Measuring Devices ( Galvanometer, Voltmeter and Ammeter & Meter Bridge ): Practice Problem & Solution
A potentiometer wire has length $4\text{ m}$ and resistance $8\text{ }\Omega$. The resistance that must be connected in series with the wire and an accumulator of e.m.f. $2\text{ V}$, so as to get a potential gradient $1\text{ mV}$ per $\text{cm}$ on the wire is: (2015)
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:
Potential drop across the wire $V_w = k \cdot L = 10^{-3}\text{ V/cm} \cdot 400\text{ cm} = 0.4\text{ V}$.
Current required $I = \frac{V_w}{R_w} = \frac{0.4}{8} = 0.05\text{ A}$.
Total resistance of circuit $R_{\text{total}} = \frac{E}{I} = \frac{2}{0.05} = 40\text{ }\Omega$, so series resistance $R = 40 - 8 = 32\text{ }\Omega$.
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