Measuring Devices ( Galvanometer, Voltmeter and Ammeter & Meter Bridge ): Practice Problem & Solution
The resistances of the four arms P, Q, R and S in a Wheatstone's bridge are $10\text{ ohm}$, $30\text{ ohm}$, $30\text{ ohm}$ and $90\text{ ohm}$, respectively. The e.m.f. and internal resistance of the cell are $7\text{ volt}$ and $5\text{ ohm}$ respectively. If the galvanometer resistance is $50\text{ ohm}$, the current drawn from the cell will be: (2013)
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:
The bridge is balanced because $\frac{P}{Q} = \frac{10}{30} = \frac{1}{3}$ and $\frac{R}{S} = \frac{30}{90} = \frac{1}{3}$. No current flows through the galvanometer.nEquivalent resistance of the bridge $R_{eq} = \frac{(10+30) \times (30+90)}{(10+30) + (30+90)} = \frac{40 \times 120}{160} = 30\Omega$.nTotal resistance $= 30\Omega + 5\Omega = 35\Omega$. Current $I = \frac{V}{R_{total}} = \frac{7}{35} = 0.2\text{ A}$.
Leave a Reply