iy

Total emf he
H,-(M,-v², (ii)EMF of
htee ells in
ne
temperature in both WhenEi-eatVL0h ellA, -6073 -
erie
2.0V
The
output
external power is
'max
V-30
20) 10
/1-
resistance maximum,
-
10v
Total emfinternal when
Total resistance resistance
4 2.0 3r+ 3r 2r
R+r 2
2 x 1.0
=1.0 A.
V3
3.31 KIRCHHOFF'S LAWS
and
Ingavetrof ductexttwoeondedry concept
Kirchhof s. In 1942, a
Ohm's law to
current in any lawS, Ger m an physi cis t
0.155 or r= R

0.155 standing these

which
part of such enable
a uscomplicated
to
detecircui
rmi nts
e
1. laws, we first definecircuit.
a fewBefore under-
60 × 0.155 Electric
for a network. The term terms
2. complic ated system of electric network is used
orJunction. Any point in an electrical conductors.
55 more electric circuit
= 22.14Q.
3. Loop orconductors are joined together is a where two
Mesh. Any closed junction.
electric network is called a loopconducting
or mesh.
path in an
4. Branch. A
x0.25 = 1.52 branch is any part of the
between two junctions. network that lies
56. State the two
circuits and explain themKirchhoff' s laws for electrical
giving suitable illustrations.
12 Also state the sign
-=4.0 A conventions used.
Kirchhoff's first law or junction rule. In an electric
circuit, the algebraic sum of currents at any junction is
Or, the sum of currents entering a zero.
Sum of currents leaving that junction.junction equal to the
is

Mathematically, this law may be expressed as

=872 VW. 2l=0
 3.S0 PHYSICS-X|

Sçn onention for applying junction rule :

1. The urrents flowing towards the junction are V=+ IR
taken as positive.
Fiq. 3.145 Positive potential
2. The currents flowing away from the junction drop acrOSs a
are taken as negative. 5. The IR product is taken
as resistor,
Figure 3.142 represents a
resistor is traversed in the
negative if the
opposite
junction Jin a ircuit where
four currents meet. The
assumed current.
direction
of
currents I, and I, flowing
towards the junction are V-IR
Fig. 3.146 Negative potential drop across a
positive, while the currents resistor.
I, and I, flowing away from Illustration. Let us consider the
circuit
the junction are negative,
therefore, by junction rule:
Fig. 3.147.
D
shown in
Fig. 3.142 Junction rule:
or
1, + I, - 1, - I, =0 R,
1, + 1, = l, + 1,
i.e., Incoming current = Outgoing current
First law is also called Kirchhoff's current law (KCL).
R,
Justification. This law is based on the law of B
A
conservation of charge. When currernts in a circuit are
steady, charges cannot accumulate or originate at any Fig. 3.147 An electrical circuit.
point of the circuit. So whatever charge flows towards In Fig. 3.147, traversing in the clockwise direction
the junction in any time interval, an equal charge must around the loop ABCFA, we find that:
flow away from that junction in the same time interval. Algebraic sum of current resistance products
Kirchhoff's second law or loop rule. Around any
closed loop of a network, the algebraic sum of changes in
potential must be zero. Or, the algebraic sum of the emfs in Algebraic sum of emfs =, -E,
any loop of a circuit is equal to the sum of the products of Applying Kirchhoffs loop rule to closed path ABCFA,
Currents and resistances in it.
we get E, -E, -1,R -I,R
Mathematically, the loop rule may be expressed as Similarly, applying Kirchhoff's second rule to mesh
AV =0 or E=E IR CDEFC, we get
Sign convention for applying loop rule : E, =I,R, +( + I,) Ry
1. We carn take any direction (cdockwise or anti- Second law is also called Kirchhoff's voltage law (KV).
clockwise) as the direction of traversal. Justification. This law is based on the law of
2. The emf of cell is taken as positive if the conservation of energy. As the electrostatic force is a
direction of traversal is from its negative to the conservative force, so the work done by it along any
positive terminal (through the electrolyte). closed path must be zero.

Examples based on
V=+ V=-E Kirchhoff's LawS
Fig. 3.143 Positive emf. Fig. 3.144 Negative emf. Formulae Used
1. E l 0 (Junction rule)
3. The emf of a cell is taken as
negative if the or Total current
direction of traversal is from its positive to the incoming current =Total outgoing
negative terminal (Fig. 3.144). 2. EE= IR (Looprule)
4. The
current-resistance (IR) product is taken as
positive if the resistor is traversed in the Units Used
and
direction of assumed current. same Current Iis in resistance Rin ohm
emf E in volt. ampere,
 CURRENT ELECTRICITY
3.41
Exomple 137. Netuork rQRSand (Fig. 3.148) is made as Solving equations (1), (2) and (3), we get
battery of 4 V negligible resistance
under: PÌ has a 48 18
I, = 31 A, I, = 21A,
-terminal connected to P, QR has a resistance of
positie I, =66 A
rth
PS has a battery of 5 Vand negligible resistance with
31
602. Example 139. Find the
sitie terminal Connected to P, RS has a resistance of potential difference across each cell
milliammeter, of 20 2 resistance is connected and the rate of energy dissipation in R. (Fig. 3.150a)].
200 2. if
hetueen Pand R, calculate thee reading of the milliammeter. [CBSE SP 111
[NCERT]
200 S2
E,-12V rË=22
S 1-1} R

R=42
5VL
60 2
I-1,A E, -6V =12
4 V
Fig. 3.150(a)
Fiq. 3.148
Solution. Applying Kirchhoff's laws,
second law to the
Solution. Applying Kirchhoff's For closed loop ADCBA
loop PRQP we get 12 = 4(1, + I,)+21 =61 +41, ...)
201, + 60I = 4 ...)
For closed loop ADEFA,
Similarly, from the loop PSRP, we get
6= 4(1, + l,)+ I, =41 +51, ...i)
200(1 - 1,) -201, = -5
401- 44I, =-1 ...(i)
or

Multiplying (1) by 2and (ii) by 3, we get E,= 12 VrË=22

C
1201 + 40I, = 8 ...i)
and 1201- 132 1, =-3 ..(iv) R=4Q
+1,)
D
Subtracting (iv) from (ii), we get A

1721,= 11 E,=6V r,=19

11 = 0.064 A
1, = 172 F

Thus the milliammeter of 20 2 will read 0.064 A.

Fig. 3.150 (b)
tample 138. Using Kirchhoff 's laus in the electrical network
Shoin in Fig. 3.149, calculate the values of I, I, and l3 Solving () and (ii), we get
[CBSE D 2000C]
18 6
B 1= 7 A and ,=-A
P.D. across R=V
= (l, + I,)R
12v4 F D
6V
-x7
4 volt =7
48
volt

Fig. 3.149 P.D. across each cell P.D. across R =

7
Soution. Applying Kirchhoff 's first law at junction B, Energy dissipated in R =4Q resistor
1, + I, =l, .(1)
and Applying Kirchhoff 's second law to loops ABEFA
BCDEB, we get
21, + 51, = 12 ..(2) 576
- 21, - 31, = J =11.75 J.
-6 ...(3) 49
 PHYSI. 6,-2V

and2.0 V and
3.82 cmfk 1.5 V connected E,-1V
of respectivelyare I-32
Tu ols direction
Example 140. 10and2 Q the sane
(CBSE OD 051
resistances current in
internal as o send
prallelso resistanceof5 2. E,-4V r,-22
in external
throuçh an dingram.
() Dranr
thecircuit calculate
Kinhho 'slaus, thecircuit. Fig. 3.152
() Usinc thruçheachbranch of
(a)current resistance. rule.
p.d. aCHSS the5N in Fig.3.151. Solution. By Kirchhoff'sjunction
(b)
ircuit diagramisshown
Solution. () The
52 D From upper loop,
=1
-3I+41--2{| 31, -41, =2,-1
12
From lower loop,
F
I,
31, +21, =4-1=3
On solving equations (1), (1) and (ii), we get
22 7 9
A
13 13
13
2.0 V=E,
Example 142. Apply Kirchhoff's rules to the loops ACBPA
Fig. 3.151 and ACBQA to write the expression for the Currents
3.153. [CBSE OD 10
the currents as shown in and I, in the network shown in Fig.
(ü) (2) Let I, and I, be law for the loop Solution. By Kirchhoff's E, =6V
Fig. 3.151. Using Kirchhoff 's second
AFCBA, we get junction rule,
0.52
21, -11, =¬,-E, =2-1.5
21, -1, =0.5 ..(1) 12/
From loop AQBPA,
For loop CFEDC, we have
0.5I -I, =6-10=-4 E, =10 V
1,+5(1 +4)=¬, =1.5 (i)
51, +61, =1.5 ..2) C

Solving equations (1) and (2), we get From loop ACBPA,

R=12 Q
1 9
A
12 1, +0.51, -6
..(ü) Fig. 3.153
34 34
.:. Current through branch BA, On solving equations (), (ü) and (ü), we get
1 84 106 22
A
34 37
A,
37
A, I,=A
37
Current through branch CF,
Example 143. Use Kirchhoffs rules to detemine the
34
potential difference between the points A and D when no
Current flows in the arm BE of the electric network shown iM
Current through branch DE, Fig. 3.154(a). [CBSE OD 15)
10
A. 32
34 D
(b)P.D. across the5 Q
resistance 1V
=(4, + L,)x 5=x5 V= 1.47 V. R, R
34 22
Example 141. Use
Kirchhoff's rules to write the
expressions for the currents
+3V

diagram shown in Fig. 3.152. I,, 1, and I, in the circuit B

|CBSE OD 10] 6V 4V

Fig. 3.154 (a)

 CURRENT ELECTRICItY

the arm BE.

3.5
No current flows through Solution.Let I,, I, and I,he the
Letlbethe current along the outer loop as shown in Fig. 3.156. Kirchhoff's serrned currents
Solution. sle for thea4 chrwn
cleed
loop ADCAgives
inFig.3.154(b). 10
3S2 D -4,-I,) + 1, *1, )- -0
71, -61, -21, - 10
R,
For the closed loop ABCA, we get
1V
10-41, -2(1, +1,)-1, -0
22 3V I +61, +21, - 10 (2)
For the closed loop BCDEB, we get
4V
6V
5-2(1, + 4)-2(, +,-)=0
Fiq. 3.154 (b) or 21, -4 1, -41,=-5 ...3)
loop rule to the loop AFEBA, (3), we get
Applying Kirchhoff On solving equations (1), (2) and
(2+3)1+ R,x0=1+3+6
I=2A
I,=2.5 A, ;g
various branches of the network
alongAFD, The currents in the
From Ato D=2x2-1+3x2 =9V. are:
Van assuming point A
circuitFig. 3.155, determinethe
Example 144.In theuseKirchhoff's rules to
potential,
tobeatzero B.
potentialatpoint 4V 3.157(a), E, F
D 3A the circuit shown in Fig.
1A Example 146. In
emf2 V, 1 V,3Vand1 V, and their
of respectively.
G and H are cells are2 2, 12, 32 and12,
internalresistances D and
$20
R
potential difference between Band ofthe
Calculate (i) the terminals of each
3A potential diferenceacross the [CBSE 18C]
2V (i) the
C cells G and H.
A
Fig. 3.155

BDCRB we get H+
Solution. From the loop 22

2x2+3 R, =4 or R=0
+0=2V.
V;==2 + V, =2 ofthe
Determine thecurrentin enchbranch
[NCERT]
Fig. 3.157 (a)
Example 145.
3.156. network has been
network shown in Fig. 3.157(), the
B Solution. In Fig. emfs and internal resistances of
redrawn showing the
42
the cells explicitly.
2V
22
22 A

5V

10 V 1VF
D 3S2

D Fig. 3.157 (b)

Fig. 3.156
 PHYSICS-XII
squares ABCD
86 Example 152. Tuvo sides are of and BEFC
Curnents in differentbranches
are side BC in common.
resistances
The
conducting wihaveres withthe
as follous : AB, BE, FC and CD each 2s
lan 0.2 A; of emf2 Vand
BC, EF each 12. AcellAD. Find the internal
S, AD,
RC -, - 0.6 A;
, - ,=0.3 A; 2 2 is joined
across
branches of the circuit.
currents in resivarstanceious
0.5 A; Solution. The current distribution in
Lap
r -l, =0.3 A. of the circuit is shown in Fig.
3.163. various banches
resistance betwcen the B
Exomple 151. Find the cquiralent Fig. 3.162.
terminels Aand B in the network shown in 2Q 22
E
Girem cnch resistor R is of 10 S2.
2V 22 12 12
4e
R R 22 I, 22
i-,

i-4
J R
D

Fig. 3.163

R N
Applying Kirchhoff 's second law to the
Containing the cell and AD, we get loop
2 x I+1x(-)=2
31-1 =2
Fig. 3.162 -1)
From the loop ABCDA, we get
Solution. Imagine a battery of emf E, having no 2xI +1x (l -1,)+2 x h -1x(I-I,) =0
internal resistance, connected between the points A
and B. The distribution of current through various or
-1+6 -, =0 ..2)
branches is as shown in Fig. 3.162. Similarly, from the loop BEFCB, we get
Applying Kirchhoff 's second law to loop KLOPK, 2x 1, +1x I, +2 x 1, -1x (!, -1,)=0
we get
, R+(1, -) R-2(1- 4) R=0
-4 +6l, =0 ..3)
Solving equations (1), (2) and (3), we get
41 -I, =2 1 ..(1) 70 12 2
Similarly, from the loop LMNOL, we have I=A,
99 1= 99 A, L, =;99 A
2 1, R-(l-) R-(4, -,) R=0 Currents in different branches are
-I, +41, =1 ..(2) 12
From the loop AKPONBEA, we have AB =cD =l,= 99 2
99
2(1-1) R+ (l -4,)R=8 ..(3) 'AD =I-I=99 A,
58
Bc =l, -l, =A
10
SoBving equations (1) and (2),we get 99
3
and
2 Current through the cell =I = A
99
Substituting these values in equation (3), we get Example 153. Two points A and Bare
constant potential difference of 110 V. Amaintained
third
at a
connected to A by two resistances of 100 and point is
or 7
IR =& parallel, and to B by asingle resistance of 300 200S2
5 ..(4) Current in each S2. Fina tne
If R' is the resistance and the potential diferenct
then equivalent resistance between A and B, between A and Cand betuween C and B.
IR'-¬
Solution. The circuit arrangement and the
From (4) and (5), IR= 7 ..5) distribution is shown in Fig. 3.164. current
IR
Applying
DEFGHID, we Kirchhoff'
get s second law to the lo0p
5
or
R= 7 7
5 R=x
5 10 =14 2. I, x100 -(l- 1,)x
or 200 =0
300 1, -200 I =0 ..(1)
 CURRENT ELECTRICITY

61
D 21

421

Do B

Fig. 3.164
61A 2/ B
ADIHGCBA, we get
Similarly, from loop = 110
(1-,)200+ lx300 ..(2)
E

500 I200 1, = 110 Fig. 3.165

get
equations (1) and (2), we . The equivalent resistance of the network is
Solving
A and A Totalemf ¿ 5 IR 5
5 R'= 61 6
10 Total current 6I
2resistance
.:Current through 100 But R=12
A R'
6
resistance network is
Current through2002 Total current in the
1 10 =12 A or I=2A
=I-,=;10 A
2 resistance
Current through 300 Ls8=a=ap =pc=Ipç.=cc=21=4A
3
A
A.
g =apla =I=2
10
resistor
and C= P.D.
across 100 S2 I8c =oc =p
P.D.between A having a resistance of r2
100 1 =20 V wires each
=1, x 100=-x
5 Example 155. Twelve
cube; find the resistance
of
askeleton
300 2resistor are connected to form corners of the same edge.
Cand B = P.D. across the cube between the
two
P.D. between junction A
3
300 = 90 V.
Solution. Let a current x +2y enter the
= Ix 300=-10 x of the
the cube ABCDEFGH. From the symmetry shown in
of
and negligible internal parallel paths, current distribution will be as
154. Abattery of 10 V
Example diagonally opposite corners
resistance is connected across the resistors each of resistance Fig. 3.166. H
G
netuork consisting of 12
ofa cubical resistance of the network and 2(y - z)
12. Determine the eguivalent cube. [NCERT] y-z
of the
the current along each edge through the cell. E F
Solution. Let 6I be the current symmetrically y-z z y-2
Since the paths AA', AD and AB are
of them is same, i.e., 21. At
placed, current through each current 2/ splits D
the junctions A', Band Dthe incomingbranches, the current
equally into the two outgoing Fig 3.165. At the
through each branch is , as shown in reunite and the B x+2y
junctions B,Cand D', these currents r+ 2y A
2l each. The total
currents along BC, D'C'and CC are Fig. 3.166
current at junction Cis 6l again. loop
Applying Kirchhoff 's second law to
the loop Applying Kirchhoff 's second law to the
ABCC' EA, we get DHGCD, we get
-2 IR -IR -2 IR +E=0 or E=5 IR (y-z)r+2(y -z)r+ (y -z)r-zr=0
4
where Ris the resistance of each edge and Eis the emf or 4yr-5zr=0 or 5z =4y or
of the battery.
 PHYSICS-XI|
4x + 5y 0
or
second law to the loop ABCDA, 4
Applving kinhhoff's
we get
4
AT-y- Zt- yr=0 2xr+ xr=2x R Or 14r
I-2y-z=0 =2R
or
Hence R=14 rl.
or r-2y-y=0
Example 157. Twelvei wires Find
each having a
14 5
areconnected to form a cube. the resistance of resistancetheof 19
14
between two corners 0f a atagondt of one jace of a cub cube
AB. Then
Let R be the resistance across Solution. Imagine a battery connected
P.D. across AB=r
12
points A and C.
'so that a current of 1 A
A. This current is
divided equally along entersAB jbetunctweenion
or +R=
10
or R=r
The distribution of Current in various
and AD.
shown in Fig. 3.168. These currents finally add so is
that a
branches
Hence R=r. 12 current of 1 A flows out of junction C.
Example 156. Eleven equal wires each of resistance r form 1-2x- z H

the edges of an incomplete cube. Find the total resistance 1-2r

from one end of the vacant edge of the cube to the other. 1Ampere
Solution. Let Aand Bbe the vacant edges of the B
cube. Let an emfE applied across AB send a current 2x 1-2% -2+y
in the circuit. Since the paths AD and AE are symme ytz X-y
trical, the current 2x at A divided into two equal G
parts x and x. At other points, the current is divided as
shown in Fig. 3.167, so that again the currents combine X-y 1-2x - 2y
at Bto give current 2:x. Let Rbe the total resistance of D
the cube between A and B.
1Ampere
2(x -y)
I-y
Fig. 3.168
E
Ax-y |x-y Applying Kirchhoff 's second law to the loop
D.
AEFDA, we get
-(1-2x) -z+y+X=0 ..1)
23
A
Similarly, from the loop BHGCB, we have
2r
-y-(1-2*-z+y)-(1-2x +2y) +(x-y) =0
Fig. 3.167 ..2)
Again, from the loop FGCDE, we
Applying have
ABCDA, we getKirchhoff second law to the loop, -(y
's
On solving +z)-(1-2x+2y)
+(x-y)-y =0 ..3)
XT+ yr +xr-¬
From Ohm's law,
equations
3
(1), (2) arnd (3), we get
1
E=2x. R X*A y=0, z==A
8
2xr+ yr =2xR Now Vc =VAB t
Applying VBC
EFGHE, we get Kirchhoffs second law to the 3 6
yr loop =1x Rly

or
-(*-y)r-2(*
y -y)r -(*-y)r=0 8 8
Equivalent resistance between Aand
4

-x+y-2I +2y-x +y=0 RYAc.

CI
3|41 3 2.
4