A galvanometer, having a resistance of 50 Ω gives a full scale deflection for a current of 0.05 A. The length in meter of a resistance wire of area of cross-section 2.97× 10^–2 cm² that can be used to convert the galvanometer into an ammeter which can read a maximum of 5 A current is (Specific resistance of the wire = 5 × 10^–7 Ωm) 

(1) 9

(2) 6

(3) 3

(4) 1.5

An ammeter reads up to 1 ampere. Its internal resistance is 0.81 ohm. To increase the range to 10 A the value of the required shunt is :

(1) 0.09 Ω

(2) 0.03 Ω

(3) 0.3 Ω

(4) 0.9 Ω

The current flowing in a coil of resistance 90 Ω is to be reduced by 90%. What value of resistance should be connected in parallel with it 

(1) 9 Ω

(2) 90 Ω

(3) 1000 Ω

(4) 10 Ω

A galvanometer of \(50~\Omega \) resistance has 25 divisions. A current of 4 × 10^–4 A gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 V, it should be connected with a resistance of:

1. 2500 Ω as a shunt

2. 2450 Ω as a shunt

3. 2550 Ω in series

4. 2450 Ω in series

A moving coil galvanometer of resistance 100Ω is used as an ammeter using a resistance 0.1Ω. The maximum deflection current in the galvanometer is 100 mA. Find the minimum current in the circuit so that the ammeter shows maximum deflection :

(1) 100.1 A

(2) 1000.1 mA

(3) 10.01 mA

(4) 1.01 mA

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 

(1) 99995

(2) 9995

(3) 10³

(4) 10⁵

The ammeter has range 1 ampere without shunt. the range can be varied by using different shunt resistances. The graph between shunt resistance and range will have the nature

(1) P

(2) Q

(3) R

(4) S

The electric charge in uniform motion produces :

(1) An electric field only

(2) A magnetic field only

(3) Both electric and magnetic field

(4) Neither electric nor magnetic field

A galvanometer with a resistance of 36 Ω is changed into an ammeter by using a shunt of 4 Ω. The fraction f0 of total current passing through the galvanometer will be:

1. $\frac{1}{40}$

2. $\frac{1}{4}$

3. $\frac{1}{140}$ 

4. $\frac{1}{10}$

If an ammeter is to be used in place of a voltmeter then we must connect with the ammeter a :

(1) Low resistance in parallel

(2) High resistance in parallel

(3) High resistance in series

(4) Low resistance in series