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

In the circuit shown below, \(E_1 = 4.0~\text{V}\), \(R_1 = 2~\Omega\), \(E_2 = 6.0~\text{V}\), \(R_2 = 4~\Omega\) and \(R_3 = 2~\Omega\). The current \(I_1\) is:

1. \(1.6\) A

2. \(1.8\) A

3. \(1.25\) A

4. \(1.0\) A

The potential difference across \(8~\Omega\) resistance is \(48~\text V\) as shown in the figure below. The value of potential difference across \(X\) and \(Y\) points will be:

   
1. \(160~\text V\)
2. \(128~\text V\)
3. \(80~\text V\)
4. \(62~\text V\)

Two resistances R₁ and R₂ are made of different materials. The temperature coefficient of the material of R₁ is α and of the material of R₂ is –β. The resistance of the series combination of R₁ and R₂ will not change with temperature, if R₁/ R₂ equals :

(1) $\frac{\alpha }{\beta }$

(2) $\frac{\alpha +\beta }{\alpha -\beta }$

(3) $\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$

(4) $\frac{\beta }{\alpha }$

An ionization chamber with parallel conducting plates as anode and cathode has $5\times {10}^7$ electrons and the same number of singly-charged positive ions per cm³. The electrons are moving at 0.4 m/s. The current density from anode to cathode is $4\mu A/m^2$. The velocity of positive ions moving towards cathode is :

(1) 0.4 m/s

(2) 16 m/s

(3) Zero

(4) 0.1 m/s

A wire of resistance 10 Ω is bent to form a circle. P and Q are points on the circumference of the circle dividing it into a quadrant and are connected to a Battery of 3 V and internal resistance 1 Ω as shown in the figure. The currents in the two parts of the circle are 

(1) $\frac{6}{23}A\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em} }\frac{18}{23}A$

(2) $\frac{5}{26}A\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}}\frac{15}{26}A$

(3) $\frac{4}{25}A\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em} }\frac{12}{25}A$

(4) $\frac{3}{25}A\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em} }\frac{9}{25}A$

In the given circuit, it is observed that the current I is independent of the value of the resistance R₆. Then the resistance values must satisfy 

(1) $R_1{R}_2{R}_5=R_3{R}_4{R}_6$

(2) $\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_1+R_2}+\frac{1}{R_3+R_4}$

(3) $R_1{R}_4=R_2{R}_3$

(4) $R_1{R}_3=R_2{R}_4=R_5{R}_6$

In the given circuit, with a steady current, the potential drop across the capacitor must be :

(1) V

(2) V / 2

(3) V / 3

(4) 2V / 3

A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to current, the temperature of the wire is raised by ΔT in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross–section but of length 2 L. The temperature of the wire is raised by the same amount ΔT in the same time t. The value of N is- 

(1) 4

(2) 6

(3) 8

(4) 9

What is the equivalent resistance between terminals \(A\) and \(B\) of the network?