + 1.7 × 10-4(t - 27)]
or t - 27 =
t = 840.35 + 27
t = 867.35°C.
18. Determine the current in each branch of the following network.
10
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•
Solution:
Let us consider the currents flowing in each branch of the network as shown in the figure below.
Using Kircoff's second rule, for the loop ADCEA,
-5(I - I1) - 10(I - I2) - 10I = -10
or
5I - I1 - 2I2 = 2
-----(1)
For the loop ABDA,
-5(I1-I2)-10I1 + 5(I - I1) = 0
or
I - 4I1 + I2 = 0
-----(2)
in
For the loop BCDB,
-5I2 + 10(I - I2) + 5(I1 - I2) = 0
or
2I + I1 - 4I2 = 0
-----(3)
Adding (2) and (3), we get
I = I1 + I2
Putting this value in (1), we get
4I1 + 3I2 = 2
and in (2), we get
I2 = (3/2) I1
Similarly, putting this value in the previous equation,
4I1 + (9/2) I1 = 2
11
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or I1 = (4/17) A
I2 = (6/17) A
Now, I = I1 + I2
=
A.
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19. In a metre bridge, the balance point is found to be at 39.5cm from the end A, when
the resistor Y is of 12.5W. Determine the resistance of X. Why are the connections between
resistors in a wheatstone or metre bridge made of thick copper strips?
b) Determine the balance point of the bridge alone 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?
•
Solution:
a) The value of resistor Y = 12.5 W
The balancing length l1 = 39.5 cm
... The length l2 = 100 - l1 = 60.5cm
Using the formula x/y = l1/l2
... X = Y
= 8.1612 W
The connections between resistors in a wheatstone or metre bridge is made of thick copper strips to
minimize resistance of the connections which are not accounted for in the bridge formula.
b) When X and Y are interchanged, the value of l1 and l2 are also interchanged
Thus l1 = 60.5 cm
... The length l2 = 39.5cm
c) The galvanometer will not record any current.
20. A storage battery of emf 8.0V and internal resistance 0.5 W is being charged by a
120V dc supply using a series resistor of 15.5W. What is the terminal voltage of the battery
during charging ? What is the purpose of having a series resistor in the charging circuit?
•
Solution:
The emf of battery E
The internal resistance r
The dc suppl voltage
The series resistor R
= 8.0 V
= 0.5 W
= 120 V
= 15.5 W
... The current I
=
= 7A
... Terminal Voltage VT = E + I × r = 8 + 7 × 0.5 = 11.5V
The series resistor limits the current drawn from the external source. In its absence, the current will
be dangerously high.
21. In a potentiometer arrangement a cell of emf 1.25V gives a balance point at 35.0cm
length of the wire. If the cell is replaced by another cell and the balance point shifts to
63.0cm, what is the emf of the second cell ?
•
Solution:
The emf of first cell E1 = 1.25V
The balance point of E1 is, l1 = 35.0 cm
12
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The balance point of cell E2 is, l2 = 63cm
... The emf of second cell =
=
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... E2 =
= 2.25V.
22. The number density of conduction electrons in a copper conductor estimated is
8.5 × 1028m-3. How long does an electron take to drift from one end of a wire 3.0m long to
its other end? The area of cross section of the wire is 2 × 10-6m2 and it is carrying a current
of 3.0A?
•
Solution:
The number density n = 8.5 × 10 28m-3
The current I = 3.0A
The area of cross section A = 2 ×10-6 m2
The electronic charge e = 1.6 × 10-19C
... The drift velocity Vd =
... Time taken to drift 3.0m length =
=
1.1029 × 10-4ms-1
= 2.72 × 104s.
23. The earth's surface has a negative surface charge density of 10-9cm-2. The potential
difference of 400kV between the top of the atmosphere and the surface results (due to the
low conductivity of the lower atmosphere) in a current of only 1800A over the entire globe.
If there were no mechanism of sustaining atmospheric electric field, how much time
roughly would be required to neutralize 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 the earth =
6.37 × 106m).
•
Solution:
Surface charge density s = 10-9cm-2
The potential difference V = 400kV
The current I = 1800A
The radius of earth Re = 6.37 × 106m
The total charge Q = total surface area × s = 4pRe2 s
Time =
... The time t =
= 283.4 s.
13
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24. a) Six lead-acid type of secondary cells each of emf 2.0V and internal resistance
0.015W are joined in series to provide a supply to a resistance of 8.5W. What is the current
drawn from the supply and its terminal voltage?
b) A secondary cell after long use has an emf of 1.9V and a large international resistance of
380 W.What maximum current can be drawn from the cell? Could the cell drive the starting
motor of a car?
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•
Solution:
a) emf of each cell 'e' = 2.0V
No. of cells in series 'N' = 6
Internal resistance of each cell r = 0.015 W
External resistance R = 8.5 W
Current, I =
Ne
R + Nr
=
=
6x2
___
8.5 + 6 x 0.015
12
_
8.5 + 0.09
12 _ @ 1.4A
8.59
Terminal voltage, V = IR
= 1.4 x 8.5
=11.9v.
=
b) The emf of cell E = 1.9V
The internal resistance R = 380W
... Maximum current drawn
= 0.005 A
It is not possible for the cell to drive the starting motor of a car because a starter motor requires a
large current of the order of 100 A.
25. 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. (rAl = 2.63 × 10-8 Wm, rCu = 1.72 × 10-8 Wm. Relative
density of Al = 2.7, of Cu = 8.9).
•
Solution:
Using the formula
R=
=
Where V = Al = volume of wire
m = mass of wire
d = density of wire
For the aluminium wire,
14
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RAl =
And for the copper wire,
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Rcu=
Since RAl = Rcu and lAL = lcu,
=
or =
=
= 2.11
Thus, the aluminium wire is lighter than the copper wire. Therefore, it is preferred in overhead
cables.
26. What conclusion can you draw from the following observations on a resistor made of
alloy manganin:
Current Voltage Current Voltage
V
A
V
c
59.2
3.0
3.94
0.2
78.8
4.0
7.87
0.4
98.6
5.0
11.8
0.6
118.5
6.0
15.7
0.8
138.2
7.0
19.7
1.0
158.0
8.0
39.4
2.0
•
Solution:
Ohm's Law is valid to a high accuracy; the resistivity of the alloy magnanin is nearly independent of
temperature.
27. Answer carefully:
a) A simple voltaic cell has an emf equal to 1.0V. When the circuit is open, is there a net
field, which would give rise to a force on a test charge, (i) inside the electrolyte of the cell,
(ii) outside the cell?
b) Is there a net field inside the cell when the circuit is closed and a steady current passes
through? Explain.
c) Consider a voltaic cell in open circuit with the Cu plate at higher potentials with respect
to the Zn plate. Since the electrolyte separating them is a conducting medium, why are the
positive and negative charges on the electrodes not neutralized immediately?
d) When a secondary cell of emf 2.0V is being charged by an external supply, is the terminal
voltage of the secondary cell greater or less than 2.0V?
•
Solution:
a) Inside the cell, there is no net field; the electrostatic field due to the plates is balanced by a field
of non-electrostatic origin. (If this were not so, charge carriers inside the cell would constitute a
15
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current even when the circuit is open). Outside the cell, the net field is just the electrostatic field of
the plates.
b) When the circuit is closed from outside, current flows in the direction of electrostatic field outside,
and opposite to the direction of electrostatic field inside the cell. The latter fact shows that there is a
net field inside the cell opposite to the electrostatic field.
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c) They are prevented from being neutralized because the electrostatic field is opposed by a field in
the source that has a complex non-electrostatic origin. The latter is related to the emf of the source.
d) The current inside the secondary cell during charging is in the direction of electrostatic field. The
electrostatic field is greater than the non-electrostatic field. The terminal voltage, therefore is greater
than emf of the secondary cell.
28. 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) Doping a semiconductor (with small traces of impurity atoms) reduces/increases its
resistivity.
d) The resistance of graphite and most non-metals increases/decreases with increasing
temperature.
e) The resistivity of a semiconductor increases/ decreases rapidly with increasing
temperature.
f) The resistivity of the alloy manganin (is nearly independent of/increases with) increase
of temperature.
g) 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).
•
Solution:
a) greater
b) lower
c) reduces
d) decreases
e) decreases
f) nearly independent of
g) 10 ^22.
29. a) Given n resistors each of resistance R, how will you combine them to get the
(i) maximum
(ii) minimum effective resistance?
16
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(iii) What is the ratio of the maximum to minimum resistance?
b) Given the resistances of 1W, 2W, 3W. How will you combine them to get an equivalent
resistance of
(i) 11/3W,
(ii) 11/5W,
(iii) 6W,
(iv) 6/11W?
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c) Determine the equivalent resistance of the following networks:
•
Solution:
a) (i) in series
(ii) all in parallel
in
(iii)
b) (i) Join 1W, 2W in parallel and the combination in series with 3W.
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(ii) parallel combination of 2Ω and 3Ω in series with 1W
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= (11/5)W
(iii) All in series
(iv) All in parallel.
= (6/11)2
c) (i)
in
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Thus, the equivalent resistances is 16/3 W .
(ii) The equivalent resistances is 5 W because the 5 resistances are all in series , the equivalent
resistance is equal to the sum of the individual resistances.
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30. Determine the current drawn from a 12V supply with internal resistance 0.5W by the
following infinite network shown in the figure: Each resistor has 1W resistance.
•
Solution:
R=
Let x be the equivalent resistance of the centre network is infinite, the addition of one more set of
three resistances each of resistance 4 = 1W will not alter the total resistance. In other words, the
resistance of the network should still be x.
The figure below shows the network when one additional set of three resistances is added. Let R be
the equivalent resistance of the new network. It is clear from the diagram that
R = r + resistance of a parallel combination of r and x+r
=
-----(1)
Now, as explained above R = x. Putting R = x in (1) gives
x=
-----(2)
It is given that r = 1 W . Putting r = 1 in (2), we have
x=2+
or x2 - 2x - 2 = 0
or x(x + 1) = x=(1 ± √3) or x = 1 + √3, x= 1- √3
x = 2.73 (or) x = -0.73
The two roots of this quadratic equation are
x=
= 2.73 W or -0.73 W
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Since a resistance cannot have a negative value, x = -0.73 W is inadmissible. Hence x = 2.73 W is
taken. When this resistance is connected to a battery of voltage V = 12 V and internal resistance R =
0.5 W, the current drawn from the battery is
I = V/(x + R) = 12/(2.73 + 0.5) = 3.71 A.
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31. Figure below shows a potentiometer with a cell of 2.0 V and internal resistance
0.40W maintaining a potential drop across the resistor wire AB. A standard cell which
maintains a constant emf of 1.02V (for very moderate currents upto a few mA) gives a
balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the
standards cell, a very high resistance of 600 kW 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 e
and the balance point found
similarly turns 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 kW 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?
in
•
Solution:
(a) The value of e is given by (e1 = 1.02 V is the emf of the standard cell C1), l2 = 0.823, l1 = 0.673
(b) The high resistance R has a two-fold purpose to keep the current drawn from the standard cell
within permissible limit when far away from the balance point and to prevent a large current to flow
through the galvanometer (so that it is not damaged) when far away from the balance point.
20
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(c) No, since at balance point no current flows through R.
(d) No, the balance point is not affected by the internal resistance of C, it depends on the terminal
voltage of C and the emf of C1
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(e) The method would not work if the potential difference across AB due to the driver cell C (which is
its terminal voltage) became less than the emf of the cell C1, for then there would be no balance
point on the wire.
(f) Suppose the emf e to be measured is, say 5 mV = 5 × 10-3 V, and the potential difference across
AB due to C is 2.0 V. Also suppose that the length AB of the wire is 100 cm and its resistance per
centimeter is k. Then the value of the balance length will be given by
5 × 10-3 V = pd across AJ due to C
= current in the main circuit ´ resistance of AJ =
=
AJ = 0.25 cm
Thus the balance point will be very close to A and there is an extremely large percentage error in its
measurement. To have a large balance length, the potential difference across AB must be reduced to
about 10 mV. To achieve this the circuit is modified as shown below by putting a suitable resistor R'.
The balance length will then be larger and the percentage error will be much smaller.
32. Figure below shows a potentiometer circuit for comparison of two resistances.
The balance point with a standard resistor R = 10.0 W 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 e?
•
Solution:
Using the formula
or r1 =
= 11.8 W (approximately)
21
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.in
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In case there is no balance point, the potential drop across R or X are greater than that across AB.
Therefore, we must suitably reduce current in the outside circuit by using a series resistor.
33. Figure below shows a 2.0 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 9.5W is used in the external circuit of the cell, the balance point shifts to 64.8 cm
length of the potentiometer wire. Determine the internal resistance of the cell.?
•
Solution:
Using formula
l1 = 76.3
l2 = 64.8
R = 9.5
r=R
r=R
= 9.5
=
.
22
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in
.
ic
at
m
he 5
at 31
.m 44
w 1
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09
23
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