Two wires of resistance R₁ and R₂ have temperature coefficient of resistance ${\alpha }_1\text{ and }{\alpha }_2$, respectively. These are joined in series. The effective temperature coefficient of resistance is :

(1) $\frac{{\alpha }_1+{\alpha }_2}{2}$

(2) $\sqrt{{\alpha }_1{\alpha }_2}$

(3) $\frac{{\alpha }_1{R}_1+{\alpha }_2{R}_2}{R_1+R_2}$

(4) $\frac{\sqrt{R_1{R}_2{\alpha }_1{\alpha }_2}}{\sqrt{R_1^2+R_2^2}}$

Two cells of equal e.m.f. and of internal resistances, r_1, and $r_2(r_1>r_2)$ are connected in series. On connecting this combination to an external resistance R, it is observed that the potential difference across the first cell becomes zero. The value of R will be :

(1) $r_1+r_2$

(2) $r_1-r_2$

(3) $\frac{r_1+r_2}{2}$

(4) $\frac{r_1-r_2}{2}$

When connected across the terminals of a cell, a voltmeter measures 5V and a connected ammeter measures 10 A of current. A resistance of 2 ohms is connected across the terminals of the cell. The current flowing through this resistance will be :

(1) 2.5 A

(2) 2.0 A

(3) 5.0 A

(4) 7.5 A

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