Question

What is potentiometer? Explain its principle. Describe the method of determining a small resistance by potentiometer. Make necessary diagram also.

Answer 1

Principle of Potentiometer:

Potentiometer: Potentiometer is a apparatus with help of which the e.m.f. of a cell and potential difference can be measured accurately.

Principle: Suppose AB is a resistance of uniform area of cross section and length LAB- If a standard cell of emf E and negligible internal resistance is connected across the ends of wire A and B with the help of connection wires of negligible resistance, then the potential difference across the wire will be E. Therefore potential gradient.
k = \(\frac{E}{L_{A B}}\) ……………. (1)
If we consider a point C at a distance l from A, then
V_AC = kl ………….. (2)
Now if a cell of unknown emf E’ is connected between A and C as shown in the figure, then the deflection in galvanometer will depend on the fact that whether V_AC > E’ or E’ > V_AC. When V_AC = E’, then galvanometer will show no deflection. This fact can be understood well with the help of following equivalent circuit between A and C. This is null position galvanometer.

E’ = V_AC
or E’ = kl ………………… (3)
This is the principle of potentiometer.
If E’ > E, then galvanometer will show the deflection in one direction only. In this situation the measurement of E’ is not possible.
If the resistance of potentiometer wire is RAB and the current is I then from the ohm’s law
E = IR_AB ……………… (4)
From the equation (1)
k = \(\frac{I R_{A B}}{L_{A B}}\)
Let \(\frac{R_{A B}}{L_{A B}}\) = Density of wire is represented by the ρ
∴ k = ρI ……………… (5)
The circuit diagram for potential gradient

The potential gradient

Value of potential gradient under different conditions:

Case 1: When only the battery of negligible internal resistance is connected to potentiometer.
∴ r = 0 and R_h = 0
From the equation (8)

Case 2: When the battery of internal resistance (r) is connected to the potentiometer

Case 3: When a some other resistance or rheostat of resistance (Rh) is connected in series with the battery in the circuit of potentiometer. Then the potential gradient is calculated by equation (8).
k \(=\left(\frac{E}{R_{A B}+r+R_{h}}\right) \times \frac{R_{A B}}{L_{A B}}\)

Factors effecting the value of potential gradient:
From the equation (5)

Measurement of Small Resistance:

In figure necessary circuit is shown.
Primary circuit is as usual. In secondary circuit, unknown small resistance r, known resistance R, rheostat Rh, key K₂ and emf ε’ all are connected in series. Higher potential end of R, connects at point A which has higher potential, while lower potential end of R and r connects with the ends 1 and 3 of two way key. Terminal 2 of two way key connects with galvanometer end. Galvanometer also connects with jockey.

Working: Key K₁ is closed, key K₂ is also closed, now plug is inserted between terminal 1 and 2

of two way key. Now find out potential difference across resistance R by potentiometer. If I current flowing through secondary circuit, potential difference V across the ends of resistance R and the balancing length is l₁ for V.
By the principle of potentiometer
V = kl₁ ………….. (1)
but V = IR (by Ohm’s law)
IR = kl₁ ………… (2)
Eliminate the plug between 1 and 2 and insert it between 2 and 3. Now R and r connect in series. Current I remains same in this circuit also now the potential difference across the ends of the (R + r) is V₁ and balancing length for it is l₂.

Putting the values l₁, l₂ and R, then internal resistance of primary cell (r) can be determined.