CBSE · Class 12 · Physics · Moving Charges and MagnetismA galvanometer has a coil of resistance $12\,\Omega$ and shows full-scale deflection for a current of $3\,\text{mA}$. (i) How will you convert this galvanometer into a voltmeter of range $0 - 18\,\text{V}$? (ii) How will you convert it into an ammeter of range $0 - 6\,\text{A}$? Also calculate the equivalent resistance of the converted instruments in both cases.

Given Data:

• Resistance of galvanometer coil ($G$) = $12,\Omega$
• Full-scale deflection current ($I_g$) = $3,\text{mA} = 3 \times 10^{-3},\text{A}$

Part (i): Conversion into a Voltmeter of range $0 - 18,\text{V}$

• Desired maximum voltage ($V$) = $18,\text{V}$
• To convert a galvanometer into a voltmeter, a high resistance ($R$) must be connected in series with the galvanometer.
• Formula for series resistance: $$R = \frac{V}{I_g} - G$$
• Substituting the given values: $$R = \frac{18}{3 \times 10^{-3}} - 12$$ $$R = 6000 - 12 = 5988,\Omega$$
• Action: A high resistance of $5988,\Omega$ must be connected in series with the galvanometer.
• Equivalent Resistance of Voltmeter ($R_v$): $$R_v = G + R = 12 + 5988 = 6000,\Omega$$ (or $6,\text{k}\Omega$)

Part (ii): Conversion into an Ammeter of range $0 - 6,\text{A}$

• Desired maximum current ($I$) = $6,\text{A}$
• To convert a galvanometer into an ammeter, a small shunt resistance ($S$) must be connected in parallel with the galvanometer.
• Formula for shunt resistance: $$S = \frac{I_g \cdot G}{I - I_g}$$
• Substituting the given values: $$S = \frac{3 \times 10^{-3} \times 12}{6 - 3 \times 10^{-3}}$$ $$S = \frac{0.036}{6 - 0.003} = \frac{0.036}{5.997}$| $$S \approx 0.006,\Omega$$
• Action: A small shunt resistance of approximately $0.006,\Omega$ must be connected in parallel with the galvanometer.
• Equivalent Resistance of Ammeter ($R_a$): $$R_a = \frac{G \cdot S}{G + S} = \frac{12 \times 0.006}{12 + 0.006} = \frac{0.072}{12.006} \approx 0.006,\Omega$$