By ammeter, which of the following can be measured 

1. Electric potential

2. Potential difference

3. Current

4. Resistance

A potentiometer is an ideal device of measuring potential difference because

1. It uses a sensitive galvanometer

2. It does not disturb the potential difference it measures

3. It is an elaborate arrangement

4. It has a long wire hence heat developed is quickly radiated

The magnitude of i in ampere unit is 

1. 0.1

2. 0.3

3. 0.6

4. None of these

In Wheatstone's bridge P = 9 ohm, Q = 11 ohm, R = 4 ohm and S = 6 ohm. How much resistance must be put in parallel to the resistance S to balance the bridge 

1. 24 ohm

2. $\frac{44}{9}$ohm

3. 26.4 ohm

4. 18.7 ohm

The resistance of \(1~\text{A}\) ammeter is \(0.018~\Omega\). To convert it into \(10~\text{A}\) ammeter, the shunt resistance required will be:
1. \(0.18~\Omega\)
2. \(0.0018~\Omega\)
3. \(0.002~\Omega\)
4. \(0.12~\Omega\)

For measurement of potential difference, the potentiometer is preferred in comparison to the voltmeter because:

1. the potentiometer is more sensitive than the voltmeter.

2. the resistance of the potentiometer is less than
    the voltmeter.

3. the potentiometer is cheaper than the voltmeter.

4. the potentiometer does not take current from the circuit.

Two sources of equal emf are connected to an external resistance R. The internal resistances of the two sources are R₁ and $R_2(R_2>R_1)$. If the potential difference across the source having internal resistance R₂ is zero, then 

1. $R=R_1{R}_2/(R_1+R_2)$

2. $R=R_1{R}_2/(R_2-R_1)$

3. $R=R_2\times (R_1+R_2)/(R_2-R_1)$

4. $R=R_2-R_1$

To draw maximum current from a combination of cells, how should the cells be grouped 

1. Series

2. Parallel

3. Mixed

4. Depends upon the relative values of external and internal resistance