Chapter 10 Current Electricity Part 3 – Physics free study material by TEACHING CARE online tuition and coaching classes
Chapter 10 Current Electricity Part 3 – Physics free study material by TEACHING CARE online tuition and coaching classes
Example: 61 In the adjoining circuit, the battery E_{1} has as emf of 12 volt and zero internal resistance, while the battery E
has an emf of 2 volt. If the galvanometer reads zero, then the value of resistance X ohm is [NCERT 1990]
(a) 10
(b) 100
(c) 500
(d) 200
Solution : (b) For zero deflection in galvanometer the potential different across X should be E = 2V
In this condition
12X = 2
500 + X
\ X = 100 W
Example: 62 In the circuit shown here E_{1} = E_{2} = E_{3} = 2V and R_{1} = R_{2} = 4 W. The current flowing between point A and B
through battery E_{2} is [MP PET 2001]
 Zero
 2 A from A to B
 2 A from B to A
 None of these
Solution : (b) The equivalent circuit can be drawn as since E_{1} & E_{3} are parallely connected
So current i = 2 + 2 = 2Amp from A to B.
2
2V
A 2V
R = (R_{1} R_{2}) = 2W
B
Example: 63 The magnitude and direction of the current in the circuit shown will be [CPMT 1986, 88]
(a)
(b)
7 A from a to b through e
3
7 A from b and a through e
3
 0 A from b to a through e
 0 A from a to b through e
Solution : (d) Current i = 10 – 4 = 1A
3 + 2 + 1
from a to b via e
Example: 64 Figure represents a part of the closed circuit. The potential difference between points A and B (V_{A} – V_{B}) is
(a) + 9 V
(b) – 9 V
(c) + 3 V
(d) + 6 V
Solution : (a) The given part of a closed circuit can be redrawn as follows. It should be remember that product of current and resistance can be treated as an imaginary cell having emf = iR.
4 V 3 V 2 V
+ 9 V
Þ
–
Þ 7 Hence V_{A}
– V_{B}
= +9 V
B A B
Example: 65 In the circuit shown below the cells E_{1} and E_{2} have emf’s 4 V and 8 V and internal resistance 0.5 ohm and
1 ohm respectively. Then the potential difference across cell E_{1} and E_{2} will be
(a) 3.75 V, 7.5 V
(b) 4.25 V, 7.5 V
(c) 3.75 V, 3.5 V
(d) 4.25 V, 4.25 V
Solution : (b) In the given circuit diagram external resistance
R = 3 ´ 6 + 4.5 = 6.5W . Hence main current through the
3 + 6
circuit i = E2 – E1 = 8 – 4 = 1 amp.
R + req 6.5 + 0.5 + 0.5 2
Cell 1 is charging so from it’s emf equation E_{1} = V_{1} – ir_{1} Þ 4 = V1 – 1 ´ 0.5
2
Þ V_{1} = 4.25 volt
Cell 2 is discharging so from it’s emf equation E_{2} = V_{2} + ir_{2} Þ 8 = V2 + 1 ´ 1 Þ V_{2} = 7.5 volt
2
Example: 66 A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to this current, the temperature of the wire is raised by DT in time t. A number N of similar cells is now connected in series with a wire of the same material and crosssection but of length 2L. The temperature of wire is raised by same amount DT in the same time t. The value of N is [IITJEE (Screening) 2001] (a) 4 (b) 6 (c) 8 (d) 9
Solution : (b) Heat = mSDT = i^{2}Rt
Case I : Length (L) Þ Resistance = R and mass = m


Case II : Length (2L) Þ Resistance = 2R and mass = 2m
So m1 S1 DT1
i ^{2} R t

= 1 1 1
Þ mSDT =
i ^{2} Rt
Þ i = i Þ (3E)^{2} = (NE)^{2}
Þ N = 6

m2 S2 DT2
2
2 2 2
2mSDT
i ^{2} 2Rt
12 2R
Kirchoff’s Laws.
 Kirchoff’s first law : This law is also known as junction rule or current law (KCL). According to it the algebraic sum of currents meeting at a junction is zero e. åi = 0.
In a circuit, at any junction the sum of the currents entering the junction must
equal the sum of the currents leaving the junction. i_{1}
Here it is worthy to note that :
 i3
= i2
 i4
 If a current comes out to be negative, actual direction of current at the
junction is opposite to that assumed, i + i1 + i2 = 0
can be satisfied only if at least one
current is negative, i.e. leaving the junction.
 This law is simply a statement of “conservation of charge” as if current reaching a junction is not equal to the current leaving the junction, charge will not be
Note : @ This law is also applicable to a capacitor through the concept of displacement current treating the resistance of capacitor to be zero during charging or discharging and infinite in steady state as shown in figure.
 Kirchoff’s second law : This law is also known as loop rule or voltage law (KVL) and according to it “the algebraic sum of the changes in potential in complete traversal of a mesh (closed loop) is zero”, e. åV = 0
e.g. In the following closed loop.
– i_{1}R_{1} + i_{2}R_{2} – E_{1} – i_{3}R_{3} + E_{2} + E_{3} – i_{4}R_{4} = 0
Here it is worthy to note that :
 This law represents “conservation of energy” as if the sum of potential changes around a closed loop is not zero, unlimited energy could be gained by repeatedly carrying a charge around a
 If there are n meshes in a circuit, the number of independent equations in accordance with loop rule will be (n – 1).
 Sign convention for the application of Kirchoff’s law : For the application of Kirchoff’s laws following sign convention are to be considered
 The change in potential in traversing a resistance in the direction of current is – iR while in the opposite direction +iR
 The change in potential in traversing an emf source from negative to positive terminal is +E while in the opposite direction – E irrespective of the direction of current in the
 The change in potential in traversing a capacitor from the negative terminal to the positive terminal is + q
C
while in opposite direction – q .
C
A – C +
q 
+ 
q C 
B 
A – C+
q – q C 
B 
 The change in voltage in traversing an inductor in the direction of current is
direction it is + L di .
dt
 Ldi
dt
while in opposite
(4) Guidelines to apply Kirchoff’s law
 Starting from the positive terminal of the battery having highest emf, distribute current at various junctions in the circuit in accordance with ‘junction rule’. It is not always easy to correctly guess the direction of current, no problem if one assumes the wrong
 After assuming current in each branch, we pick a point and begin to walk (mentally) around a closed loop. As we traverse each resistor, capacitor, inductor or battery we must write down, the voltage change for that element according to the above sign
 By applying KVL we get one equation but in order to solve the circuit we require as many equations as there are So we select the required number of loops and apply Kirchhoff’s voltage law across each such loop.
 After solving the set of simultaneous equations, we obtain the numerical values of the assumed currents. If any of these values come out to be negative, it indicates that particular current is in the opposite direction from the assumed
Note : @ The number of loops must be selected so that every element of the circuit must be included in at least one of the loops.
@ While traversing through a capacitor or battery we do not consider the direction of current.
@ While considering the voltage drop or gain across as inductor we always assume current to be in increasing function.
 Determination of equivalent resistance by Kirchoff’s method : This method is useful when we are not able to identify any two resistances in series or in parallel. It is based on the two Kirchhoff’s laws. The method may be described in the following
 Assume an imaginary battery of emf E connected between the two terminals across which we have to calculate the equivalent
 Assume some value of current, say i, coming out of the battery and distribute it among each branch by applying Kirchhoff’s current
 Apply Kirchhoff’s voltage law to formulate as many equations as there are It should be noted that at least one of the equations must include the assumed battery.
 Solve the equations to determine
E ratio which is the equivalent resistance of the network.
i
e.g. Suppose in the following network of 12 identical resistances, equivalent resistance between point A and C
is to be calculated.
According to the above guidelines we can solve this problem as follows
Step (1) Step (2)
An imaginary battery of emf E is assumed across the terminals A and C
The current in each branch is distributed by assuming 4i current coming out of the battery.
Step (3) Applying KVL along the loop including the nodes A, B, C and the battery E. Voltage equation is
 2iR – iR – iR – 2iR + E = 0
Step (4) After solving the above equation, we get 6iR = E Þ equivalent resistance between A and C is
R = E = 6iR = 3 R
4i 4i 2
Example: 67 In the following circuit E_{1} = 4V, R_{1} = 2W [MP PET 2003]
E_{2} = 6V, R_{2} = 2W and R_{3} = 4W. The current i_{1} is
 6 A
 8 A
(c) 2.25 A
(d) 1 A
Solution : (b) For loop (1) 2i_{1}
– 2(i1
 i2
) + 4 = 0 Þ 2i_{1}
– i2 = 2
…… (i)
4V 2W
For loop (2) 4i_{2}
+ 2(i1
 i2
) + 6 = 0 Þ 3i_{2}
– i1 = 3
i1 i1
…… (ii)
After solving equation (i) and (ii) we get i1 = 1.8 A and i2 = 1.6 A i2 i2
6V 4W
Example: 68 Determine the current in the following circuit
 1 A
 5 A
 4 A
 3 A
Solution : (a) Applying KVL in the given circuit we get 2i + 10 – 5 – 3i = 0 Þ i = 1A
Second method : Similar plates of the two batteries are connected together, so the net emf = 10 – 5 = 5V Total resistance in the circuit = 2 + 3 = 5W
\ i = SV
SR
= 5 = 1A 5
Example: 69 In the circuit shown in figure, find the current through the branch BD
 5 A
 0 A
 3 A
 4 A
Solution : (a) The current in the circuit are assumed as shown in the fig.
Applying KVL along the loop ABDA, we get A C
– 6i_{1} – 3 i_{2} + 15 = 0 or 2i_{1} + i_{2} = 5 …… (i)
Applying KVL along the loop BCDB, we get
– 3(i_{1} – i_{2}) – 30 + 3i_{2} = 0 or – i_{1} + 2i_{2} = 10 …… (ii) Solving equation (i) and (ii) for i_{2}, we get i_{2} = 5 A
15 V
i1 D
30 V
Example: 70 The figure shows a network of currents. The magnitude of current is shown here. The current i will be [MP PMT 1995]
 3 A
 13 A
 23 A
 – 3 A
Solution : (c)
i = 15 + 3 + 5 = 23 A
Example: 71 Consider the circuit shown in the figure. The current i_{3} is equal to [AMU 1995]
 5 amp
 3 amp
 – 3 amp
 – 5/6 amp
Solution : (d) Suppose current through different paths of the circuit is as follows.
After applying KVL for loop (1) and loop (2)
28W
54W
We get
28i_{1} = 6 – 8
Þ i1 = – 1 A
2
and
54i_{2} = 6 – 12 Þ i_{2} = – 1 A
3
8 V 12 V
Hence i3 = i1 + i2 = – 5 A
6
Example: 72 A part of a circuit in steady state along with the current flowing in the branches, with value of each resistance is shown in figure. What will be the energy stored in the capacitor C_{0} [IITJEE 1986]
(a) 6 ´ 10^{–4} J
(b) 8 ´ 10^{–4} J
(c) 16 ´ 10^{–4} J
(d) Zero
Solution : (b) Applying Kirchhoff’s first law at junctions A and B respectively we have 2 + 1 – i_{1} = 0 i.e., i_{1} = 3A and i_{2} + 1 – 2 – 0 = 0 i.e., i_{2} = 1A
Now applying Kirchhoff’s second law to the mesh ADCBA treating capacitor as a seat of emf V in open circuit
– 3 ´ 5 – 3 ´ 1 – 1 ´ 2 + V = 0 i.e. V(= V_{A} – V_{B}) = 20 V
So, energy stored in the capacitor U = 1 CV ^{2} = 1 (4 ´ 10 ^{–}^{6} ) ´ (20)^{2} = 8 ´ 10 ^{–}^{4} J
2 2
Example: 73 In the following circuit the potential difference between P and Q is
 15 V
 10 V
 5 V
 5 V
Solution : (c) By using KVL
5 ´ 2 – VPQ + 15 = 0 Þ V_{PQ} = 5V
Different Measuring Instruments.
 Galvanometer : It is an instrument used to detect small current passing through it by showing deflection. Galvanometers are of different types g. moving coil galvanometer, moving magnet galvanometer, hot wire galvanometer. In dc circuit usually moving coil galvanometer are used.
 It’s symbol : ; where G is the total internal resistance of the
 Principle : In case of moving coil galvanometer deflection is directly proportional to the current that passes through it e. i µ θ .
 Full scale deflection current : The current required for full scale deflection in a galvanometer is called full scale deflection current and is represented by i_{g}.
 Shunt : The small resistance connected in parallel to galvanometer coil, in order to control current flowing through the galvanometer is known as
 Ammeter : It is a device used to measure current and is always connected in series with the ‘element’ through which current is to be
 The reading of an ammeter is always lesser than actual current in the
 Smaller the resistance of an ammeter more accurate will be its An ammeter is said to be ideal if its resistance r is zero.
 Conversion of galvanometer into ammeter : A galvanometer may be converted into an ammeter by connecting a low resistance (called shunt S) in parallel to the galvanometer G as shown in
 Equivalent resistance of the combination =
GS
G + S
 G and S are parallel to each other hence both will have equal potential difference e. ig G = (i – ig )S ; which gives
Required shunt
S = ig G
(i – ig )
 To pass nth part of main current (e. ig
= i ) through the galvanometer, required shunt
n
S = G .
(n – 1)
 Voltmeter : It is a device used to measure potential difference and is always put in parallel with the ‘circuit element’ across which potential difference is to be
 The reading of a voltmeter is always lesser than true
 Greater the resistance of voltmeter, more accurate will be its reading. A voltmeter is said to be ideal if its resistance is infinite, e., it draws no current from the circuit element for its operation.
 Conversion of galvanometer into voltmeter : A galvanometer
may be converted into a voltmeter by connecting a large resistance R in series with the galvanometer as shown in the figure.
 Equivalent resistance of the combination = G + R
 According to ohm’s law V = i_{g} (G + R); which gives


V æ V ö
Required series resistance R =
– G = ç – 1÷ G
ig ç Vg ÷
 If n^{th} part of applied voltage appeared across galvanometer (e.
R = (n – 1) G .
V = V ) then required series resistance
g n
 Wheatstone bridge : Wheatstone bridge is an arrangement of four resistance which can be used to measure one of them in terms of Here arms AB and BC are called ratio
arm and arms AC and BD are called conjugate arms
 Balanced bridge : The bridge is said to be balanced when deflection in galvanometer is zero e. no current flows through the galvanometer or in
other words V = V . In the balanced condition P = R , on mutually changing
B D Q S
the position of cell and galvanometer this condition will not change.
 Unbalanced bridge : If the bridge is not balanced current will flow from D to B if V_{D} > V_{B} e.
(VA – VD ) < (VA – VB ) which gives PS > RQ.
 Applications of wheatstone bridge : Meter bridge, post office box and Carey Foster bridge are instruments based on the principle of wheatstone bridge and are used to measure unknown
 Meter bridge : In case of meter bridge, the resistance wire AC is 100 cm Varying the position of
tapping point B, bridge is balanced. If in balanced position of bridge AB = l, BC (100 – l) so that
Q = (100 – l) .
Also
P l
P = R Þ S = (100 – l) R
Q S l
Example: 74 The scale of a galvanometer of resistance 100 W contains 25 divisions. It gives a deflection of one division on passing a current of 4 ´ 10^{–4} A. The resistance in ohms to be added to it, so that it may become a voltmeter of range 2.5 volt is [EAMCET 2003]
(a) 100 (b) 150 (c) 250 (d) 300
Solution : (b) Current sensitivity of galvanometer = 4 ´ 10^{–}^{4} Amp/div.
So full scale deflection current (i_{g}) = Current sensitivity ´ Total number of division = 4 ´ 10^{–4} ´ 25 = 10^{–2} A
To convert galvanometer in to voltmeter, resistance to be put in series is R = V
ig
 G =
2.5
10 ^{–}^{2}
– 100 = 150 W
Example: 75 A galvanometer, having a resistance of 50 W gives a full scale deflection for a current of 0.05 A. the length in meter of a resistance wire of area of crosssection 2.97 ´ 10^{–2} cm^{2} that can be used to convert the galvanometer into an ammeter which can read a maximum of 5A current is : (Specific resistance of the wire = 5 ´ 10^{–7} Wm) [EAMCET 2003]
(a) 9 (b) 6 (c) 3 (d) 1.5
Solution : (c) Given G = 50 W, i_{g} = 0.05 Amp., i = 5A, A = 2.97 ´ 10^{–2} cm^{2} and r = 5 ´ 10^{–7}Wm
By using
i = 1 + G
Þ S = G.ig
Þ r l = Gig Þ l = Gig A
on putting values l = 3 m.
ig S
(i – ig )
A (i – ig )
(i – ig )r
Example: 76 100 mA current gives a full scale deflection in a galvanometer of resistance 2 W. The resistance connected with the galvanometer to convert it into a voltmeter of 5 V range is
[KCET 2002; UPSEAT 1998; MNR 1994 Similar to MP PMT 1999]
(a) 98 W (b) 52 W (c) 80 W (d) 48 W
Solution : (d)
R = V I g
 G =
5
100 ´ 10 ^{–}^{3}
– 2 = 50 – 2 = 48 W .
Example: 77 A milliammeter of range 10 mA has a coil of resistance 1 W. To use it as voltmeter of range 10 volt, the resistance that must be connected in series with it will be [KCET (Engg./Med.) 2001]
(a) 999 W (b) 99 W (c) 1000 W (d) None of these
Solution : (a) By using
R = V – G ig
Þ R =
10
10 ´ 10 ^{–}^{3}
– 1 = 999W
Example: 78 In the following figure ammeter and voltmeter reads 2 amp and 120 volt respectively. Resistance of voltmeter is
(a) 100 W
(b) 200 W
(c) 300 W
(d) 400 W
Solution : (c) Let resistance of voltmeter be R_{V}. Equivalent resistance between X and Y is RXY
= 75RV
75 + RV
Reading of voltmeter = potential difference across X and Y = 120 = i ´ R_{XY}
= 2 ´
75RV
75 + RV
Þ R_{V}
= 300W
Example: 79 In the circuit shown in figure, the voltmeter reading would be
 Zero
 5 volt
 1 volt
 2 volt
Solution : (a) Ammeter has no resistance so there will be no potential difference across it, hence reading of voltmeter is zero.
Example: 80 Voltmeters V_{1} and V_{2} are connected in series across a d.c. line. V_{1}reads 80 V and has a per volt resistance of 200 W, V_{2} has a total resistance of 32 kW. The line voltage is [MNR 1992]
(a) 120 V (b) 160 V (c) 220 V (d) 240 V
Solution : (d) Resistance of voltmeter V_{1} is R_{1} = 200 ´ 80 = 16000 W and resistance of voltmeter V_{2} is R_{2} = 32000 W
æ R‘ ö
By using relation V ‘ = ç ÷ V;
ç Req ÷
where V¢ = potential difference across any resistance R¢ in a series grouping.
è ø R_{1} R_{2}
So for voltmeter V_{1} potential difference across it is
æ R ö
80 = ç ^{1} ÷ .V
è R1 + R2 ø
Þ V = 240 V
V
Example: 81 The resistance of 1 A ammeter is 0.018 W. To convert it into 10 A ammeter, the shunt resistance required will be [MP PET 1982]
(a) 0.18 W (b) 0.0018 W (c) 0.002 W (d) 0.12 W
Solution : (c) By using
i = 1 + 4
Þ 10 = 1 + 0.018
Þ S = 0.002 W
i_{g} S 1 S
Example: 82 In meter bridge the balancing length from left and when standard resistance of 1 W is in right gas is found to be 20 cm. The value of unknown resistance is [CBSE PMT 1999]
(a) 0.25 W (b) 0.4 W (c) 0.5 W (d) 4 W
Solution: (a) The condition of wheatstone bridge gives
X = 20 r , r– resistance of wire per cm, X– unknown resistance
\ X = 20 ´ R = 1 ´ 1 = 0.25 W
R 80 r
X R = 1W
80 4
P = 20 r Q = 80 r
20 cm 80 cm
Example: 83 A galvanometer having a resistance of 8 W is shunted by a wire of resistance 2 W. If the total current is 1 amp, the part of it passing through the shunt will be [CBSE PMT 1998]
(a) 0.25 amp (b) 0.8 amp (c) 0.2 amp (d) 0.5 amp
Solution: (b) Fraction of current passing through the galvanometer
ig = S or ig
i S + G i
= 2 = 0.2
2 + 8
So fraction of current passing through the shunt
is = 1 – ig
i i
= 1 – 0.2 = 0.8 amp
Example: 84 A moving coil galvanometer is converted into an ammeter reading upto 0.03 A by connecting a shunt of resistance 4r across it and into an ammeter reading upto 0.06 A when a shunt of resistance r connected across it. What is the maximum current which can be through this galvanometer if no shunt is used
[MP PMT 1996]
(a) 0.01 A (b) 0.02 A (c) 0.03 A (d) 0.04 A
Solution: (b) For ammeter, S = ig G
(i – ig )
Þ ig
G = (i – ig )S
So ig G = (0.03 – ig )4r
…… (i) and
ig G = (0.06 – ig )r
…… (ii)
Dividing equation (i) by (ii)
Þ 3i_{g} = 0.06 Þ i_{g} = 0.02 A
1 = (0.03 – ig ) 4 Þ 0.06 – i
0.06 – ig g
= 0.12 – 4ig
Potentiometer.
Potentiometer is a device mainly used to measure emf of a given cell and to compare emf’s of cells. It is also used to measure internal resistance of a given cell.
 Superiority of potentiometer over voltmeter : An ordinary voltmeter cannot measure the emf accurately because it does draw some current to show the As per definition of emf, it is the potential difference when a cell is in open circuit or no current through the cell. Therefore voltmeter can only measure terminal voltage of a give n cell.
Potentiometer is based on no deflection method. When the potentiometer gives zero deflection, it does not draw any current from the cell or the circuit i.e. potentiometer is effectively an ideal instrument of infinite resistance for measuring the potential difference.
 Circuit diagram : Potentiometer consists of a long resistive wire AB of length L (about 6m to 10 m long) made up of mangnine or A battery of known voltage e and internal resistance r called supplier battery or driver cell. Connection of these two forms primary circuit.
One terminal of another cell (whose emf E is to be measured) is connected at one end of the main circuit and the other terminal at any point on the resistive wire through a galvanometer G. This forms the secondary circuit. Other details are as follows
J = Jockey
K = Key
R = Resistance of potentiometer wire,
r = Specific resistance of potentiometer wire.
R_{h} = Variable resistance which controls the current through the wire AB
(3) Points to be remember
 The specific resistance (r) of potentiometer wire must be high but its temperature coefficient of resistance (a) must be
 All higher potential points (terminals) of primary and secondary circuits must be connected together at point A and all lower potential points must be connected to point B or
 The value of known potential difference must be greater than the value of unknown potential difference to be
 The potential gradient must remain For this the current in the primary circuit must remain constant and the jockey must not be slided in contact with the wire.
 The diameter of potentiometer wire must be uniform
 Potential gradient (x) : Potential difference (or fall in potential) per unit length of wire is called potential
V volt
æ e ö
V iR iq e R


gradient i.e. x = L m
where V = iR = ç
è R + Rh
 r ÷.R . So
x = L =
L = A = (R + R
 r) . L
 Potential gradient directly depends upon
 The resistance per unit length (R/L) of potentiometer
 The radius of potentiometer wire (e. Area of crosssection)
 The specific resistance of the material of potentiometer wire (e. r)
 The current flowing through potentiometer wire (i)
 x indirectly depends upon
 The emf of battery in the primary circuit (e. e)
 The resistance of rheostat in the primary circuit (e. R_{h})
Note : @When potential difference V is constant then
x1 = L2 x 2 L1
@ Two different wire are connected in series to form a potentiometer wire then
x1 = R1 . L2
x 2 R2 L1
@ If the length of a potentiometer wire and potential difference across it’s ends are kept constant and if it’s diameter is changed from d_{1} ® d_{2} then potential gradient remains unchanged.
@ The value of x does not change with any change effected in the secondary circuit.
 Working : Suppose jocky is made to touch a point J on wire then potential difference between A and J
will be V = xl
At this length (l) two potential difference are obtained
 V due to battery e and
 E due to unknown cell
If V > E then current will flow in galvanometer circuit in one direction If V < E then current will flow in galvanometer circuit in opposite direction
If V = E then no current will flow in galvanometer circuit this condition to known as null deflection position, length l is known as balancing length.


V iR æ e ö R

In balanced condition E =
xl or
E = xl =
L l =
L l = ç R + R
 r ÷. L ´ l
Note : @ If V is constant then L µ l Þ
L1 = l1 L2 l2
 Standardization of potentiometer : The process of determining potential gradient experimentally is known as standardization of
Let the balancing length for the standard emf E_{0} is l_{0} then by the principle of
potentiometer E_{0}
= xl_{0} Þ
x = E0
l
0
 Sensitivity of potentiometer : A potentiometer is said to be more sensitive, if it measures a small potential difference more
 The sensitivity of potentiometer is assessed by its potential The sensitivity is inversely proportional to the potential gradient.
 In order to increase the sensitivity of potentiometer
 The resistance in primary circuit will have to be
 The length of potentiometer wire will have to be increased so that the length may be measured more
 Difference between voltmeter and potentiometer
Application of Potentiometer.
(1) To determine the internal resistance of a primary cell
 Initially in secondary circuit key K’ remains open and balancing length (l_{1}) is Since cell E is in open circuit so it’s emf balances on length l_{1} i.e. E = xl_{1} ……. (i)
 Now key K¢ is closed so cell E comes in closed If the process is repeated again then potential difference V balances on
length l_{2}
i.e. V = xl_{2}
……. (ii)
 By using formula internal resistance r = æE – 1ö. R‘

ç ÷
è ø
æ l1 – l2 ö
r = ç
è l2
÷. R‘
ø
 Comparison of emf’s of two cell : Let l_{1} and l_{2} be the balancing lengths with the cells E_{1} and E_{2}
respectively then E_{1}
= xl_{1}
and E_{2}
= xl_{2} Þ
E1 = l1 E2 l2
Note : @ Let E_{1}
> E_{2}
and both are connected in series. If balancing length
is l_{1} when cell assist each other and it is l_{2} when they oppose each other as shown then :
+ E1– + E2– + E1– – E2+
(E1 + E2 ) = xl1 (E1 – E2 ) = xl 2
Þ E1 + E2
E1 – E2
= l1 or
l 2
E1 = l1 + l2
E2 l1 – l2
 Comparison of resistances : Let the balancing length for resistance R_{1} (when XY is connected) is l_{1} and let balancing length for resistance R_{1} + R_{2} (when YZ is connected) is l_{2}.
Then iR_{1} = xl_{1} and i(R_{1} + R_{2}) = xl_{2}
Þ R2
R1
= l2 – l1 l1
(4) To determine thermo emf
 The value of thermoemf in a thermocouple for ordinary temperature difference is very low (10–6 volt). For this the potential gradient x must be also very low (10^{–4} V/m). Hence a high resistance (R) is connected in series with the potentiometer wire in order to reduce
 The potential difference across R must be equal to the emf of
standard cell i.e. iR = E_{0}
\ i = E0
R
 The small thermo emf produced in the thermocouple e = xl
x = ir = iR
L
\ e = iRl where L = length of potentiometer
L
wire, r = resistance per unit length, l = balancing length for e
(5) To calibrate ammeter and voltmeter
Example: 85 A battery with negligible internal resistance is connected with 10m long wire. A standard cell gets balanced on 600 cm length of this wire. On increasing the length of potentiometer wire by 2m then the null point will be displaced by
(a) 200 cm (b) 120 cm (c) 720 cm (d) 600 cm
Solution : (b) By using
L1 = l1
Þ 10 = 600 Þ l
= 720 cm .
L2 l 2
12 l 2 2


Hence displacement = 720 – 600 = 120 cm
Example: 86 In the following circuit a 10 m long potentiometer wire with resistance 1.2 ohm/m, a resistance R_{1} and an accumulator of emf 2 V are connected in series. When the emf of thermocouple is 2.4 mV then the deflection in galvanometer is zero. The current supplied by the accumulator will be
(a) 4 ´ 10^{–4} A (b) 8 ´ 10^{–4} A (c) 4 ´ 10^{–3} A (d) 8 ´ 10^{–3} A
E E 2.4 ´ 10 ^{–}^{3} _{–}_{4}
Solution : (a)
E = x l = ir l
\ i = rl = rl =
1.2 ´ 5
= 4 ´ 10 A .
Example: 87 The resistivity of a potentiometer wire is 40 ´ 10^{–8} Wm and its area of cross section is 8 ´ 10^{–6} m^{2}. If 0.2
amp. Current is flowing through the wire, the potential gradient will be [MP PET/PMT 1998]
(a) 10^{–2} volt/m (b) 10^{–1} volt/m (c) 3.2 ´ 10^{–2} volt/m (d) 1 volt/m
V iR
irL ir
0.2 ´ 40 ´ 10^{–}^{8} _{–}_{2}
Solution : (a) Potential gradient = =
L
= = =
L AL A
8 ´ 10^{–}^{6}
= 10
V/ m
Example: 88 A deniel cell is balanced on 125 cm length of a potentiometer wire. When the cell is short circuited with a 2 W resistance the balancing length obtained is 100 cm. Internal resistance of the cell will be [RPMT 1998]
(a) 1.5 W (b) 0.5 W (c) 1.25 W (d) 4/5 W
Solution: (b) By using r = l1 – l2 ´ R‘ Þ r = 125 – 100 ´ 2 = 1 = 0.5 W
l2 100 2
Example: 89 A potentiometer wire of length 10 m and a resistance 30 W is connected in series with a battery of emf
2.5 V and internal resistance 5 W and an external resistance R. If the fall of potential along the potentiometer wire is 50 mV/mm, the value of R is (in W) [KCET 1998]
(a) 115 (b) 80 (c) 50 (d) 100
Solution : (a) By using
x = e . R
(R + Rh + r) L
Þ 50 ´ 10 ^{–}^{6}
= 2.5 ´ 30
Þ R = 115
10 3
(30 + R + 5) 10
Example: 90 A 2 volt battery, a 15 W resistor and a potentiometer of 100 cm length, all are connected in series. If the resistance of potentiometer wire is 5 W, then the potential gradient of the potentiometer wire is [AIIMS 1982]
(a) 0.005 V/cm (b) 0.05 V/cm (c) 0.02 V/cm (d) 0.2 V/cm
Solution : (a) By using
x = e . R
(R + Rh + r) L
Þ x =
2
(5 + 15 + 0)
´ 5 = 0.5 V/m = 0.005 V/cm 1
Example: 91 In an experiment to measure the internal resistance of a cell by potentiometer, it is found that the balance point is at a length of 2 m when the cell is shunted by a 5 W resistance; and is at a length of 3 m when the cell is shunted by a 10 W resistance. The internal resistance of the cell is, then [Haryana CEE 1996]
(a) 1.5 W (b) 10 W (c) 15 W (d) 1 W
æ l_{1} – l _{2} ö æ l_{1} – 2 ö
Solution : (b) By using r = ç
è l 2
÷ R‘
ø
Þ r = ç
è
÷ ´ 5

ø
…… (i)
and
r = æ l1 – 3 ö ´ 10
……. (ii)

ç ÷
è ø
On solving (i) and (ii) r = 10 W
Example: 92 A resistance of 4 W and a wire of length 5 metres and resistance 5 W are joined in series and connected to a cell of emf 10 V and internal resistance 1 W. A parallel combination of two identical cells is balanced across 300 cm of the wire. The emf E of each cell is [RPET 2001; MP PMT 1997]
(a) 1.5 V (b) 3.0 V (c) 0.67 V (d) 1.33 V
Solution : (b) By using
Eeq = e . R ´ l
Þ E = 10 ´ 5 ´ 3 Þ E = 3 volt
(R + Rh + r) L
(5 + 4 + 1) 5
Example: 93 A potentiometer has uniform potential gradient across it. Two cells connected in series (i) to support each other and (ii) to oppose each other are balanced over 6 m and 2 m respectively on the potentiometer wire. The emf’s of the cells are in the ratio of [MP PMT 2002; RPMT 2000]
(a) 1 : 2 (b) 1 : 1 (c) 3 : 1 (d) 2 : 1
Solution : (d) If suppose emf’s of the cells are E_{1} and E_{2} respectively then
E_{1} + E_{2} = x ´ 6 ……. (i) [x = potential gradient) ] and E_{1} – E_{2} = x ´ 2 …….(ii)
Þ E1 + E2
= 3 Þ
E_{1} = 2
E1 – E2 1 E2 1
Example: 94 In the following circuit the potential difference between the points B and C is balanced against 40 cm length of potentiometer wire. In order to balance the potential difference between the points C and D, where should jockey be pressed
 32 cm (b) 16 cm (c) 8 cm (d) 4 cm
Solution : (a)
1 = 1 + 1 = 2 = 1
or R_{1} = 5 W
R 10 10 10 5
R_{2} = 4W, l_{1} = 40 cm, l_{2} = ?
l 2 = l
R2
1 R1
or l 2
= 40 ´ 4 = 32 cm
5
Example: 95 In the following circuit diagram fig. the lengths of the wires AB and BC are same but the radius of AB is three times that of BC. The ratio of potential gradients at AB and BC will be
(a) 1 : 9
(b) 9 : 1
(c) 3 : 1
(d) 1 : 3
1 x_{1}
r 2 æ r ö ^{2} 1
Solution : (a)
x µ Rp µ Þ
= ^{ } ^{2} = ç ÷ =


r 2 x 2
2 è 3r ø 9
Example: 96 With a certain cell the balance point is obtained at 0.60 m from one end of the potentiometer. With another cell whose emf differs from that of the first by 0.1 V, the balance point is obtained at 0.55 m. Then, the two emf’s are
(a) 1.2 V, 1.1 V (b) 1.2 V, 1.3 V (c) – 1.1 V, – 1.0 V (d) None of the above
Solution : (a) E_{1}
= x (0.6) and E_{2}
= E_{1}
– 0.1 = x (0.55) Þ
E1 =
E1 – 0.1
0.6
0.55
or 55 E_{1} = 60 E_{1} – 6 Þ
E1 = 6 = 1.2 V
5
thus E_{2} = 1.1 V