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Step-by-Step Solution
Step 1: Identify the given circuit parameters
• The galvanometer has resistance $50\,\Omega$.
• An external resistance of $2950\,\Omega$ is connected in series with the galvanometer.
• The battery voltage is $3\,\text{V}$.
• This arrangement produces a full-scale deflection of $30$ divisions in the galvanometer.
Step 2: Calculate the initial total series resistance and current
Total resistance in series initially, $R_{\text{initial}} = R_G + R_1 = 50\,\Omega + 2950\,\Omega = 3000\,\Omega.$
The current through the circuit in this initial configuration is
$ I_{\text{initial}} = \dfrac{3\,\text{V}}{3000\,\Omega} = 1 \times 10^{-3}\,\text{A} = 1\,\text{mA}.$
Since $1\,\text{mA}$ produces $30$ divisions, one division corresponds to $ \dfrac{1\,\text{mA}}{30} = \dfrac{1}{30} \,\text{mA}.$
Step 3: Relate the desired deflection (20 divisions) to the required current
For $20$ divisions, the current must be
$ I_{\text{desired}} = 20 \times \dfrac{1}{30} \,\text{mA} = \dfrac{2}{3}\,\text{mA}.$
Step 4: Find the new total resistance to achieve the reduced current
Using Ohm’s law with the same battery ($3\,\text{V}$), the required total resistance $R_{\text{total,new}}$ in series to give $\dfrac{2}{3}\,\text{mA}$ is
$ 3\,\text{V} = R_{\text{total,new}} \times \dfrac{2}{3}\,\text{mA}.$
So,
$ R_{\text{total,new}} = \dfrac{3\,\text{V}}{\frac{2}{3}\,\text{mA}} = 3 \times \dfrac{3}{2} \times \dfrac{1}{\text{mA}} = 4500\,\Omega.$
Step 5: Subtract the galvanometer’s own resistance to find the additional series resistance
We must remember that this total $4500\,\Omega$ includes the galvanometer’s own resistance $(50\,\Omega)$, so the added series resistance $R_{\text{added}}$ should be
$ R_{\text{added}} = 4500\,\Omega - 50\,\Omega = 4450\,\Omega.$
Step 6: Conclude the required resistance
Therefore, to reduce the deflection from $30$ divisions to $20$ divisions, the resistance needed in series is $4450\,\Omega.$
