CBSE · Class 12 · Physics · Moving Charges and MagnetismA galvanometer has a resistance of $15\;\Omega$ and the meter gives full-scale deflection for a current of $4\;\text{mA}$. (a) How can you convert this galvanometer into a voltmeter of range $0$ to $6\;\text{V}$? (b) How can you convert it into an ammeter of range $0$ to $6\;\text{A}$?

Given Data:

• Resistance of the galvanometer ($G$) = $15;\Omega$
• Current for full-scale deflection ($I_g$) = $4;\text{mA} = 4 \times 10^{-3};\text{A}$

Part (a): Conversion into a Voltmeter of range $0$ to $6;\text{V}$

\nTo convert a galvanometer into a voltmeter, a high resistance (multiplier) $R$ must be connected in series with the galvanometer.

• Desired maximum voltage range ($V$) = $6;\text{V}$

Formula: $$V = I_g (G + R)$$

Step-by-step Calculation:

1. Substitute the given values into the formula: $$6 = (4 \times 10^{-3}) (15 + R)$$
2. Rearrange to solve for $(15 + R)$: $$15 + R = \frac{6}{4 \times 10^{-3}}$$ $$15 + R = \frac{6000}{4} = 1500;\Omega$$
3. Calculate the resistance $R$: $$R = 1500 - 15 = 1485;\Omega$$

Conclusion for (a): A high resistance of $1485;\Omega$ must be connected in series with the galvanometer.

Part (b): Conversion into an Ammeter of range $0$ to $6;\text{A}$

\nTo convert a galvanometer into an ammeter, a small resistance (shunt) $S$ must be connected in parallel with the galvanometer.

• Desired maximum current range ($I$) = $6;\text{A}$

Formula: $$S = \frac{I_g \cdot G}{I - I_g}$$

Step-by-step Calculation:

1. Substitute the given values into the numerator and denominator: $$\text{Numerator} = (4 \times 10^{-3}) \times 15 = 60 \times 10^{-3} = 0.06;\text{V}$$ $$\text{Denominator} = 6 - (4 \times 10^{-3}) = 6 - 0.004 = 5.996;\text{A}$|
2. Calculate the shunt resistance $S$: $$S = \frac{0.06}{5.996} \approx 0.010;\Omega$$

Conclusion for (b): A low resistance (shunt) of approximately $0.010;\Omega$ must be connected in parallel with the galvanometer.