A galvanometer coil has a resistance of 15 Ω and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?

Solution

Given, a galvanometer coil.
Resistance on galvanometer, G = 15 Ω
Current pasing through the galvanometer,I_g = 4 mA = 4 x 10^–3 A
Current across the ammeter, I = 6 A

Using formula,

S = \frac{I_{g} . G}{I - I_{g}}

Putting values,

S = \frac{4 \times 10^{- 3} \times 15}{6 - 0.004} = \frac{60 \times 10^{- 3}}{5.996} = 10 \times 10^{- 3} Ω

S = 10 m Ω

Therefore, a resistance of 10mΩ should be connected so as to convert the galvanometer into an ammeter.

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.

Answer the following questions:
A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?

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Moving Charges And Magnetism

A galvanometer coil has a resistance of 15 Ω and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?

Some More Questions From Moving Charges And Magnetism Chapter

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