Measuring Devices ( Galvanometer, Voltmeter and Ammeter & Meter Bridge ): Practice Problem & Solution
A potentiometer wire is $100\text{ cm}$ long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at $50\text{ cm}$ and $10\text{ cm}$ from the positive end of the wire in the two cases. The ratio of emf's is: (2016 - I)
Solution Explained:
To solve this problem, we apply the core principles of Measuring Devices ( Galvanometer, Voltmeter and Ammeter & Meter Bridge ). Understanding the underlying formula is key to arriving at the correct answer below:
When cells support each other, $E_1 + E_2 = k \cdot 50$. When they oppose each other, $E_1 - E_2 = k \cdot 10$.
Taking the ratio gives $\frac{E_1 + E_2}{E_1 - E_2} = 5$, which simplifies to $6 E_2 = 4 E_1$.
Thus, the ratio of their e.m.f.'s is $\frac{E_1}{E_2} = \frac{3}{2}$.
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