A voltmeter has resistance of 2000 ohms and it can measure upto 2V. If we want to increase its range to 10 V, then the required resistance in series will be 

1. 2000 Ω

2. 4000 Ω

3. 6000 Ω

4. 8000 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)
            

In the given figure, battery E is balanced on 55 cm length of potentiometer wire but when a resistance of 10 Ω is connected in parallel with the battery then it balances on 50 cm length of the potentiometer wire then internal resistance r of the battery is 

1. 1 Ω

2. 3 Ω

3. 10 Ω

4. 5 Ω

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

In the circuit shown, a meter bridge is in its balanced state. The meter bridge wire has a resistance 0.1 ohm/cm. The value of unknown resistance X and the current drawn from the battery of negligible resistance is 

1. 6 Ω, 5 amp

2. 10 Ω, 0.1 amp

3. 4 Ω, 1.0 amp

4. 12 Ω, 0.5 amp

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. $V\text{'}<2V$

The measurement of the voltmeter in the following circuit is :

1. 2.4 V

2. 3.4 V

3. 4.0 V

4. 6.0 V

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

The potential difference across the 100Ω resistance in the following circuit is measured by a voltmeter of 900 Ω resistance. The percentage error made in reading the potential difference is :

1. $\frac{10}{9}$

2. 0.1

3. 1.0

4. 10.0