If each resistance in the figure is \(9~\Omega\), then the reading of the ammeter is:

1. \(5~\text{A}\)
2. \(8~\text{A}\)
3. \(2~\text{A}\)
4. \(9~\text{A}\)

Two wires of equal diameters, of resistivities ρ₁ and ρ₂ and lengths l₁ and l₂, respectively, are joined in series. The equivalent resistivity of the combination is :

1. $\frac{{\rho }_1{l}_1+{\rho }_2{l}_2}{l_1+l_2}$

2. $\frac{{\rho }_1{l}_2+{\rho }_2{l}_1}{l_1-l_2}$

3. $\frac{{\rho }_1{l}_2+{\rho }_2{l}_1}{l_1+l_2}$

4. $\frac{{\rho }_1{l}_1-{\rho }_2{l}_2}{l_1-l_2}$ 

Four resistances of 100 Ω each are connected in the form of square. Then, the effective resistance along the diagonal points is :

1. 200 Ω

2. 400 Ω

3. 100 Ω

4. 150 Ω

Equivalent resistance between the points A and B is (in Ω) 

1. $\frac{1}{5}$

2. $1\frac{1}{4}$

3. $2\frac{1}{3}$

4. $3\frac{1}{2}$

In the circuit shown here, what is the value of the unknown resistor R so that the total resistance of the circuit between points P and Q is also equal to R 

1. 3 ohms

2. $\sqrt{39}ohms$

3. $\sqrt{69}ohms$

4. 10 ohms

The resistors of resistances 2 Ω, 4 Ω and 8 Ω are connected in parallel, then the equivalent resistance of the combination will be :

1. $\frac{8}{7}\Omega$

2. $\frac{7}{8}\Omega$

2. $\frac{7}{4}\Omega$

4. $\frac{4}{9}\Omega$ 

In the circuit, the potential difference across PQ will be nearest to 

1. 9.6 V

2. 6.6 V

3. 4.8 V

4. 3.2 V

Three resistors are connected to form the sides of a triangle ABC, the resistance of the sides AB, BC and CA are 40 ohms, 60 ohms and 100 ohms respectively. The effective resistance between the points A and B in ohms will be 

1. 32

2. 64

3. 50

4. 200

Equivalent resistance across terminals \(A\) and \(B\) will be:

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. $2({\rho }_1+{\rho }_2)$