A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?

Solution

Given, a galvanometer coil.
Resistance on the galvanometer, R = 12 Ω
Current on galvanometer, I_g =3 mA = 3 x 10^–3A
Voltage ,V = 18 V

By using formula,

   $V = I_{g} (R + R_{g})$

$\frac{V}{I_{g}} = R + R_{g}$
or,
   $R = \frac{V}{I_{g}} - R_{g} = \frac{18}{3 \times 10^{- 3}} - 12 R = 6 \times 10^{3} - 12 = 5988 Ω$

Some More Questions From Moving Charges And Magnetism Chapter

Two moving coil meters, M₁ and M₂ have the following particulars:
R₁ = 10 Ω N₁ = 30,
A₁ = 3.6 x 10^–3 m², B₁ = 0.25 T
R₂ = 14 Ω; N₂ = 42,
A₂ = 1.8 x 10^–3 m², B₂ = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M_1.

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Moving Charges And Magnetism

A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?

Some More Questions From Moving Charges And Magnetism Chapter

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