Science Answer the questions.[Phy-2M]

$\text{A}=\pi\times\text{r}^{2}$

$\text{r}=\frac{\text{d}}{2}=\frac{0.3}{2}\text{mm}=0.15\text{mm}=0.15\times10^{-3}\text{m}$

$\rho=\frac{\text{R}\times\text{A}}{\text{I}}=\frac{\text{R}\times\pi\text{r}^2}{\text{I}}$

$\frac{26\times3.14\times(0.15\times10^{-3})^2}{1}$

$-1.83\times10^{-6}\Omega\text{m}$

$\text{R}=\frac{240}{5}=48\text{ohm}.$

$\Rightarrow\text{E}=2\times10\times\text{x}\dots{2}$

$\Rightarrow\text{x}=\frac{200}{20}=10\text{m}$

$\text{H}=\text{i}^{2}\text{rt}$

$\text{V}=\text{I}_{1}\times\text{R}_{1}$

$\text{V}=\text{I}_{2}\times\text{R}_{2}$

$\text{I}_{1}\times\text{R}_{1}=\text{I}_{2}\times\text{R}_{2}=\text{I}_{2}\times\frac{\text{R}_{1}}{2}$

$\text{I}_{2}\frac{\text{I}_{1}\times\text{R}_{1}}{\frac{\text{R}_{1}}{2}}=2\times\text{I}_{1}$

$\therefore\text{I}=2\text{I}$

Thus, from this we can conclude that if the resistance of the circuit is halved then the current gets doubled.

$\text{I}=\frac{\text{V}}{\text{R}}$

1. When $1\Omega$ and $10^6\Omega$ are connected in parallel:

Let R be the equivalent resistance.

$\therefore \ \frac{1}{\text{R}}=\frac{1}{1}+\frac{1}{10^6}$

$\text{R}=\frac{10^6}{1+10^6}\approx\frac{10^6}{10^6}=1\Omega$

Therefore, equivalent resistance $\approx1\Omega$

2. When $1\Omega$, $103\Omega$ and $106\Omega$ are connected in parallel:

Let R be the equivalent resistance.

$\frac{1}{\text{R}}=\frac{1}{1}+\frac{1}{10^3}+\frac{1}{10^6}\frac{10^6+10^3+1}{10^6}$

$\text{R}=\frac{1000000}{1001001}=0.999 \ \Omega$

Therefore, equivalent resistance = $0.999\Omega$.

$R_{1} = \frac{1.0 \text{ V}}{0.1 \text{ A}} = \text{10 }\Omega \text{ ; }$

$ R_{2} = \frac{2.2 \text{ V}}{0.2 \text{ A}} = \text{11 }\Omega \text{ ; }$

$R_{3} = \frac{4.0 \text{ V}}{0.4 \text{ A}} = \text{ 10 } \Omega \text{ ; }$

$R_{4} = \frac{6.4 \text{ V}}{0.6 \text{ A}} = \text{10.6 } \Omega$

$\therefore \text{Mean } R = \text{(10 + 11 + 10 +10.6) } \Omega$

$ = \text{10.4 } \Omega$

$\text{W}=\text{Q}\times\text{V}$

$\therefore\text{V}=\frac{\text{W}}{\text{V}}$

$\text{V}=\frac{15\text{J}}{20\times10^{-3}}$

Since, $\text{Q}=20\text{mC}=20\times10^{-3}\text{C}$

$\text{V}=7.5\times10^{2}\text{C}$ 

$\text{I}=\frac{\text{P}}{\text{V}}=\frac{12}{12}=1\text{A}$

$\text{R}=\frac{\text{V}}{\text{I}}=\frac{12}{1}=12\Omega$

$\text{I}=\frac{\text{Q}}{\text{t}}$

the conductor will rise. As most conductors have a positive temperature coefficient of resistance, the resistance of the conductor will increase according to the following relation.

R = R_ref [1 + a( T - T_ref)]

Where

R = Conductor resistance at temperture ''T''

R_ref = Conductor resistance at reference temperature.

T_ref, usually 20°C, but sometimes 0°C.

$\alpha$ = Temperature cofficient of resistance for the conductor material.

T_ref = Reference temperature that a is specified at for the conductor material.

$\frac{1}{\text{R}}=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{4}{16}$