class xii physics chapter-3 current electricity module - 4...
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Atomic Energy Education Society, Mumbai
CLASS XII PHYSICS Chapter-3
CURRENT ELECTRICITY Module - 4
By Girish Kumar
PGT (Physics) AECS Narora
1) Kirchhoff’s junction rule
2) Kirchhoff’s loop rule
3) Wheatstone bridge
4) Metrebridge
5) Determination of resistivity by metrebridge
6) Potentiometer
7) Sensitivity of potentiometer
8) Difference between voltmeter and potentiometer
9) Application of potentiometer
Current Electricity
Topics covered in Module - 4
KIRCHHOFF’S LAWS:
I Law or Current Law or Junction Rule:
The algebraic sum of electric currents at a junction in any electrical
network is always zero.
Sign Conventions:
1. The incoming currents towards the junction are taken positive.
2. The outgoing currents away from the junction are taken negative.
Note: The charges cannot accumulate at a junction.The number of
charges that arrive at a junction in a given time must leave in the same
time in accordance with conservation of charges.
I1 - I2 - I3 + I4 - I5 = 0
II Law or Voltage Law or Loop Rule:
The algebraic sum of all the potential drops and emf’s along any closed
path in an electrical network is always zero.
Loop ABCA:
- E1 + I1.R1 + (I1 + I2).R2 = 0
Loop ACDA:
- (I1 + I2).R2 - I2.R3 + E2 = 0
Sign Conventions:
1. The emf is taken negative when we traverse from positive to negative
terminal of the cell through the electrolyte.
2. The emf is taken positive when we traverse from negative to positive
terminal of the cell through the electrolyte.
The potential falls along the direction of current in a current path
and it rises along the direction opposite to the current path.
3. The potential fall is taken negative.
4. The potential rise is taken positive.
Note: The path can be traversed
in clockwise or anticlockwise
direction of the loop.
Wheatstone Bridge: Currents through the arms are assumed by
applying Kirchhoff’s Junction Rule.
Applying Kirchhoff’s Loop Rule for:
Loop ABDA:
-I1.P - Ig.G + (I - I1).R = 0
Loop BCDB:
- (I1 - Ig).Q + (I - I1 + Ig).S + Ig.G = 0
When Ig = 0, the bridge is said to balanced.
By manipulating the above equations, we get P
Q
R
S
Metre Bridge:
When the galvanometer
current is made zero by
adjusting the jockey
position on the metre-
bridge wire for the given
values of known and
unknown resistances,
Metre Bridge is based
on the principle of
Wheatstone Bridge.
(Since,
Resistance α
length)
Therefore, X = R (100 – l) ⁄ l
R R R RAJ
X X X
AJ l
RJB 100 - l JB
Determination of resistivity If r is radius of the wire and l’ it's length, then resistivity of it's
material will be
ρ = X πr2/l'
Important points :
1) If the connections of galvanometer and cell are replaced the null point will be at the same positions.
2) When the resistance taken out from the resistance box have equal value to X the null point is obtained at mid-point of wire AC.
3) The wire AC is made up of an alloy nichrome or constantan so as to provide a sufficient resistance in it's small length.
4) In order to avoid the error in the reading dur to neglected resistance of copper strip the balance point should be preferably obtained at point of wire AC.
Potentiometer:
Principle:
V = I R
= I ρl/A
If the constant current flows
through the potentiometer wire
of uniform cross sectional area
(A) and uniform composition
of material (ρ), then
V = Kl or V α l
V /l is a constant.
The potential difference across any length of a
wire of uniform cross-section and uniform
composition is proportional to its length when a
constant current flows through it.
Sensitivity of a potentiometer: A potentiometer is sensitive if
1) It is capable of measuring very small potential differences 2) It shows a significant change in balancing length for a small change in potential
difference being measured. Sensitivity of a potentiometer is measure of the reciprocal of the potential gradient k along the length of it's wire Sensitivity = 1/k Sensitivity of a potentiometer can be increased by decreasing the potential gradient k along the length of wire, which can be achieved by
i) Increasing the resistance of series resistor like that of rheostat
ii) Decreasing the applied voltage of the battery connected across the potentiometer wire, provided that potential difference across the wire remains more than emf of cell to be measured
iii) For a given potential difference, the sensitivity can be increased by increasing the length of the potentiometer wire.
iv) For a potentiometer wire of fixed length, the potential gradient can be decreased by reducing the current in the circuit with the help of rheostat.
Difference between voltmeter and potentiometer
Voltmeter Potentiometer
Its resistance is high but finite. Its resistance is infinite.
It draws some current from source of emf. It does not draw any current from the source of unknown emf.
The potential difference measured by it is lesser than the actual potential difference.
The potential difference measured by it is equal to actual potential difference.
Its sensitivity is low. Its sensitivity is high.
It is based on deflection method. It is based on non-deflection method.
1. Comparison of emf’s using
Potentiometer:
The balance point is obtained
for the cell when the potential at
a point on the potentiometer
wire is equal and opposite to
the emf of the cell.
E1 = VAJ1 = I ρl1 /A
E2 = VAJ2 = I ρl2 /A
Note:
The balance point will not be obtained on the potentiometer wire if the fall
of potential along the potentiometer wire is less than the emf of the cell to
be measured.
The working of the potentiometer is based on null deflection method. So
the resistance of the wire becomes infinite. Thus potentiometer can be
regarded as an ideal voltmeter.
E1 / E2 = l1 /l2
Applications of Potentiometer
2) Internal resistance of a primary cell by potentiometer
When key K2 kept open emf of cell will be, E = KL1
with the help of resistance R and close key K2 The potential difference of cell will be V = KL2 -----2
From eq 1 & 2, E/V= L1/L2 ----- 3
Since E = I(R+r) V = IR Now, eq 3 becomes
r = R[(L1-L2)/L2]
----- 1
3) Let E1 and E2 are connected in series (E1 > E2)
when cells assist each other, let balancing length is L1,
(E1+E2 = KL1)
when cells oppose each other, let balancing length is L2,
(E1-E2 – KL2)
4) Comparison of resistance by potentiometer
Let the balancing length for resistance R1
(YX connected ) is L1, Let the balancing length for resistance R1+R2 (when YZ connected ) be L2,
IR1= KL1
and i(R1 + R2) = KL2
R2/R1= (L2-L1)/L1
E1+E2/E1-E2 = L1/L2
E1/E2 = L1+L2/L1-L2 Or