The resistance of an ideal voltmeter is 

(1) Zero

(2) Very low

(3) Very large

(4) Infinite

The net resistance of a voltmeter should be large to ensure that :

A galvanometer having a resistance of 8 ohms is shunted by a wire of resistance 2 ohms. If the total current is 1 amp, the part of it passing through the shunt will be :

(1) 0.25 amp

(2) 0.8 amp

(3) 0.2 amp

(4) 0.5 amp

A voltmeter of resistance 1000 Ω gives full-scale deflection when a current of 100 mA flow through it. The shunt resistance required across it to enable it to be used as an ammeter reading 1 A at full-scale deflection is :

(1) 10000 Ω

(2) 9000 Ω

(3) 222 Ω

(4) 111 Ω

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 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. $\frac{1}{10}$