A galvanometer of resistance 50 Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. To reduce this

Question

A galvanometer of resistance 50 Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. To reduce this deflection to 20 divisions, the resistance in series should be

TestHub · Accepted Answer

Initial current in the circuit is given by:

$I_{1} = \frac{V}{R _{1} + G} = \frac{3}{2950 + 50} = \frac{3}{3000} = 1 0^{- 3} A$

Since the deflection in the galvanometer is directly proportional to the current ( $I \propto θ$ ), we have:

$\frac{I _{1}}{I _{2}} = \frac{θ _{1}}{θ _{2}}$

$\frac{1 0 ^{- 3}}{I _{2}} = \frac{30}{20} = \frac{3}{2}$

$I_{2} = \frac{2}{3} \times 1 0^{- 3} A$

Let the new series resistance be $R_{2}$ . The new current is:

$I_{2} = \frac{V}{R _{2} + G}$

$\frac{2}{3} \times 1 0^{- 3} = \frac{3}{R _{2} + 50}$

$R_{2} + 50 = \frac{9}{2 \times 1 0 ^{- 3}} = 4500$

$R_{2} = 4500 - 50 = 4450 Ω$

Answer: $4450 Ω$

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