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Chapter 10 - Electrical Circuits

Kirchoff's Law

1. Kirchoff's 1st law/ Kirchoff's current law/ Kirchoff's junction law:

The algebric sum of current at a junction is always zero.
i.e. ΣI=0

Sign convention
If the current is towards the junction point, then it is taken as +ve and if the current is away from the junction point, then it is taken as -ve.

In figure at junction A,
Applying kirchoff's 1st law
I 1+(-I2)+I3+(-I4) = 0
I 1-I2+I3-I4=0
I 1+I3=I2+I4

This means the sum of total current incoming to junction is equal to the sum of total current moving away from the junction.
This law is accordance with the conservation of charge.

2. Kirchoff's 2nd law/ Kirchoff's voltage law/ Kirchoff's loop law:

The algebric sum of EMF's and P.D.'s at a loop is always zero.
ΣE + ΣIR = 0
This law follows the conservation of energy.

Sign convention
If the direction of taken loop touches the +ve terminal of the cell then it's emf is taken as +ve otherwise -ve. If the direction of current is in the direction of taken loop, then the current is taken as +ve otherwise -ve.

In the figure: taking the loop FCBAF
ΣE + ΣIR = 0
[E 2 + (-E1)] + [-I2R2 + I 1R1]=0
E 2-E1= I2R2 - I 1R1
E 1 - E2= I1R1 - I 2R2

Wheatstone Bridge Circuit

It is the combination of four resistors in which three of them is known and one is unknown. The combination of resistor are as shown in figure. In the figure P,Q and R are known resistance and X is unknown.
At the balance condition, that is for null deflection of galvanometer, we can write,
P/Q = X/R
This condition is called wheatstone bridge condition.

Wheatstone Bridge Condition:

The wheatstone bridge circuit consist of four resistance P,Q,X and R as shown in figure. Let I 1,I1,I1,I1 are the current through resistor P,X,Q,R respectively. A galvanometer of resistance Rg is connected between the terminal B and D through which a current I g flow.

Applying the kirchoff's current law at a junction point B and D. The value of current through a resistance Q and R is:
I 3= I1- Ig
I 4= I2 + Ig
and taking the loop ABDA and applying kirchoff's voltage law:
ΣE + ΣIR = 0
or, 0 + I1 x P + Ig x Rg - I 2 x X = 0
at a balance condition (null deflection condition)
I g = 0, we get,
I 1P - I2 x X = 0
I 1P = I2X <-------- (i)
Taking the loop BCDB and applying kirchoff's voltage law,
ΣE + ΣIR = 0
0 + I3 x Q - I4 x R - I x Rg = 0
at balance condition (I g=0) we get,
I 3 x Q = I4 x R
I 1Q = I2R <------ (ii)
dividing equation (i) and equation (ii) we get,
P/Q = X/R
which is the required wheatstone bridge condition.

It is a device which is used to measure the emf of the cell. Compare the EMF of the two cell and determine the internal resistance of a cell.
It consist of 10 m long uniform cross section wire made up of constant or Eureka which is divided to equal segment. Each segment is 1 m long is placed over a wooden board by the help of copper stripes. A meter scale is placed in the lower copper's part of the wooden board provided two way scale.

Principle of Potentiometer:

On sliding the jockey over a potentiometer wire AB a null point is obtained at a point 'C' of length 'lac' .
from the ohm's law,
V= IRac <-------- (i)
I= current through the potentiometer wire.
R ac= (ρ x lac)/A
where, ρ= resistivity of the material of the potentiometer wire.
A= cross sectional area.
from equation (i) and (ii),
V= I. (ρ x l ac)/A
V= k. lac
where, k= Iρ/A = constant
V ∝ l ac
V ∝ l

When a constant current is passed through the potentiometer then the potential dropped across any portion is directly proportional to the length of that portion.

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