A capacitor is connected to a cell of emf E having some internal resistance r. The potential difference in steady state across the 

(1) Cell is < E

(2) Cell is E

(3) Capacitor is > E

(4) Capacitor is < E

The maximum power drawn out of the cell from a source is given by (where r is internal resistance)  

(1) $E^2/2r$

(2) $E^2/4r$

(3) $E^2/r$

(4) $E^2/3r$

Find out the value of current through 2Ω resistance for the given circuit 

(1) 5 A

(2) 2 A

(3) Zero

(4) 4 A

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) $R=R_2-R_1$

The magnitude of i in ampere unit is 

(1) 0.1

(2) 0.3

(3) 0.6

(4) None of these

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

By ammeter, which of the following can be measured 

(1) Electric potential

(2) Potential difference

(3) Current

(4) Resistance

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.