Calculate the equivalent resistance between A and B 

1. $\frac{9}{2}\Omega$

2. 3 Ω

3. 6 Ω

4. $\frac{5}{3}\Omega$

The total current supplied to the circuit by the battery is:

1. \(1~\text{A}\)
2. \(2~\text{A}\)
3. \(4~\text{A}\)
4. \(6~\text{A}\)

For the network shown in the figure the value of the current i is 

1. $\frac{9V}{35}$

2. $\frac{5V}{18}$

3. $\frac{5V}{9}$

4. $\frac{18V}{5}$

The magnitude and direction of the current in the circuit shown will be 

1. $\frac{7}{3}$A from a to b through e

2. $\frac{7}{3}$A from b to a through e

3. 1A from b to a through e

4. 1A from a to b through e

In the circuit shown, A and V are ideal ammeter and voltmeter respectively. Reading of the voltmeter will be

1. 2 V

2. 1 V

3. 0.5 V

4. Zero

Four identical cells each having an electromotive force (e.m.f.) of 12V, are connected in parallel. The resultant electromotive force (e.m.f.) of the combination is :

1. 48 V

2. 12 V

3. 4 V

4. 3 V

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

A resistance of 4 Ω and a wire of length 5 metres and resistance 5 Ω are joined in series and connected to a cell of e.m.f. 10 V and internal resistance 1 Ω. A parallel combination of two identical cells is balanced across 300 cm of the wire. The e.m.f. E of each cell is:

1. 1.5 V

2. 3.0 V

3. 0.67 V

4. 1.33 V

In the circuit shown here, E₁ = E₂ = E₃ = 2 V and R₁ = R₂ = 4 ohms. The current flowing between points A and B through battery E₂ is 

1. Zero

2. 2 amp from A to B

3. 2 amp from B to A

4. None of the above

The equivalent resistance between the points P and Q in the network given here is equal to (given $r=\frac{3}{2}\Omega$) :

1. $\frac{1}{2}\Omega$

2. 1 Ω

3. $\frac{3}{2}\Omega$

4. 2 Ω