MP Board · Class 12 · Physics · Current ElectricityExplain the working principle of a potentiometer. How is it used to compare the EMFs of two primary cells? Draw the necessary circuit diagram and derive the working formula.

Working Principle of a Potentiometer

\nA potentiometer is a versatile instrument used to measure the potential difference, EMF of a cell, and internal resistance accurately. It does not draw any current from the circuit being measured when balanced, effectively acting as an ideal voltmeter with infinite resistance.

• Principle: The fundamental principle of a potentiometer is that the potential drop across any portion of the potentiometer wire is directly proportional to the length of that portion, provided the wire has a uniform cross-sectional area and a constant current flows through it.
• Mathematical Expression: If $V$ is the potential difference across a length $l$ of the wire, then: $$V \propto l \implies V = Kl$$ where $K$ is the potential gradient (potential drop per unit length of the wire, $K = \frac{V_{\text{total}}}{L}$). When no current is drawn from the cell at the balance point, the EMF of the cell equals the potential drop across the balancing length.

Comparison of EMFs of Two Primary Cells

Circuit Description and Connections:

1. A uniform wire $AB$ of length $L$ is connected in series with a driver cell (accumulator) of constant EMF $E$ and a rheostat ($Rh$) through a plug key ($K$). This forms the primary circuit.
2. The positive terminals of both primary cells whose EMFs ($E_1$ and $E_2$) are to be compared are connected to the high-potential terminal $A$ of the potentiometer wire.
3. The negative terminals of the cells $E_1$ and $E_2$ are connected to a two-way key. The common terminal of the two-way key is connected through a galvanometer ($G$) and a jockey ($J$) to slide along the wire $AB$. This forms the secondary circuit.

Step-by-Step Derivation:

1. Close the key $K$ in the primary circuit so that a steady current flows through wire $AB$.
2. Connect cell $E_1$ into the circuit by inserting a plug between the terminals connected to $E_1$ in the two-way key.
3. Slide the jockey along wire $AB$ until the galvanometer shows a null deflection (zero reading). Let this balancing length be $l_1$.
4. According to the potentiometer principle, the EMF $E_1$ of the first cell is proportional to its balancing length $l_1$: $$E_1 = Kl_1 \quad \text{--- (Equation 1)}$$
5. Now, disconnect cell $E_1$ and connect cell $E_2$ into the circuit by operating the two-way key.
6. Find the new balancing length $l_2$ for cell $E_2$ where the galvanometer again shows null deflection.
7. The EMF $E_2$ of the second cell is proportional to its balancing length $l_2$: $$E_2 = Kl_2 \quad \text{--- (Equation 2)}$$
8. Dividing Equation 1 by Equation 2: $$\frac{E_1}{E_2} = \frac{Kl_1}{Kl_2}$$ $$\frac{E_1}{E_2} = \frac{l_1}{l_2}$$ \nThus, by measuring the balancing lengths $l_1$ and $l_2$, the ratio of the EMFs of the two cells can be accurately compared without drawing any current from them.