WHEATSTONE BRIDGE
For a certain adjustment of Q, VBD = 0,
then no current flows through the
galvanometer.
VB = VD or VAB= VAD
I1.P = I2.R
Likewise, VBC = VDC
I1.Q = I2.S
I
  r  IR  0
n
n
I
r  nR
3. Mixed grouping
Number of rows is m and number of cells in
each row is n
Applying Kirchhoff’s law
P R

Q S
Dividing, we get,
I
r  IR  0
m
mn
I=
mR  nr
GROUPING OF IDENTICAL CELLS
1. Series Grouping

r 

n  n
For current through R to be maximum,
mR = nr
r
R=
I
resistance/No. of Line
R
emf of the cell is  and internal
resistance is r. R is the external
resistance, I is the current passing
through the circuit and n is the total
number of cells.
Applying Kirchhoff’s law
  ir +   ir + ........ (to n times ) 
iR = 0
i
n
R  nr
2. Parallel grouping

nr
= one line total internal
m
RC-CIRCUIT
Charging: Let us assume that the
capacitor in the shown network is
uncharged for t < 0. The switch is
connected to position 1 at t = 0.
Now, 'C' is getting charged.
If the charge on capacitor at time 't' is
q.
writing the loop rule,
q
+ IR  E = 0
C

r

r

r


R
dq
q
E
dt
c
dq
 EC  q
dt
dq
1

dt
EC  q RC
RC
Integrating
R
Applying Kirchhoff’s law
dq
1 t

dt
o EC  q
RC o
1
 ln | EC  q |oq 
.t
RC

q

EC  q  t

EC
RC

ln

q  EC 1  e  t / RC


At t = 0, q = 0 and
t =  , q = E C (the maximum charge.)
= qmax
Thus,
t


q  qmax 1  e RC 


dq qmax t / RC E  t / RC
i

e
 e
dt
RC
R
E
i  imax e t / RC where imax 
R
Discharging
Consider the same arrangement as we
had in the previous case with one
difference that the capacitor has
charge qo for t<0 and the switch is
connected to position 2 at t = 0. If the
charge on capacitor is q at any later
moment t then the loop equation is
given as
q
 IR  0
c
dq q
R

dt
c
dq 1

dt
q RC
dq
1
t
o
q
t

qo
RC
q  q0 .e  t / RC
the
range
of
an
Suppose the ammeter gives full
scale deflection when a current Ig
flows through it. Now if we want to
convert the reading of the ammeter
in such a manner that it gives full
scale deflection for a higher current
I in the branch of the circuit, we
connect a small resistance S in
parallel to the coil of the
galvanometer,
which
has
a
resistance G.
The resistance value is so chosen
that out of the total current I only Ig
flows through the coil and the
remaining current flows through S.
As potential difference across S =
potential difference across G.
 IG 

 I  Ig 


 q   RC  dt
ln
Changing
ammeter
S=  g
and t = t, q = q
q0
An Ammeter is an instrument used
for measuring current in electrical
circuits. A galvanometer is a low
resistance instrument. A large
current passing through it may
damage the instrument
(I – Ig)S = IgG
Integrating, at t = 0, q = q0
q
AMMETER
So, effectively this ammeter will
measure current up to I ampere and its
effective resistance
 GS 
.
S  G
= 
In practice G is large as compared to
S. Therefore the effective resistance
of the ammeter equals RA  S, which is
small. An ideal ammeter has zero
resistance.