JEE Main · 2005 · 08 May 2005 · Offline · Current Electricity

WhatsApp Email

A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 volt, the resistance in ohms needed to be connected in series with the coil will be

(A) $1 0^{3}$
(B) $1 0^{5}$
(C)99995
(D)9995✓ Correct

Answer: (D)

Solution

The full-scale current is

I_{g} = \frac{150}{10} = 15

mA and the full-scale voltage is

V_{g} = \frac{150}{2} = 75

mV, so the galvanometer's own resistance is

G = \frac{V _{g}}{I _{g}} = \frac{0.075}{0.015} = 5 Ω

. For each of the

150

divisions to read

1

V, the meter must reach

150

V at its full-scale current, needing a total resistance

\frac{150}{0.015} = 10000 Ω

. The resistance to add in series is

10000 - 5 = 9995 Ω

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

Practise more: all JEE Main PYQs on Current Electricity →

JEE Main · 2005 · 08 May 2005 · Offline · Current Electricity

Solution

More JEE Main questions on Current Electricity