(a) State the working principle of a potentiometer. With the help of the circuit diagram; explain how a potentiometer is used to compare the emf’s of two primary cells. Obtain the required expression used for comparing the emf’s.

(b) Write two possible causes for one sided deflection in a potentiometer experiment.

OR

(a) State Kirchhoff’s rules for an electric network. Using Kirchhoff’s rules, obtain the balance condition in terms of the resistances of four arms of Wheatstone bridge.

(b) In the meter bridge experimental set up, shown in the figure, the null point ’D’ is obtained at a distance of 40 cm from end A of the meter bridge wire. If a resistance of 10W is connected in series with R₁, null point is obtained at AD=60 cm. Calculated the values of R₁ and R₂. 

(a) Working Principle of Potentiometer

Principle:
Consider a long resistance wire AB of uniform cross-section. It’s one end A is connected to the positive terminal of battery B₁ whose negative terminal is connected to the other end B of the wire through key K and a rheostat (Rh).
The battery B₁ connected in circuit is called the driver battery and this circuit is called the primary circuit. By the help of this circuit a definite potential difference is applied across the wire AB; the potential falls continuously along the wire from A to B. The fall of potential per unit length of wire is called the potential gradient. It is denoted by ‘k’.
A cell e is connected such that it’s positive terminal is connected to end A and the negative terminal to a jockey J through the galvanometer G. This circuit is called the secondary circuit. In primary circuit the rheostat (Rh) is so adjusted that the deflection in galvanometer is on one side when jockey is touched on wire at point A and on the other side when jockey is touched on wire at point B.
The jockey is moved and touched to the potentiometer wire and the position is found where galvanometer gives no deflection. Such a point P is called null deflection point. V_AB is the potential difference between points A and B and L meter be the length of wire, then the potential gradient is given by, 
                                                 

If the length of wire AP in the null deflection position be l, then the potential difference

between points A and P,

                                                      V_{AP =} kl

Therefore, Emf of the cell  = V_{AP= kl}

In this way the emf of a cell may be determined by a potentiometer.

Comparison of two emfs’ of a cell:

First of all the ends of potentiometer are connected to a battery B₁, key K and rheostat Rh such that the positive terminal of battery B₁ is connected to end A of the wire. This completes the primary circuit.

Now the positive terminals of the cells C1 and C2 whose emfs’ are to be compared are connected to A and the negative terminals to the jockey J through a two-way key and a galvanometer (fig). This is the secondary circuit.

Distance if P₁ from A is l_1.

So, AP_{1 =} l₁

Emf of cell C₁ is given by, ₁ = kl₁                                                    … (1)

Now, plug is taken from terminals 1 and 3 and inserted between the terminals 2 and 3 to bring cell C₂ in the circuit. Null deflection point is obtained at P₂.

Distance from P₂ to A is l₂.

Emf of cell C₂, ₂ = kl₂                                                                        … (2)

Now, dividing equation (1) by (2), we get
                                                  
Out of these cells if one is standard cell. 

OR

Kirchhoff’s rule states that,

(i) At any junction, the sum of the currents entering the junction is equal to the sum of the currents leaving the junction.

(ii) The algebraic sum of the charges in potential around any closed loop involving resistors and cells in the loop is zero.

Conditions of balance of a Wheatstone bridge:

P, Q, R and S are four resistance forming a closed bridge, called Wheatstone bridge.
                                          

A battery is connected across A and C, while a galvanometer is connected B and D. Current is absent in the galvanometer’s balance point.

Derivation of Formula: Let the current given by battery in the balanced position be I. This current on reaching point A is divided into two parts I₁ and I₂. At the balanced point, current is zero.

Applying Kirchhoff’s I law at point A,

I - I₁ - I ₂ = 0 or

 I = I₁ + I ₂                                                              ...(i)

Applying Kirchhoff’s II law to closed mesh ABDA,

- I₁P + I ₂R = 0 or

  I₁P = I ₂ R                                                            ...(ii)

Applying Kirchhoff’s II law to mesh BCDB,

- I₁Q + I ₂S = 0 or

 I₁Q = I ₂S                                                              ...(iii)

Dviding equation (ii) by (iii), we get, 

                                          

 ; which is the required condition of balance for Wheatstone bridge.
b)