(3)

As the two conducting wires are of the same material, equal lengths and equal diameters. They will have the same resistance.

Let the resistance of each wire be R and applied potential difference, V

$$\begin{gathered} R_{series}=R_1+R_2 \\ = R+R=2R \\ {P}_{series}=\frac{V^2}{R_{series}} \\ = \frac{V^2}{\left(2R\right)} \left(i\right) \end{gathered}$$

$$\begin{gathered} \frac{1}{R_{parallel}}=\frac{1}{R_1}+\frac{1}{R_2} \\ = \frac{1}{R}+\frac{1}{R} \\ {R}_{parallel} = \frac{R}{2} \\ {P}_{parallel} = \frac{R}{2} \\ {P}_{parallel} = \frac{V^2}{R_{parallel}} \\ = \frac{2V^2}{R} \\ Dividing equation \left(i\right) by equation \left(ii\right) \\ \frac{P_{series}}{P_{parallel}}=\frac{1}{4} \end{gathered}$$

(4)

As it is clear From the graph, the current for conductor$A₂$is less than that for $A₁, and A₁$ is less than $A₃$.
we can say , $Iₐ₂ < Iₐ₁ < Iₐ₃.$

$We know, R = V/I or R \propto 1/I.$

If I is less, then R will be more

$$\begin{gathered} Iₐ₂ < Iₐ₁ < Iₐ₃ \\ R₂ > R₁ > R₃. \end{gathered}$$

(2)

Ammeter is a device used for measuring electric current in amperes.

(4)

For a given continuous piece of uniform wire, the resistance is directly proportional to its length. Thus, silver which is a good conductor of electricity and the resistance in it is directly proportional to its length.

(4)

Hence $$\begin{gathered} r_1 = R_1 + R_2 \left(i\right) \\ and r_2 = \frac{R_1 R_2}{R_1 + R_2} \left(ii\right) \\ \Rightarrow R_1 R_2 = r_2 r_1 \\ Now, {\left(R_1 - R_2\right)}^2 = {\left(R_1 + R_2\right)}^2 - 4 R_1 R_2 \\ = r_1^2 - 4r_2 r_1 \\ {R}_1 - R_2 = \sqrt{r_1^2 - 4r_2 r_1} \dots \left(iii\right) \\ and R_1 + R_2 = r_1 \dots \left(iv\right) \\ solving equation \left(iii\right) and \left(iv\right) for R_1 and R_2. \\ {R}_1 = \frac{r_1 + \sqrt{r_1^2 - 4r_1 r_2}}{2} \\ and R_2 = \frac{r_1 - \sqrt{r_1^2 - 4r_1 r_2}}{2} \\ \frac{R_1}{R_2}=\frac{r_1 + \sqrt{r_1^2 - 4r_1 r_2}}{r_1 - \sqrt{r_1^2 - 4r_1 r_2}} \end{gathered}$$

(1)

It is given that the three resistors are connected in parallel and the resistance of each resistor is $3 \Omega .$

$R_1=R_2=R_3=3\Omega$

From the formula given below, we can calculate the total resistance of all the three resistors:

$$\begin{gathered} \frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} \\ =\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{3}{3} \\ \therefore R_p=1\Omega \end{gathered}$$

Hence, the total resistance is  1$\Omega$

(1)

Resistance of one resistor = $\frac{1}{5}\Omega$

Number of resistors = 5

Maximum resistance can be obtained by combining the resistors in series: 

$$\begin{gathered} R_s=R_1+R_2+R_3+R_4+R_5 \\ =\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5} \\ =\frac{1+1+1+1+1}{5}=\frac{5}{5} \\ =1\Omega \end{gathered}$$

Hence, a person on combining five resistors in series gets resistance  $1 \Omega$

(4)

A-2; B-3; C-4; D-1

 (1)

The electrical resistance of a wire can be expressed as:$R=\rho \frac{L}{A}$

Where, A = Area of cross-section of the conductor, L = Length of the conductor, $\rho$ = Resistivity

From this relation, it is clear that the resistance is directly proportional to the length and inversely proportional to area of cross-section.

If length becomes 2L, then

$$\begin{gathered} R^{\text{'}}=\frac{\rho 2L}{A}=2\frac{\rho L}{A} \\ So, R\text{'} = 2R \end{gathered}$$

Thus, the resistance becomes 2 times if the length of the wire is doubled.

(4)

The correct way of connecting ammeter and voltmeter in the circuit to determine the equivalent resistance of two resistors is connecting ammeter in series and voltmeter in parallel. Ammeter is connected in series, so that the whole current passes through it and voltmeter is connected in parallel so that it could measure the complete voltage of the circuit.