Increasing the current sensitivity may not necessarily increase the voltage sensitivity of galvanometer. Justify.

Solution

Let the deflection produced in applying voltage V is

φ

then,

voltage sensitivity = \frac{φ}{v} = \frac{NBA}{kR}

The voltage sensitivity may be increased by
(i) increasing N, B, A and,
(ii) decreasing k.

Current sensitivity is deflection per unit current whic is , given by

\frac{φ}{I}

current sensitivity = \frac{NBA}{k}

It can be increased by
(i) increasing NBA
(ii) decreasing k.

There is a factor of resistance R in voltage sensitivity.
Hence, increasing the current sensitivity may not necessarily increase the voltage sensitivity of a galvanometer.

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

Increasing the current sensitivity may not necessarily increase the voltage sensitivity of galvanometer. Justify.

Some More Questions From Moving Charges And Magnetism Chapter

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