Talha's Physics Academy
Potentiometer and Comparison of EMFs
Video Lecture
Watch the complete video lecture below to understand the working principle of a potentiometer and how it measures potential difference and compares EMFs.
Potentiometer
A potentiometer typically consists of a uniform resistance wire $AB$ stretched along a meter scale, with lengths commonly ranging from 1 meter up to 5 or 10 meters. The larger the length of the wire, the smaller the potential drop per unit length, leading to a greater accuracy of measurement.
Explanation and Measurement of EMF
To compare the EMF of an unknown cell ($E_x$) with a standard cell of known EMF ($E_s$):
- The positive terminal of the driving battery (EMF $E$) is connected to terminal $A$ of the potentiometer wire.
- The positive terminals of both the unknown cell ($E_x$) and the standard cell ($E_s$) are also connected to terminal $A$.
- Their negative terminals are connected through a two-way key and a sensitive galvanometer to a sliding jockey.
Using the two-way key, the unknown cell $E_x$ is first introduced into the galvanometer circuit, and the sliding jockey is adjusted to find the balance point $C_1$ corresponding to balancing length $l_x$. At this balance point, no current flows through the galvanometer, and we have:
$$E_x = I r l_x \quad \text{--- (i)}$$
where $r$ is the resistance per unit length of the wire and $I$ is the working current.
Next, disconnecting $E_x$, the standard cell $E_s$ is introduced into the circuit, and its new balance length $l_s$ is determined:
$$E_s = I r l_s \quad \text{--- (ii)}$$
Comparing the Two EMFs
Dividing equation (i) by equation (ii):
$$\frac{E_x}{E_s} = \frac{I r l_x}{I r l_s}$$
$$\frac{E_x}{E_s} = \frac{l_x}{l_s}$$
$$E_x = E_s \left(\frac{l_x}{l_s}\right)$$
Conclusion: This equation gives the ratio of the two EMFs in terms of the ratio of their respective balancing lengths. If the standard cell EMF ($E_s$) is known, the unknown EMF ($E_x$) can be easily computed.
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