The electric intensity \(E,\) current density \(j\) and specific resistance \(k\) are related to each other by the relation:
1. \(E = j/k\)
2. \(E = jk\)
3. \(E = k/j\)
4. \(k = j E\)

If \(n\), \(e\), \(\tau\) and \(m\) respectively represent the density, charge relaxation time and mass of the electron, then the resistance of a wire of length \(l\) and area of cross-section \(A\) will be:
1. \(\frac{ml}{ne^2\tau A}\)
2. \(\frac{m\tau^2A}{ne^2l}\)
3. \(\frac{ne^2\tau A}{2ml}\)
4. \(\frac{ne^2 A}{2m\tau l}\)

Two wires of the same dimensions but resistivities ρ₁ and ρ₂ are connected in series. The equivalent resistivity of the combination is 

1. ρ₁ + ρ₂

2. $\frac{{\rho }_1+{\rho }_2}{2}$

3. $\sqrt{{\rho }_1{\rho }_2}$

4. $2({\rho }_1+{\rho }_2)$

In the network shown in the figure, each of the resistance is equal to 2 Ω. The resistance between the points A and B is 

1. 1 Ω

2. 4 Ω

3. 3 Ω

4. 2 Ω

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