2. CURRENT ELECTRICITY
Electric charge set in motion
through a conductor constitutes
electric current…
A Source of e.m.f connected through a
conductor develops electric field at
every point of conductor
Direction of electric current is
conventionally opposite to
flow of electron
3. HOW DO WE DETERMINE
CURRENTS AND POTENTIAL
DIFFERENCE IN COMPLICATED
CIRCUITS??????
4. KIRCHHOFF’S LAWS : FIRST LAW (CURRENT LAW
OR JUNCTION LAW)
STATEMENT:
The algebraic
sum of electric
current at any
junction is equal
to zero.
S I=0
Current flowing towards
junction is positive .
Current flowing away from
junction is negative.
• In accordance with law of
conservation of charge.
• I1 +I4 –I2 –I3-I5=0
• I1+I4 = I2+I3+I5
5. STATEMENT OF KVL:
STATEMENT:
In a closed loop of electrical
network ,algebraic sum of
potential difference for all
components plus algebraic sum
of all E.M.F.s is equal to zero
S IR+S E=0
Direction of tracing
is same as
conventional
current flow, P.D
across resistance
is -ve otherwise
+ve
E.M.F is positive :we
traverse from –ve
terminal to +ve of
cell.
E.M.F is negative:we
traverse from +ve
terminal to -ve
6. Let us assume current flows
from higher potential A to
lower potential B.
For loop ABEFA:
+I1R1 +R2 (I1-I2)+R6I1 -V1 =0
For loop BCDEB
R3I2 +R4I2 +R5I2+R2 (I1 -I2)=0
KIRCHOFF’S SECOND LAW
IS IN ACCORDANCE WITH
LAW OF CONSERVATION
OF ENERGY.
A B C
DE
F
9. WHEATSTONE’S NETWORK
RES
ISTA
NCE
S
E.M.F AND
GALVANOM
ETER
NULL POINT BALANCED
CONDITION
4
RES
ISTA
NCE
S
R1
R2
R3
R4
E.M.F E IS
CONNECTED
DIAGONALLY
BET Pnts AAND
C through key K.
Sensitive
GALVANOMETE
R G is connected
between vertices
B and D.
‘Along loop
‘ADC’ there
will be pnt D
where
potential
=potential pnt
B along loop
ABC
Network is
balanced if
Vb =V d
If so Ig=0
In
balanced
condition
current flows
in rest ofckt,
Galvanomet
er will not
show any
deflection .It
shows a null
point.
R1/R2=R3/R4
10. PROOF:
I=Current supplied.
Current I is divided into 2
parts.I1 through R1 and I2
through R3
• I=I1 +I2
• When network is balanced
• Vb =V d ,Ig=0,
• Through resistance R2 along
BC is I1 and the current
through resistance R4 is along
DC.
• By applying Kirchoff’s 2nd
law ‘ABGDA’
• -I1R1+0xG+I2R3=0
• I1R1=I2R3
• ..eq1
By applying
Kirchoff’s 2nd law
to closed loop
‘BCDGB’,We get
-I1R2+I2R4+0xG=0
I1R2=I2R4….eq2
Dividing eq2 by eq1
we get
R1/R2=R3/R4…eq3
R1/R3=R2/R4
THIS IS
BALANCING
COND
If position of cell and
galvanometer are
interchanged the
balancing cond is same.
• So branches AC
and BD are called
conjugate arms.
12. X/R =σ lx/σ lr
X/R = lx/lr
X =R(lx/lr)
X=R(lx/100-lx)
13. ERRORS AND MINIMIZATION OF ERRORPOSSIBLEERRORS
1)IF THE WIRE IS NOT
UNIFORM THEN ITS
RESISTANCE WILL NOT
BE PROPORTIONAL TO
LENGTH.SO ERROR IN
VALUE OF UNKNOWN
RESISTANCE.
2)AT ENDS OF WIRE
CONTACT RESISTANCES
ARE DEVELOPED.
3)THE ENDS OF WIRE
MAY NOT COINCIDE
WITH ZERO AND 100 CM
MARK ON METERSCALE.
MINIMIZATION
1)WIRE MUST BE
UNIFORM.
2) VALUE OF R
SHOULD BE CHOSEN
SUCH THAT NULL
POINT IS OBTAINED AS
NEAR CLOSE TO
CENTRE OF WIRE AS
POSSIBLE.
3)EXP SHOULD BE
REPEATED BY
INTERCHANGING
POSITION OF X AND R
TO MINIMIZE ERROR .
14. PROBLEMS:
Two resistances of values 20 Ω and 30 Ω are connected in
left and right gap of a meterbridge.Find the shift in null point
when resistance of 20 Ω is shunted by another resistance of
20 Ω.
Solution=
16. RESISTANCE OF GALVANOMETER BY KELVIN’S METHOD
G/R=σlg/σlr .: G=R( lg/100-lg)
Thus Kelvin’s method is an
EQUAL DEFLECTION
METHOD. D IS CALLED
BALANCE POINT.
Null point is not necessary and deflection
in galvanometer in the unknown
resistance arm itself indicate equipotential
point where bridge is balanced
1)Adjust rheostat to get
suitable deflection in G.
2) Obtain point of contact for
which galvanometer shows
same deflection as before.
