For the network shown in the figure below, the value of the current i is:

   

1. $\frac{18\mathrm{V}}{5}$

2. $\frac{5\mathrm{V}}{9}$

3. $\frac{9\mathrm{V}}{35}$

4. $\frac{5\mathrm{V}}{18}$

A car battery of emf \(12~\text{V}\) and internal resistance \(5\times 10^{-2}~\Omega\) receives a current of \(60~\text{A}\) from an external source. The terminal voltage of the battery is:

If there are two bulbs of (\(40~\text{W},200~\text{V}\)), and (\(100~\text{W},200~\text{V}\)), then the correct relation for their resistance is:

When three identical bulbs are connected in series, the consumed power is 10 W. If they are now connected in parallel then the consumed power will be:
1.  30 W
2.  90 W
3.  \(\frac{10}{3}\) W
4.  270 W

The current in 8 Ω resistance is (in the figure below):

1.  0.69 A
2.  0.92 A
3.  1.30 A
4.  1.6 A

The terminal potential difference of a cell is greater than its emf when:

A battery is charged at a potential of 15 V for 8 hours when the current flowing is 10 A. The battery on discharge supplies a current of 5 A for 15 hours. The mean terminal voltage during discharges is 14 V. The "Watt hour" efficiency of the battery is:

1. 80%

2. 90%

3. 87.5%

4. 82.5% 

Five equal resistances each of resistance R are connected as shown in the figure below. A battery of V volts is connected between A and B. The current flowing in AFCEB will be:

1. V/R

2. V/2R

3. 2V/R

4. 3V/R

The specific resistance of a conductor increases with:

For a cell, the terminal potential difference is 2.2 V when the circuit is open and reduces to 1.8 V when the cell is connected to the resistance of R = 5 Ω. The internal resistance of cell (r) is:

1. \(\frac{10}{9}~ \Omega\)
2. \(\frac{9}{10}~ \Omega\)
3. \(\frac{11}{9}~ \Omega\)
4. \(\frac{5}{9}~ \Omega\)