If an ammeter $A$ reads $2$ A and the voltmeter $V$ reads $20$ V, what is the value of resistance $R$? (Assuming finite resistances of ammeter and voltmeter)
            

A galvanometer has a resistance of 25 ohm and a maximum of 0.01 A current can be passed through it. In order to change it into an ammeter of range 10 A, the shunt resistance required is 

1. 5/999 ohm

2. 10/999 ohm

3. 20/999 ohm

4. 25/999 ohm

A galvanometer has 30 divisions and a sensitivity 16$\mu A/\text{div.}$ It can be converted into a voltmeter to read 3 V by connecting (approximately):

1. Resistance nearly 6 k Ω in series

2. 6 k Ω in parallel

3. 500 Ω in series

4. It cannot be converted

A voltmeter has a range 0-V with a series resistance R. With a series resistance 2R, the range is 0-V'. The correct relation between V and V' is :

1. $V^{\text{'}}=2V$

2. $V^{\text{'}}>2V$

3. $V^{\text{'}}>>2V$

4. $V\text{'}<2V$

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

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 $90~\Omega$ is to be reduced by $90\%$. What value of resistance should be connected in parallel with it?
1. $9~\Omega$
2. $90~\Omega$
3. $1000~\Omega$
4. $10~\Omega$