Chapter 9 Current Electricity Part 2 – Physics free study material by TEACHING CARE online tuition and coaching classes
Chapter 9 Current Electricity Part 2 – Physics free study material by TEACHING CARE online tuition and coaching classes
(iii) Points lying on the parallel axis of symmetry can never have same potential.
 The network can be folded about the parallel axis of symmetry, and the overlapping nodes have same Thus as shown in figure, the following points have same potential
(a) 5 and 6 (b) 2, 0 and 4 (c) 7 and 8
Þ
Note : @ Above network may be split up into two equal parts about the parallel axis of symmetry as shown
in figure each part has a resistance R¢, then the equivalent resistance of the network will be
2
R = R‘ .
2
R R
R R Þ
R ¢ = 3R

1 R R 3 1 3
A A¢
Some Standard Results for Equivalent Resistance.
Example: 27 In the figure a carbon resistor has band of different colours on its body. The resistance of the following body is
[Kerala PET 2002]
(a) 2.2 kW
(b) 3.3 kW
(c) 5.6 kW
(d) 9.1 kW
Solution : (d) R = 91 ´ 10^{2} ± 10% » 9.1 kW
Example: 28 What is the resistance of a carbon resistance which has bands of colours brown, black and brown [DCE 1999]
(a) 100 W (b) 1000 W (c) 10 W (d) 1 W
Solution : (a) R = 10 ´ 10^{1} ± 20% » 100 W
Example: 29 In the following circuit reading of voltmeter V is [MP PET 2003]
 12 V
 8 V
 20 V
 16 V
Solution : (a) P.d. between X and Y is V_{XY} = V_{X} – V_{Y} = 1 ´ 4 = 4 V …. (i)
and p.d. between X and Z is V_{XZ} = V_{X} – V_{Z} = 1 ´ 16 = 16 V …. (ii) On solving equations (i) and (ii) we get potential difference between Y
4 W Y
16 W
and Z i.e., reading of voltmeter is VY – VZ
= 12V
16 W Z 4 W
Example: 30 An electric cable contains a single copper wire of radius 9 mm. It’s resistance is 5 W. This cable is replaced by six insulated copper wires, each of radius 3 mm. The resultant resistance of cable will be [CPMT 1988]
(a) 7.5 W (b) 45 W (c) 90 W (d) 270 W
Solution : (a) Initially : Resistance of given cable
R = r l
p ´ (9 ´ 10 ^{3} )^{2}
….. (i) l
Finally : Resistance of each insulated copper wire is
R ‘ = r l
p ´ (3 ´ 10 ^{3} )^{2}
Hence equivalent resistance of cable
9 mm
l
R ‘ 1 æ l ö
Req = 6
= 6 ´ ç r p ´ (3 ´ 10 ^{3} )^{2} ÷
….. (ii)
è ø
On solving equation (i) and (ii) we get R_{eq} = 7.5 W
Example: 31 Two resistance R_{1} and R_{2} provides series to parallel equivalents as
n then the correct relationship is
1
æ R ö ^{2}
æ R ö ^{2}
æ R ö 3 / 2
æ R ö 3 / 2
(a) ç^{ } ^{1} ÷
è R2 ø
+ ç^{ } ^{2} ÷
è R1 ø
= n^{2}
(b)
ç 1 ÷
è R2 ø
+ ç^{ } ^{2} ÷
è R1 ø
= n3 / 2
æ R ö æ R ö
æ R ö1 / 2
æ R ö1 / 2
(c)
ç^{ } ^{1} ÷ + ç^{ } ^{2} ÷ = n
(d)
ç 1 ÷
+ ç^{ } ^{2} ÷
= n1 / 2
è R2 ø è R1 ø
è R2 ø è R1 ø



R1 R2 RS
(R1 + R2 )^{2}
Solution : (d) Series resistance RS = R1 + R2 and parallel resistance RP
= R + R Þ R =
= n
R1 R2
R + R
Þ 1 2 =
Þ + = Þ + =
Example: 32 Five resistances are combined according to the figure. The equivalent resistance between the point X and Y
will be [UPSEAT 1999; AMU 1995; CPMT 1986]
(a) 10 W
(b) 22 W
(c) 20 W
(d) 50 W
Solution : (a) The equivalent circuit of above can be drawn as Which is a balanced wheatstone bridge.
So current through AB is zero.
10 W
Y
So 1 = 1 + 1 = 1
Þ R = 10 W
X 10 W B
R 20 20 10
Example: 33 What will be the equivalent resistance of circuit shown in figure between points A and D [CBSE PMT 1996]
(a) 10 W
(b) 20 W
(c) 30 W
(d) 40 W
Solution : (c) The equivalent circuit of above fig between A and D can be drawn as
Balanced wheatstone
A
D
So Req = 10 + 10 + 10 = 30W
bridge A
Þ
Series
D
Example: 34 In the network shown in the figure each of resistance is equal to 2W. The resistance between A and B is
[CBSE PMT 1995]
 1 W
 2 W
 3 W
 4 W
Solution : (b) Taking the portion COD is figure to outside the triangle (left), the above circuit will be now as resistance of each is 2 W the circuit will behaves as a balanced wheatstone bridge and no current flows through CD. Hence
R_{AB} = 2W A C O
D E
B
Example: 35 Seven resistances are connected as shown in figure. The equivalent resistance between A and B is [MP PET 2000]
 3 W
 4 W
(c) 4.5 W
(d) 5 W
Solution : (b)

 R
=
 S
Parallel (66) = 3W
So the circuit is a balanced wheatstone bridge.
So current through 8W is zero Req = (5 + 3)(5 + 3) = 8 8 = 4W
Example: 36 The equivalent resistance between points A and B of an infinite network of resistance, each of 1 W, connected as shown is [CEE Haryana 1996]
 Infinite
 2 W
1 + 5 W
2
 Zero
Solution : (c) Suppose the effective resistance between A and B is R_{eq}. Since the network consists of infinite cell. If we exclude one cell from the chain, remaining network have infinite cells i.e. effective resistance between C and D
will also R_{eq}
R Req 1
 R
A C
Req

So now Req = Ro + (R Req ) = R + R + R
Þ Req = [1 + 5]

B D
Example: 37 Four resistances 10 W, 5 W, 7 W and 3 W are connected so that they form the sides of a rectangle AB, BC, CD and DA respectively. Another resistance of 10 W is connected across the diagonal AC. The equivalent resistance between A & B is [EAMCET Med. 2000]
(a) 2 W (b) 5 W (c) 7 W (d) 10 W
Solution : (b)
C
Þ 5W
A B
Series C
(5 S 5) = 10
5W
Þ 5W
10W
A B
ß

So Req = 10 ´ 10 = 5W
10 10
Ü
A
10W
B
Example: 38 The equivalent resistance between A and B in the circuit shown will be
 5 r
4
 6 r
5
 7r
6
 8 r
7
Solution : (d) In the circuit, by means of symmetry the point C is at zero potential. So the equivalent circuit can be drawn as
2r
r r Þ r
A r r B
ß
2 8
æ 8r ö 8
r + r + 3 r = 3 r
R_{eq} = ç 3  2r ÷ = 7 r Ü
è ø Ü
A 2r B A 2r B
Example: 39 In the given figure, equivalent resistance between A and B will be [CBSE PMT 2000]
(a)
(b)
(c)
(d)
14 W
3
3 W
14
9 W
14
14 W
9
Solution : (a) Given Wheatstone bridge is balanced because P = R . Hence the circuit can be redrawn as follows
Q S
Series 3 + 4 = 7 W
Parallel
7W
Req = 14 W
3
A A
Þ
Series 6 + 8 = 14 W
B
14W
Example: 40 In the combination of resistances shown in the figure the potential difference between B and D is zero, when unknown resistance (x) is [UPSEAT 1999; CPMT 1986]
 4 W
 2 W
 3 W
 The emf of the cell is required
Solution : (b) The potential difference across B, D will be zero, when the circuit will act as a balanced wheatstone bridge and
P = R Þ 12 + 4 = 1 + 3
Þ x = 2W
Q S x 1 / 2
Example: 41 A current of 2 A flows in a system of conductors as shown. The potential difference (V_{A} – V_{B}) will be
[CPMT 1975, 76]
(a) + 2V
(b) + 1V
 – 1 V
 – 2 V
Solution : (b) In the given circuit 2A current divides equally at junction D along the paths DAC and DBC (each path carry 1A current).
Potential difference between D and A, V_{D} – V_{A} = 1 ´ 2 = 2 volt …. (i) Potential difference between D and B, V_{D} – V_{B} = 1 ´ 3 = 3 volt ….. (ii) On solving (i) and (ii) V_{A} – V_{B} = + 1 volt
Example: 42 Three resistances each of 4 W are connected in the form of an equilateral triangle. The effective resistance
between two corners is [CBSE PMT 1993]
Solution : (d)
(a) 8 W (b) 12 W (c) 3 W
8
4 + 4 = 8W
 8W
3
Þ On Solving further we get equivalent resistance is 8 W
3
4W
Example: 43 If each resistance in the figure is of 9 W then reading of ammeter is [RPMT 2000]
 5 A
 8 A
 2 A
 9 A
Solution : (a) Main current through the battery
i = 9 = 9 A . Current through each resistance will be 1A and only 5
1
resistances on the right side of ammeter contributes for passing current through the ammeter. So reading of ammeter will be 5A.
Example: 44 A wire has resistance 12 W. It is bent in the form of a circle. The effective resistance between the two points on any diameter is equal to [JIPMER 1999]
(a) 12 W (b) 6 W (c) 3 W (d) 24 W
Solution : (c) Equivalent resistance of the following circuit will be
Req = 6 = 3W 2
Example: 45 A wire of resistance 0.5 Wm^{–1} is bent into a circle of radius 1 m. The same wire is connected across a diameter AB as shown in fig. The equivalent resistance is
 p ohm
(b) p (p + 2) ohm
(c) p / (p + 4) ohm
(d) (p + 1) ohm
Solution : (c) Resistance of upper semicircle = Resistance of lower semicircle
= 0.5 ´ (pR) = 0.5 pW
Resistance of wire AB = 0.5 ´ 2 = 1 W
Hence equivalent resistance between A and B
0.5p W
1 = 1 + 1 + 1 Þ R = p W
RAB
0.5p
1 0.5p
^{AB} (p + 4)
0.5p W
Example: 46 A wire of resistor R is bent into a circular ring of radius r. Equivalent resistance between two points X and Y on its circumference, when angle XOY is a, can be given by
(a)
Ra
4p ^{2}
(2p – a)
(b)
R (2p – a) 2p
(c) R (2p – a)
(d)
4p (2p – a)
Ra
Solution : (a) Here R
= R ´ (ra) = Ra
æ a = l ö
and R
= R ´ r(2p – a) = R (2p – a)
XWY
2pr
çQ ÷

2p è ø
XZY
2pr 2p
R R
Ra ´ R (2p – a) a
Req =
XWY XZY
= 2p 2p = R (2p – a)
RXWY + RXZY
Ra + R(2p – a)
4p ^{2}
2p 2p
Example: 47 If in the given figure i = 0.25 amp, then the value R will be [RPET 2000]
(a) 48 W
(b) 12 W
(c) 120 W
(d) 42 W
Solution : (d) i = 0.25 amp V = 12 V
Req = V
i
= 12 = 48 W
0.25
Now from the circuit Req = R + (60  20 10)
= R + 6
Þ R = R_{eq} – 6 = 48 – 6 = 42 W
Example: 48 Two uniform wires A and B are of the same metal and have equal masses. The radius of wire A is twice that of wire B. The total resistance of A and B when connected in parallel is [MNR 1994]
 4 W when the resistance of wire A is 25 W (b) 5 W when the resistance of wire A is 4 W
(c) 4 W when the resistance of wire B is 4.25 W (d) 5 W when the resistance of wire B is 4 W
Solution : (a) Density and masses of wire are same so their volumes are same i.e. A_{1}l_{1} = A_{2}l_{2}
R l A æ A ö ^{2} æ r ö ^{4}
Ratio of resistances of wires A and B ^{A} = ^{1} ´ ^{2} = ç ^{ } ^{2} ÷ = ç ^{2} ÷
Since r_{1}
= 2r_{2} so
R_{A} = 1
RB 16
Þ R_{B}
RB l 2 A1
= 16 R_{A}
è A1 ø
è r1 ø
Resistance R
and R are connected in parallel so equivalent resistance R = RA RB = 16RA , By checking

A B
A
+ RB 17


correctness of equivalent resistance from options, only option (a) is correct.
Cell.
The device which converts chemical energy into electrical energy is known as electric cell.
 A cell neither creates nor destroys charge but maintains the flow of charge present at various parts of the circuit by supplying energy needed for their organised
(2) Cell is a source of constant emf but not constant current.
 Mainly cells are of two types :
(i) Primary cell : Cannot be recharged
 Secondary cell : Can be recharged
(4) The direction of flow of current inside the cell is from negative to positive electrode while outside the cell is form positive to negative electrode.
 A cell is said to be ideal, if it has zero internal
(6) Emf of cell (E) : The energy given by the cell in the flow of unit charge in the whole circuit (including the
cell) is called it’s electromotive force (emf) i.e. emf of cell
E = W , It’s unit is volt
q
or
The potential difference across the terminals of a cell when it is not given any current is called it’s emf.
(7) Potential difference (V) : The energy given by the cell in the flow of unit charge in a specific part of electrical circuit (external part) is called potential difference. It’s unit is also volt
or
The voltage across the terminals of a cell when it is supplying current to external resistance is called potential difference or terminal voltage. Potential difference is equal to the product of current and resistance of that given part i.e. V = iR.
 Internal resistance (r) : In case of a cell the opposition of electrolyte to the flow of current through it is called internal resistance of the The internal resistance of a cell depends on the distance between electrodes (r
µ d), area of electrodes [r µ (1/A)] and nature, concentration (r µ C) and temperature of electrolyte [r µ (1/temp.)]. Internal resistance is different for different types of cells and even for a given type of cell it varies from to cell.
Cell in Various Position.
(1) Closed circuit (when the cell is discharging)
(i) Current given by the cell i =
E
R + r
(ii) Potential difference across the resistance V = iR
 Potential drop inside the cell = ir
(iv) Equation of cell
E = V + ir
(E > V)
(v) Internal resistance of the cell r = æ E – 1ö× R

ç ÷
è ø
V ^{2} æ
E ö ^{2}
(vi) Power dissipated in external resistance (load) P = Vi = i ^{2} R = = ç ÷ .R
R è R + r ø
E 2
Power delivered will be maximum when
R = r
so Pmax = 4r .
This statement in generalised from is called “maximum power transfer theorem”.
(vii) Short trick to calculate E and r : In the closed circuit of a cell having emf E and internal resistance r_{.} If external resistance changes from R_{1} to R_{2} then current changes from i_{1} to i_{2} and potential difference changes from V_{1} to V_{2}. By using following relations we can find the value of E and r.
i1i2
æ i2 R2 – i1 R1 ö
V2 – V1
E =

2
 i1
(R1 – R2 )
r = ç
è
i1 – i2
÷ =
ø i1
 i2
Note : @ When the cell is charging i.e. current is given to the cell then E = V – ir and E < V.
+ V –
(2) Open circuit and short circuit
Open circuit R
C D A B
E, r  Short circuit
R = 0
E, r 
(i) Current through the circuit i = 0  (i) Maximum current (called short circuit current) flows momentarily isc = E r 
(ii) Potential difference between A and B, V_{AB} = E  (ii) Potential difference V = 0 
(iii) Potential difference between C and D, V_{CD} = 0  
Note : @ Above information’s can be summarized by the following graph V V_{max} =E; i = 0
i_{max} =E/r ; V = 0 i 
Example: 49 A new flashlight cell of emf 1.5 volts gives a current of 15 amps, when connected directly to an ammeter of resistance 0.04 W. The internal resistance of cell is [MP PET 1994]
(a) 0.04 W (b) 0.06 W (c) 0.10 W (d) 10 W
Solution : (b) By using i =
E
R + r
Þ 15 =
1.5
 + r
Þ r = 0.06 W
Example: 50 For a cell, the terminal potential difference is 2.2 V when the circuit is open and reduces to 1.8 V, when the cell is connected across a resistance, R = 5W. The internal resistance of the cell is [CBSE PMT 2002]
(a)
10 W
9
 9 W
10
11 W
9
 5 W
9
Solution : (a) In open circuit, E = V = 2.2 V, In close circuit, V = 1.8 V, R = 5W




So internal resistance, r = æ E – 1ö R = æ 2.2 – 1ö ´ 5 Þ r = 10 W
ç ÷ ç 1.8 ÷ 9
Example: 51 The internal resistance of a cell of emf 2V is 0.1 W. It’s connected to a resistance of 3.9 W. The voltage across the cell will be [CBSE PMT 1999; AFMC 1999; MP PET 1993; CPMT 1990]
 5 volt (b) 1.9 volt (c) 1.95 volt (d) 2 volt
Solution : (c) By using r = æ E – 1ö R Þ 0.1 = æ 2 – 1ö´ 3.9
Þ V = 1.95 volt


ç ÷ ç ÷
è ø è ø
Example: 52 When the resistance of 2 W is connected across the terminal of the cell, the current is 0.5 amp. When the resistance is increased to 5 W, the current is 0.25 amp. The emf of the cell is [MP PMT 2000]
(a) 1.0 volt (b) 1.5 volt (c) 2.0 volt (d) 2.5 volt
Solution : (b) By using
E = i1i2 (R
– R ) = 0.5 ´ 0.25 (2 – 5) = 1.5 volt
(i2 – i1 ) ^{1}
^{2} (0.25 – 0.5)
Example: 53 A primary cell has an emf of 1.5 volts, when shortcircuited it gives a current of 3 amperes. The internal resistance of the cell is [CPMT 1976, 83]
(a) 4.5 ohm (b) 2 ohm (c) 0.5 ohm (d) 1/4.5 ohm
Solution : (c)
isc = E
r
Þ 3 = 1.5
r
Þ r = 0.5 W
Example: 54 A battery of internal resistance 4 W is connected to the network of resistances as shown. In order to give the maximum power to the network, the value of R (in W) should be [IITJEE 1995]
(a) 4/9
(b) 8/9
 2
 18
Solution : (c) The equivalent circuit becomes a balanced wheatstone bridge
R 2R 3R
Þ Þ
For maximum power transfer, external resistance should be equal to internal resistance of source
Þ (R + 2R)(2R + 4 R) = 4 (R + 2R) + (2R + 4 R)
i.e.
3R ´ 6R = 4
3R + 6R
or R = 2W
Example: 55 A torch bulb rated as 4.5 W, 1.5 V is connected as shown in the figure. The emf of the cell needed to make the bulb glow at full intensity is [MP PMT 1999]
 5 V
 5 V
(c) 2.67 V
(d) 13.5 V
Solution : (d) When bulb glows with full intensity, potential difference across it is 1.5 V. So current through the bulb and resistance of 1W are 3 A and 1.5 A respectively. So main current from the cell i = 3 + 1.5 = 4.5 A. By using E = V + iR Þ E = 1.5 + 4.5 ´ 2.67 = 13.5 V.
Grouping of cell.
Group of cell is called a battery.
 Series grouping : In series grouping anode of one cell is connected to cathode of other cell and so on.
 n identical cells are connected in series
(a) Equivalent emf of the combination
Eeq = nE
(b) Equivalent internal resistance req = nr
 Main current = Current from each cell = i = nE
R + nr
(d) Potential difference across external resistance V = iR
 Potential difference across each cell V ‘ = V
n
æ nE ö ^{2}
(f) Power dissipated in the circuit P = ç R + nr ÷ . R
è
(g) Condition for maximum power R = nr
ø
and
Pmax
æ 2 ö

= n ç ÷
4r
è ø
 This type of combination is used when nr <<
(ii) If nonidentical cell are connected in series
Cells are connected in right order  Cells are wrongly connected  
E_{1}, r_{1} E2, r2
i R (a) Equivalent emf Eeq = E1 + E2 (b) Current i = Eeq R + req (c) Potential difference across each cell V_{1} and V2 = E2 – ir2 
= E 
1 
– ir1 
(a)
(b)
(c)  E1, r1 E_{2}, r_{2} (E_{1} > E_{2}) 1 2 i R
Equivalent emf E_{eq} = E_{1} – E_{2} Current i = E1 – E2 R + req in the above circuit cell 1 is discharging so it’s equation is E1 = V1 + ir1 Þ V1 = E1 – ir1 and cell 2 is charging so it’s equation E2 = V2 – ir2 Þ V2 = E2 + ir2 
 Parallel grouping : In parallel grouping all anodes are connected at one point and all cathode are connected together at other
(i) If n identical cells are connected in parallel
 Equivalent emf E_{eq} = E
(b) Equivalent internal resistance
Req
= r / n
(c) Main current i =
E
R + r / n
(d) P.d. across external resistance = p.d. across each cell = V = iR
i
æ E ö ^{2}
(e) Current from each cell i‘ = n (f) Power dissipated in the circuit P = ç R + r / n ÷ . R
è ø
æ E 2 ö
(g) Condition for max power
R = r / n
and
Pmax
= n ç ÷
4r
(h) This type of combination is used when nr >> R
è ø
 If nonidentical cells are connected in parallel : If cells are connected with right polarity as shown below then
(a) Equivalent emf E
= E1r2 + E2 r1
(b) Main current i =
eq
Eeq
r + Req
r1 + r2
(c) Current from each cell i
= E1 – iR
and i
= E2 – iR
1 2
1 2


Note : @In this combination if cell’s are connected with reversed polarity as shown in figure then :
Equivalent emf E
= E1r2 – E2 r1
i1 E_{1},r_{1}


_{1} + r2 i
R
(3) Mixed Grouping : If n identical cell’s are connected in a row and such m row’s are connected in parallel as shown.
 Equivalent emf of the combination Eeq= nE
(ii) Equivalent internal resistance of the combination req
nE
= nr m
mnE
(iii) Main current flowing through the load i =
 Potential difference across load V = iR
(v) Potential difference across each cell V ‘ = V
n
R + nr
m
=
mR + nr
(vi) Current from each cell i ‘ = i
n
(vii) Condition for maximum power
 Total number of cell = mn
R = nr
m
and
Pmax
= (mn) E 2
4r
Example: 56 A group of N cells whose emf varies directly with the internal resistance as per the equation E_{N} = 1.5 r_{N} are connected as shown in the following figure. The current i in the circuit is [KCET 2003]
 51 amp
 1 amp
 15 amp
 5 amp
Solution : (d)
i = Eeq = 1.5r1 + 1.5r2 + 1.5r3 +……………. = 1.5 amp
req
r_{1} + r_{2} + r_{3} + ……
Example: 57 Two batteries A and B each of emf 2 volt are connected in series to external resistance R = 1 W. Internal resistance of A is 1.9 W and that of B is 0.9 W, what is the potential difference between the terminals of battery A [MP PET 2001]
 2 V
 8 V
 0
 None of these
Solution : (c)
i = E1 + E2
= 2 + 2 = 4 . Hence V
= E – ir
= 2 – 4 1.9 = 0
R + r1 + r2
1 + 1.9 + 0.9 3.8
A A A
3.8
Example: 58 In a mixed grouping of identical cells 5 rows are connected in parallel by each row contains 10 cell. This combination send a current i through an external resistance of 20 W. If the emf and internal resistance of each cell is 1.5 volt and 1 W respectively then the value of i is [KCET 2000]
(a) 0.14 (b) 0.25 (c) 0.75 (d) 0.68
Solution : (d) No. of cells in a row n = 10; No. of such rows m = 5
i = nE = 10 ´ 1.5 = 15 = 0.68 amp
æ R + nr ö æ 20 + 10 ´ 1 ö 22

ç ÷ ç ÷
è ø è 5 ø
Example: 59 To get maximum current in a resistance of 3 W one can use n rows of m cells connected in parallel. If the total no. of cells is 24 and the internal resistance of a cell is 0.5 then
(a) m = 12, n = 2 (b) m = 8, n = 4 (c) m = 2, n = 12 (d) m = 6, n = 4
Solution : (a) In this question R = 3W, mn = 24, r = 0.5W and R = mr . On putting the values we get n = 2 and m = 12.
n
Example: 60 100 cells each of emf 5V and internal resistance 1 W are to be arranged so as to produce maximum current in a 25 W resistance. Each row contains equal number of cells. The number of rows should be [MP PMT 1997]
(a) 2 (b) 4 (c) 5 (d) 100
Solution : (a) Total no. of cells, = mn = 100………………………………………………………………………… (i)
Current will be maximum when R = nr ; 25 = n ´ 1
Þ n = 25 m……………………………………… (ii)
m m
From equation (i) and (ii) n = 50 and m = 2