Current Electricity

Point A and B are at same potential so they can be merged/folded.

For loop 1

$ \begin{aligned} & -\frac{1}{2} x-\frac{1}{2}(x+y)-x+6=0 \\\\ &\Rightarrow -2 x-\frac{y}{2}=-6 \\\\ &\Rightarrow -4 x-y=-12 \\\\ &\Rightarrow 4 x+y=12 .......(1) \end{aligned} $

For loop 2

$ \begin{aligned} & \frac{1}{2} y+1 y+12+\frac{1}{2}(x+y)=0 \\\\ &\Rightarrow \frac{3}{2} y+\frac{y}{2}+\frac{x}{2}=-12 \\\\ &\Rightarrow 2 y+\frac{x}{2}=-12 \\\\ & \begin{array}{l} \Rightarrow4 y+x=-24 \\\\ \Rightarrow4 y+x-16 x-4 y =-24-48 \end{array} \\\\ &\Rightarrow -15 x=-72 \end{aligned} $

$ \Rightarrow $ $x=\frac{72}{15}$

$ \Rightarrow $ $4\left(\frac{72}{15}\right)+y=12$

$ \begin{aligned} \Rightarrow y & =12-\frac{288}{15} \\\\ & =\frac{180-288}{15} \\\\ & =\frac{-108}{15}=-7.2 \mathrm{~A} \end{aligned} $

Current in $R_1=7.2 \mathrm{~A}$

Current in $R_2=\frac{x+y}{2}=\left(\frac{72}{15}-\frac{108}{15}\right) \frac{1}{2}$

$ \begin{aligned} & =\frac{1}{2}\left(\frac{36}{15}\right)=\frac{2.4}{2} \mathrm{~A} \\\\ & =1.2 \mathrm{~A} \end{aligned} $

Current in $R_3=x=4.8 \mathrm{~A}$

Current in $R_5=\frac{1}{2} x=2.4 \mathrm{~A}$

Electron mobility.

$ \begin{aligned} \mu_e & =\frac{\text { Drift velocity }}{\text { Electric field }} \\ \mu_e & =\frac{v_d}{\mathbf{E}} \\ {\left[\mu_e\right] } & =\frac{\left[v_d\right]}{[E]}=\frac{\left[\mathrm{LT}^{-1}\right]}{\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{AT}]}} \\ \Rightarrow \quad\left[\mu_e\right] & =\left[\mathrm{M}^{-1} \mathrm{~T}^2 \mathrm{~A}\right] \end{aligned} $

So, SI units of mobility is $\mathrm{kg}^{-1} \mathbf{s}^2 \mathbf{A}$

Given, density of water,

$ \rho_w=1000 \mathrm{~kg} / \mathrm{m}^3 $

Specific heat capacity of water,

$ s_w=400 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1} $

Volume of water, $V=31=3 \times 10^{-3} \mathrm{~m}^3$

∴ Mass of water, $m=V \times \rho_w$

$ =3 \times 10^{-3} \times 1000=3 \mathrm{~kg} $

Change of temperature of water,

$ \begin{aligned} \Delta T & =[(273+80)-(273+0)] \mathrm{K} \\ & =80 \mathrm{~K} \end{aligned} $

Heat required, $Q=m s_w \Delta T$

$ \begin{aligned} & =3 \times 4200 \times 80=1008000 \\ & =1.008 \times 10^6 \mathrm{~J} \end{aligned} $

Heat produced by heater, $H=\frac{V^2}{R} t$

$ \begin{array}{ll} \Rightarrow & 1.008 \times 10^6=\frac{(200)^2}{50} \times t \\ \Rightarrow & t=1260 \mathrm{~s}=\frac{1260}{60} \mathrm{~min}=21 \mathrm{~min} \end{array} $

Current density in circular wire,

$ J=1 \times 10^5 r \mathrm{~A} / \mathrm{m}^3 $

∴ Current, $I=J A$

$ \begin{aligned} & =1 \times 10^5 r \times \pi r^2=10^5 \pi r^3 \\ & =10^5 \times 314 \times\left(2 \times 10^{-4}\right)^3 \\ & \quad\left[\because r=2 \mathrm{~mm}=2 \times 10^{-4} \mathrm{~m}\right] \\ & =2512 \times 10^{-7} \mathrm{~A} \end{aligned} $

The rate of thermal energy produced,

$ \begin{aligned} P & =V I \\ & =70 \times 2512 \times 10^{-7}=1.76 \times 10^{-4} \mathrm{~W} \end{aligned} $

Energy consumed, $E=P \times t$

$ \begin{aligned} & =1.76 \times 10^{-4} \times 1000 \quad[\because t=1000 \mathrm{~s}] \\ & =0.176 \mathrm{~J} \end{aligned} $

Specific resistance depends on the nature of material and also depends on temperature of material. For conductor, it increases with increase of temperature. Hence, statement $I$ is incorrect. When wire of resistance $R$ is increased to $n$ times of its original length, then its new resistance $R^{\prime}$ is given as

$ R^{\prime}=n^2 R $

Here, $n=4, R=6 \Omega$

$ \therefore R^{\prime}=4^2 \times 6=96 \Omega $

Hence, statement II is incorrect and statement III is correct.

Average collision time,

$ \begin{aligned} \tau & =91 \times 10^{-15} \mathrm{~s} \\ e & =1.6 \times 10^{-19} \mathrm{C} \\ m_e & =9.1 \times 10^{-31} \mathrm{~kg} \end{aligned} $

∴ Mobility of electron in a wire,

$ \begin{aligned} \mu & =\frac{e \tau}{m_e} \\ & =\frac{1.6 \times 10^{-19} \times 9.1 \times 10^{-15}}{9.1 \times 10^{-31}} \\ & =1.6 \times 10^{-3} \mathrm{~m}^2 / \mathrm{V}-\mathrm{s} \end{aligned} $

The temperature coefficient of resistance for most of metals in pure form is positive because on increasing temperature of metals, its resistance increases. Hence, statement I is true.

In the statement II,

Given, diameter of wire $=2 \mathrm{~mm}$

∴ radius, $r=1 \mathrm{~mm}=10^{-3} \mathrm{~m}$

Charge, $q=360 \pi \mathrm{C}$

Time, $t=2 \mathrm{~h}=2 \times 60 \times 60 \mathrm{~s}=7.2 \times 10^3 \mathrm{~s}$

Charge density, $n=5 \times 10^{22}$

electrons $/ \mathrm{cm}^3=5 \times 10^{28}$ electrons $/ \mathrm{m}^3$

Current flowing through the wire.

$ I=\frac{q}{t}=\frac{360 \pi}{7.2 \times 10^3}=\frac{\pi}{20} \mathrm{~A} $

We know that, $I=n e A v_d$

$ \begin{aligned} & \Rightarrow \quad v_d=\frac{I}{n e A} \\ & =\frac{\pi / 20}{5 \times 10^{28} \times 1.6 \times 10^{-19} \times \pi \times\left(10^{-3}\right)^2} \\ & =6.25 \times 10^{-6} \mathrm{~m} / \mathrm{s} \end{aligned} $

Hence, statement II is also true.

Semiconductors like pure germanium are operated in non-linear region, hence these do not obey Ohm's law. Hence, statement III is also true.

Radius of cylindrical resistor,

$ r=7 \mathrm{~mm}=7 \times 10^{-3} \mathrm{~m} $

Length, $l=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}$

Resistivity, $\rho=10^{-6} \Omega-\mathrm{m}$

$ \begin{aligned} \therefore \quad R & =\rho \cdot \frac{l}{A} \\ & =10^{-6} \times \frac{4 \times 10^{-2}}{\pi r^2} \\ & =\frac{4 \times 10^{-8}}{\frac{22}{7} \times\left(7 \times 10^{-3}\right)^2} \\ & =\frac{4 \times 10^{-8}}{22 \times 7 \times 10^{-6}} \\ & =\frac{2}{77} \times 10^{-2} \Omega \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \text { Given, } \frac{E}{t}=1.54 \mathrm{~W} \\ & \Rightarrow \quad \frac{I^2 R T}{t}=1.54 \\ & \Rightarrow \quad I=\sqrt{\frac{1.54}{R}}=\sqrt{\frac{1.54}{\frac{2}{77} \times 10^{-2}}}=77 \mathrm{~A} \end{aligned}\\ &\text { ∴ Current density, }\\ &\begin{aligned} I=\frac{I}{A} & =\frac{I}{\pi r^2}=\frac{77}{\frac{22}{7} \times\left(7 \times 10^{-3}\right)^2} \\ & =5 \times 10^5 \mathrm{~A} / \mathrm{m}^2 \end{aligned} \end{aligned} $

We know that, $\tau=\frac{m}{n e^2 \rho}$

∴ Mean free path, $\lambda=\tau \cdot v_d=\frac{m v_d}{n e^2 \rho}$

$ \begin{aligned} & \text { Here, } n=9 \times 10^{28} / \mathrm{m}^3, \rho=1 \times 10^{-8} \Omega-\mathrm{m} \\ & \qquad \begin{aligned} v_d & =1.6 \times 10^6 \mathrm{~m} / \mathrm{s} \\ & m=9.1 \times 10^{-31} \mathrm{~kg}, e=1.6 \times 10^{-19} \mathrm{C} \end{aligned} \\ & \therefore \lambda=\frac{9.1 \times 10^{-31} \times 1.6 \times 10^6}{9 \times 10^{28} \times\left(1.6 \times 10^{-19}\right)^2 \times 1 \times 10^{-8}} \\ & \\ & =62.5 \times 10^{-9} \mathrm{~m} \\ & = 62.5 \mathrm{~nm} \end{aligned} $

Resistivity of metal,

$ \rho=1 \times 10^{-8} \Omega-\mathrm{m} $

Number of electrons per unit volume,

$ \begin{aligned} & n=9 \times 10^{28} \text { electrons } / \mathrm{m}^3 \\ & m=9 \times 10^{-31} \mathrm{~kg} \end{aligned} $

We know that, $\rho=\frac{m}{n e^2 \tau}$

$ \begin{aligned} & \Rightarrow \quad \tau=\frac{m}{n e^2 \rho} \\ & =\frac{9 \times 10^{-31}}{9 \times 10^{28} \times\left(1.6 \times 10^{-19}\right)^2 \times 1 \times 10^{-8}} \\ & =3.9 \times 10^{-14} \mathrm{~s} \simeq 4 \times 10^{-14} \mathrm{~s} \end{aligned} $

A wire is stretched such that its resistance increase by $6 \%$.

If we assume initial resistance to be 100 $\Omega$. Then the new resistance will be given by

$ R_{\text {new }}=100 \Omega+\frac{6}{100} \times 100 \Omega=106 \Omega $

We know that, the resistance is given by

$ R=\rho \frac{l}{A} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Where, area $(A)$ can be expressed in terms of volume $(V)$ is

$ A=\frac{V}{l} $

Equation (i) then gives us, $R=\rho \frac{l^2}{V}$

Since, the resistivity and volume are going to remain the same, only change in length will be observed.

So, $\quad l^2=\frac{R V}{\rho}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

(Original length)

Let new length be $l_{\text {new }}$, so

$ l_{\text {new }}^2=\frac{R_{\text {new }} V}{\rho} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Dividing Eq. (iii) by Eq. (ii), we get

$ \begin{aligned} & \frac{l_{\text {new }}^2}{l^2}=\frac{R_{\text {new }}}{R}=\frac{106}{100} \\ \Rightarrow \quad & l_{\text {new }}=1.029 l \end{aligned} $

\% change in length

$ \begin{aligned} & =\left(\frac{l_{\mathrm{new}}-l}{l}\right) \times 100 \% \\ & =\left(\frac{1.029 l-l}{l}\right) \times 100 \\ & =0.029 \times 100 \\ & =29 \% \approx 3 \% \end{aligned} $

The given Circuit diagram is shown below.

Applying KVL in loop $A B C D A$.

$ 2+i_1 R-2=0 \Rightarrow i_1=0 \mathrm{~A} $

Applying KVL in loop $B E F C B$,

$ 2+i_2 R-2-i_1 R=0 \Rightarrow i_2=i_1=0 \mathrm{~A} $

Applying KVL in loop $E G H F E$, we get

$ \begin{aligned} & 2+i_3 R-2-i_2 R=0 \Rightarrow i_3=i_2=0 \mathrm{~A} \\ & \therefore \quad i_1=i_2=i_3=0 \mathrm{~A} \end{aligned} $

The resistivity of material is too much high as $10^8 \Omega$ - m which is equivalent to the value of resistivity of insulator.

$ \begin{aligned} &\text { We know that, resistance of wire, }\\ &\begin{aligned} & R=\rho \frac{L}{A}=\rho \frac{L}{\pi r^2} \\ \Rightarrow \quad & R \propto \frac{L}{r^2} \Rightarrow \frac{R_2}{R_1}=\frac{L_2}{L_1} \times\left(\frac{r_1}{r_2}\right)^2 \end{aligned}\\ &\begin{aligned} \text { Given, } L_1 & =L, L_2=2 L \\ r_1 & =r, r_2=3 r \\ \frac{R_2}{R_1} & =\frac{2 L}{L} \times\left(\frac{r}{3 r}\right)^2=2 \times \frac{1}{9}=\frac{2}{9} \\ \Rightarrow \quad R_2 & =\frac{2}{9} R_1=\frac{2}{9} R \end{aligned} \end{aligned} $

Given, $l_1=60 \mathrm{~cm}, l_2=40 \mathrm{~cm}$,

$ R=4 \Omega $

∴ Internal resistance of the cell,

$ \begin{aligned} r & =R\left(\frac{l_1}{l_2}-1\right)=4\left(\frac{60}{40}-1\right) \\ & =4 \times \frac{1}{2}=2 \Omega \end{aligned} $

The given circuit diagram is shown as :

Applying KVL in loop $A B D A$, we get

$\begin{aligned} 2 I_1+4-I_2 & =0 \\ I_2 & =2 I_1+4\quad \text{... (i)} \end{aligned}$

Applying KVL in loop $BCDB$, we get

$\begin{array}{rlrl} & 1\left(I_1+I_3\right)-4\left(I_2-I_3\right)-4 =0 \\ \Rightarrow & I_1+I_3-4 I_2+4 I_3-4 =0 \\ \Rightarrow & I_1+5 I_3-4\left(2 I_1+4\right)-4 =0 \\ \Rightarrow & -7 I_1+5 I_3 =20 \\ \Rightarrow & I_3= \frac{20+7 I_1}{5}\quad \text{... (ii)} \end{array}$

By applying KVL in loop $ADCA$, we get

$ \begin{aligned} & & I_2+4\left(I_2-I_3\right)-10 & =0 \\ \Rightarrow & & 5 I_2-4 I_3 & =10 \\ \Rightarrow & & 5\left(2 I_1+4\right)-4\left(\frac{20+7 I_1}{5}\right) & =10 \quad \text{[From Eqs. (i) and (ii)]} \end{aligned}$

$\Rightarrow \quad I_1=1.364 \mathrm{~A}$

From Eq. (i), $I_2=2 \times 1.364+4=6.278 \mathrm{~A}$

From Eq (ii), $I_3=\frac{20+7 \times 1.364}{5}=5.91 \mathrm{~A}$

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

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