Current Electricity

Resistance of wire at $0^{\circ} \mathrm{C}, R_0=20 \Omega$

Temperature coefficient,

$\begin{aligned} & \alpha=5 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1} \\ & R_t=2 R_0=2 \times 20=40 \Omega \end{aligned}$

we know that,

$\begin{aligned} R_t & =R_0(1+\alpha t) \\ 40 & =20\left(1+5 \times 10^{-3} t\right) \Rightarrow 2=1+5 \times 10^{-3} t \end{aligned}$

$\Rightarrow \quad t=\frac{1}{5 \times 10^{-3}}=\frac{1000}{5}=200^{\circ} \mathrm{C}$

Given, length of wire, $l=2 \mathrm{~m}$

Time taken to drift electron, $t=40 \times 10^3 \mathrm{~s}$

Current, $I=1.6 \mathrm{~A}, A=4 \mathrm{~mm}^2=4 \times 10^{-6} \mathrm{~m}^2$

$\therefore$ Drift velocity, $v_d=\frac{l}{t}$

$=\frac{2}{40 \times 10^3}=5 \times 10^{-5} \mathrm{~m} / \mathrm{s}$

We know that,

$\begin{aligned} I & =n e A v_d \\ \Rightarrow \quad n & =\frac{I}{e A v_d} \\ & =\frac{1.6}{1.6 \times 10^{-19}} \times 4 \times 10^{-6} \times 5 \times 10^{-5} \\ & =0.5 \times 10^{29} \\ & =5 \times 10^{28} \text { electrons } / \mathrm{m}^3 \end{aligned}$

The circuit diagram is shown as,

Applying KVL in loop 3, we get

$I_3 R+R\left(I_3-I_2\right)=0$

$\begin{aligned} 2 I_3 R-R I_2 & =0 \\ I_3 & =\frac{I_2}{2} \quad \text{... (i)} \end{aligned}$

Applying KVL in loop 2, we get

$\begin{aligned} & \text { Applying } K V L \text { in } \\ & R I_2-3 \varepsilon+R\left(I_2-I_1\right)+2 \varepsilon+R I_2-R\left(I_2-I_3\right)=0 \\ & \Rightarrow 2 R I_2-R I_1+R I_3-\varepsilon=0 \\ & \Rightarrow 2 R I_2-R I_1+R \frac{I_2}{2}-\varepsilon=0 \\ & \Rightarrow \quad 5 R I_2-2 R I_1-2 \varepsilon=0 \quad \text{... (ii)} \end{aligned}$

By applying KVL in loop 1, we get

$\begin{gathered} R I_1+\varepsilon-2 \varepsilon+R\left(I_1-I_2\right)=0 \\ 2 R I_1-R I_2-\varepsilon=0 \\ I=\frac{R I_2+\varepsilon}{2 R} \quad \text{... (iii)} \end{gathered}$

From Eqs. (ii) and(iii), we get

$\begin{aligned} 5 R I_2-2 R \frac{\left(R I_2+\varepsilon\right)}{2 R}-2 \varepsilon & =0 \\ 5 R I_2-R I_2-\varepsilon-2 \varepsilon & =0 \\ \Rightarrow \quad I_2 & =\frac{3 \varepsilon}{4 R} \end{aligned}$

Given, current density,

$J=\beta\left(r+r_0\right)^2=\beta\left(r^2+2 \pi \pi_0+r_0^2\right)$

We know that, current density, $J=\frac{\operatorname{Current}(I)}{\operatorname{Area}(A)}$

$\Rightarrow$ Current, $I=J \cdot A$

or $d I=J \cdot d A$

$\begin{aligned} & \Rightarrow \quad \int d I=\int J \cdot d A \\ & \Rightarrow \quad I=\int \beta\left(r^2+2 \pi_0+r_0^2\right) \cdot d A \end{aligned}$

Where, $d A$ is area element in cartesian coordinate.

In polar coordinates $d A=r d r d \theta$

$\begin{aligned} & \therefore \quad I=2 \beta \int_0^{\pi / 6} d \theta \int_0^R\left(r^2+2 \pi_0+r_0^2\right) r d r \\ & (\because \text { for } \theta, 0 \text { to } \pi \backslash 6 \text { and } 5 \pi \backslash 6 \text { to } \pi \text { is symmetric) } \\ & =2 \beta[\theta]_0^{\pi / 6}\left[\int_0^R r^3 d r+2 r_0 \int_0^R r^2 d r+r_0^2 \int_0^R r d r\right] \\ & 2 \beta\left(\frac{\pi}{6}\right)\left[\left(\frac{r^4}{4}\right)_0^R+2 r_0\left(\frac{r_3}{3}\right)_0^R+r_0^2\left(\frac{r^2}{2}\right)_0^R\right] \\ & =\beta \frac{\pi}{3}\left[\frac{R^4}{4}+2 r_0 \frac{R^3}{3}+\frac{r_0^2 R^2}{2}\right] \\ & =\pi \beta\left[\frac{R^4}{12}+\frac{2 r_0 R^3}{9}+\frac{r_0^2 R^2}{6}\right] \end{aligned}$

Let $r$ be the internal resistance of the cell.

for case -1

$\begin{aligned} & I=1 \mathrm{~A}, R=25 \Omega \text {, then } \\ & I=\frac{E}{r+R} \\ & \Rightarrow \quad E=I(r+R)=1(r+25) \\ & \Rightarrow \quad E=r+25 \quad \text{... (i)}\\ \end{aligned}$

For the second condition,

$\begin{aligned} & I^{\prime}=0.5 \mathrm{~A}, R^{\prime}=10 \Omega \\ & I^{\prime}=\frac{E}{R^{\prime}+r} \\ & \Rightarrow \quad E=I^{\prime}\left(R^{\prime}+r\right) \\ & =0.5(10+r) \\ & \Rightarrow \quad E=5+0.5 r \quad \text{... (ii)}\\ \end{aligned}$

Now, from Eqs. (i) and (ii), we get

$\begin{aligned} \Rightarrow & & r+25 & =5+0.5 r \\ \Rightarrow & & r-0.5 r & =5-25 \Rightarrow 0.5 r=25 \\ \Rightarrow & & r & =\frac{25}{0.5}=5 \Omega \end{aligned}$

When S₂ open.

Assume resistance of XY = R.

Resistance of wire per unit length, x = $\frac{R}{L}$ = R $\Omega$ m^$-$1

$\because$ I = E₀ / R

Now, the potential drop across 50 cm length is 6 V, so

$\frac{E_0}{R}$ $\times$ R $\times$ $\frac{50}{100}=6$

$\Rightarrow$ E₀ = 12 V

When S₂ closed, potential drop across 0.416 cm length,

V₁ = $\frac{E_0}{R}\times$ R $\times\,0.416=12\times0.416\approx 5$ V

Hence, E $-$ Ir = 5V

$\Rightarrow$ 6 $-$ Ir = 5

$\because$ I = $\frac{5}{10}$

$\therefore$ 6 $-$ = $\frac{5}{10}$r

$\Rightarrow$ r = 2$\Omega$

Given all resistances have same resistance R.

Now, we can redraw the circuit as below

Let resistances be R₁, R₂, R₃ and R₄.

$\because$ ${{{R_1}} \over {{R_3}}} = {{{R_2}} \over {{R_4}}}$

So, circuit will behave as a Wheatstone bridge and no current will flow through middle resistor.

$\therefore$ ${R_{eq}} = {{({R_1} + {R_2})({R_3} + {R_4})} \over {({R_1} + {R_2}) + ({R_3} + {R_4})}}$

$ = {{(R + R)(R + R)} \over {(R + R) + (R + R)}}$

$ = R$