Measuring Devices ( Galvanometer, Voltmeter and Ammeter & Meter Bridge ): Practice Problem & Solution
A galvanometer of resistance $50 \Omega$ is connected to a battery of $3 \text{ V}$ along with a resistance of $2950 \Omega $ in series. A full scale deflection of $30$ divisions is obtained in the galvanometer. In order to reduce this deflection to $20$ divisions, the resistance in series should be: (2008)
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:
Deflection is inversely proportional to total circuit resistance. Using $frac{\theta_1}{\theta_2} = \frac{R_2 + G}{R_1 + G}$, we get $\frac{30}{20} = \frac{R_2 + 50}{2950 + 50}$, giving $\R_2 = 4450 \Omega$.
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