Current Electricity

Since, conductors are material which can conduct electricity at room temperature due to presence of valence electrons in their outermost shell and on heating valence electrons gain energy, get excited and leave the conduction band. Now, due to absence / lack of free electrons in conduction band, on heating conductors do not conduct electricity.

According to given figure and by using Kirchhoff's current law,

Input current at P = Output current from P

$\therefore 4+5=I+3+5$

$\Rightarrow 9=8+I$

$\Rightarrow I=9-8$

= 1A from P to Q

Let length of each wire be $l$.

$\therefore$ Total length $=10 l$

At, lst balance point, $l_1: l_2=6 l: 4 l$

and as in case of potentiometer, $\frac{\mathrm{emf}_1}{\mathrm{emf}_2}=\frac{l_1}{l_2}$

i.e. $\varepsilon \propto l$

If $l$ increases, emf also increases and emf is only increased by increasing resistance $(R)$ in main circuit.

Given, length of germanium, $l=0.925 \mathrm{~cm} =0.925 \times 10^{-2} \mathrm{~m} $

Area of cross-section,

$A=1 \mathrm{~mm}^2$

$\begin{aligned} & =\left(10^{-3} \mathrm{~m}\right)^2 \\ & =10^{-6} \mathrm{~m}^2 \\ n_i & =2.5 \times 10^{19} \mathrm{~m}^{-3} \\ \mu_n & =0.19 \mathrm{~m}^2 / \mathrm{V}-\mathrm{s} \\ \mu_e & =0.39 \mathrm{~m}^2 / \mathrm{V}-\mathrm{s} \end{aligned}$

$\text { Conductivity, } \begin{aligned} \sigma & =n_i e\left(\mu_e+\mu_n\right) \\ & =2.5 \times 10^{19} \times 1.6 \times 10^{-19}(0.39+0.19) \\ & =2.32 \mathrm{~mho~m}^{-1} \end{aligned}$

Resistivity, $\rho=\frac{1}{\sigma}=0.431 \Omega-\mathrm{m}$

$\text { and resistance, } \begin{aligned} R & =\frac{\rho l}{A} \\ & =\frac{0.431 \times 0.925 \times 10^{-2}}{10^{-6}} \\ & =0.3986 \times 10^4=3.9 \times 10^3 \\ & =4 \times 10^3 \Omega=4 \mathrm{k} \Omega \end{aligned}$

Given, emf of cell $=1.8 \mathrm{~V}$

Current, $I=17 \mathrm{~A}$

Resistance of ammeter, $R=0.06 \Omega$

Let internal resistance be $r$.

$\begin{aligned} & \text { Since, } \mathrm{emf}=I(R+r) \\ & \therefore \quad 1.8=17(0.06+r) \\ & \Rightarrow \quad \frac{1.8}{17}=0.06+r \\ & \Rightarrow \quad r=\frac{1.8}{17}-0.06=0.046 \Omega \\ & \end{aligned}$

A shunt is a low resistance which is connected in parallel with a galvanometer (or ammeter) to protect it from large current and to increase the range of ammeter.

Using Kirchhoff's loop law, in loop BDCR₁ B, we get

2 $\times$ 2 + 3R₁ = 4 or R₁ = 0

$\therefore$ Potential drop at B = Potential drop at C

$\Rightarrow$ V_B = V_C = 2 + V_A

= 2 + 0 [$\because$ V_A = 0, given]

$\Rightarrow$ V_B = 2 V

According to the question, the circuit diagram is as shown below

Power dissipated in circuit,

$ \begin{equation*} P=\frac{E^2}{R_{\mathrm{eff}}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

The given circuit is a balanced Wheatstone bridge.

So, its equivalent resistance is given by

$ \begin{aligned} R_{\mathrm{eff}} & =(4+4)\|(4+4)=8\| 8 \\ & =\frac{8 \times 8}{8+8}=4 \Omega \end{aligned} $

From Eq. (i), $P=\frac{(12)^2}{4}=36 \mathrm{~W}$

The equivalent resistance between point $A$ and $B$ is calculated as

According to given situation, circuit diagram of meter bridge is given as

At balance condition of meter bridge,

$ \begin{aligned} & \frac{S}{R} =\frac{25}{(100-25)} \\ \Rightarrow & \frac{S}{1} =\frac{25}{75} \\ \Rightarrow & S =\frac{1}{3} \Omega=0.33 \Omega \end{aligned} $

Given, length of wire, $P=l$

Length of wire, $Q=\frac{l}{2}$

Diameter of wire, $P=d$

Diameter of wire, $Q=\frac{d}{2}$

Resistance $R=\rho \cdot \frac{l}{A}=\rho \cdot \frac{l}{\pi d^2 / 4}$

$ \begin{aligned} & \begin{aligned} \therefore \frac{\text { Resistance of } Q}{\text { Resistance of } P} & =\left(\frac{\rho \cdot \frac{l}{2}}{\frac{\pi}{4}\left(\frac{d}{2}\right)^2}\right) / \frac{\rho l}{\frac{\pi}{4} d^2} \\ =\frac{\frac{1}{2}}{\frac{1}{4}}= & \frac{4}{2}=\frac{2}{1} \\ \Rightarrow \text { Resistance of } Q & =\text { Resistance of } P \times 2 \\ & =10 \times 2=20 \Omega \end{aligned} \end{aligned} $

The given circuit diagram is

Equivalent resistance of given circuit

$ \begin{aligned} & =(100 \Omega| | 100 \Omega)+100 \Omega+100 \Omega \\ & =\frac{100 \times 100}{100+100}+200=50+200=250 \Omega \end{aligned} $

Equivalent emf of circuit

$ =20-10=10 \mathrm{~V} $

Current in circuit,

$ I=\frac{E}{R_{\mathrm{eq}}}=\frac{10}{250}=0.04 \mathrm{~A} $

Current density is related to electric field as

$ \mathbf{J}=\frac{\mathbf{E}}{\rho} $

where, $\rho=$ resistivity.

Here, $I=5 \mathrm{~A}, A=1 \mathrm{~mm}^2=1 \times 10^{-6} \mathrm{~m}^2$ and $\rho=3 \times 10^{-8} \Omega-\mathrm{m}$

So, electric field inside the conductor is

$ \begin{aligned} \mathbf{E} & =\mathbf{J} \cdot \rho=\frac{I}{A} \cdot \rho \\ & =\frac{5 \times 3 \times 10^{-8}}{1 \times 10^{-6}}=0.15 \mathrm{~V} / \mathrm{m} \end{aligned} $

For balance condition of Wheatstone bridge,

$ \frac{R_2}{R_4}=\frac{R_1}{R_3} \quad \text { or } \quad \frac{R_2}{R_1}=\frac{R_4}{R_3} $

We are given with following ammeter

Now, $\quad V_{A B}=I_g G=\left(I-I_g\right) S$

$ \begin{aligned} \left(100 \times 10^{-6}\right) 100 & =\left(I-100 \times 10^{-6}\right) 0.1 \\ I & -100 \times 10^{-6}=10^{-1} \\ I & =0.1+100 \times 10^{-6} \\ & =0.1+0.0001=0.1001 \mathrm{~A} \\ & =100.1 \mathrm{~mA} \end{aligned} $

So, circuit current is 100.1 mA .

From the following circuit diagram, ${{{R_1}} \over {{R_3}}} = {{{R_2}} \over {{R_4}}}$, so the given Wheatstone bridge is balanced.

Hence, there is no current in the wire of 10 $\Omega$ resistance.

As,

Now, if the 10 $\Omega$ wire is replaced with 15 $\Omega$ wire, the bridge still remains balanced. Hence, current drawn from battery will be 10A.

Given, resistance of galvanometer, R_g = 60$\Omega$

Current, I_g = 4.5 mA = 4.5 $\times$ 10^$-$3 A

Resistance used in converting a galvanometer into voltmeter, R = 4.5 k$\Omega$ = 4.5 $\times$ 10³ $\Omega$

To make the voltmeter, the galvanometer and the high resistance (R) have to be connected in series combination as shown below.

Image

$\therefore$ Maximum current in galvanometer is

${I_g} = {E \over {R + {R_g}}}$

$ \Rightarrow E = {I_g}(R + {R_g})$

$ = 4.5 \times {10^{ - 3}} \times (4.5 \times {10^3} + 60)$

$ = 20.5$ V

Given, potential gradient = 0.4 mV cm^$-$1

= 0.4 $\times$ 10^$-$3 V cm^$-$1

Length of wire, l = 10 m = 1000 cm

Electromagnetic force of cell, E = 4 V

Potential gradient along wire = ${{potential\,difference\,along\,wire} \over {length\,of\,wire}}$

$ \Rightarrow 0.4 \times {10^{ - 3}} = {{l \times 40} \over {1000}}$ V cm^$-$1

Let R be the required resistance unplugged in the box.

$\therefore$ Current in wire, $l = {4 \over {400}}A$ ...... (i)

Also, $l = {E \over {R + R'}} = {4 \over {40 + R}}$ ....... (ii)

Equating Eq. (i) and (ii), we get

${4 \over {40 + R}} = {4 \over {400}} \Rightarrow R = 400 - 40 = 360\Omega $