In an exercise to increase current sensitivity of a galvanometer by 25%, its resistance is also increased 1.5 times. How will the voltage sensitivity of the meter be affected ?

Solution

Here given,
Resistance is increased to 1.5 times.
i.e, R' = 1.5 R

Current sensitivity is increased by 25%.

i.e., $I'_{s} = I_{s} + \frac{25}{100} I_{s} = \frac{125}{100} I_{s} = \frac{5}{4} I_{s}$

$Initially, voltage sensitivity, V_{s} = \frac{I_{s}}{R} and, V'_{s} = \frac{I_{s}'}{R^{'}} = \frac{(5 / 4) I_{s}}{1.5 R} = \frac{5}{6} V_{s}$

Therefore,

% increase in voltage sensitivity= $(1 - \frac{V'_{s}}{V_{s}}) \times 100$ = $(1 - \frac{5}{6}) \times 100 = 16.7 %$

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.

For Daily Free Study Material Join wiredfaculty

Moving Charges And Magnetism

In an exercise to increase current sensitivity of a galvanometer by 25%, its resistance is also increased 1.5 times. How will the voltage sensitivity of the meter be affected ?

Some More Questions From Moving Charges And Magnetism Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Location