Balancing lengths are
1₁ = 35 cm, l₂ = 63 cm
E₁ = 1.25v E₂ = ?
At the following point \(E \propto 1\)
\(\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}} \)
\(\therefore \frac{1.25}{E_{2}}=\frac{35}{63} \quad \therefore E_{2}=\frac{1.25 \times 63}{35}=2.25 \mathrm{v} \)
∴ EMF of the second cell = 2.25 V
V_d = 0.03 x 10^-3 m s^-1

 

Circuit Diagram:
Here, 2 cells are in series,
\(\therefore \varepsilon_{\mathrm{tot}} =\varepsilon+\varepsilon=2 \varepsilon  \)
\(\varepsilon_{\mathrm{tot}}  =10 \mathrm{~V}\)
Here 4, 6 and 12 are in parallel
\(\therefore \frac{1}{\mathrm{R}_{\mathrm{p}}}  =\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_1}=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}  \)
\(\mathrm{R}_{\mathrm{p}}  =2 \Omega\)
Now, the circuit becomes,
(i) current drawn from the cell (through the circuit) is,
\(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{s}}}=\frac{10}{8+2}=1 \mathrm{~A}\)
Potential drop across the parallel combination of 3 resistors is \(\mathrm{V}^{\prime}=1 R_P=1 \times 2=2 \mathrm{~V}\)
(ii) Current through 8 resistor is I =1 A
\((\because 8 \Omega, 2 \Omega \text { in series) }\)
Current through \(\mathrm{R}=4 \Omega \ is, \mathrm{I}=\frac{\mathrm{V}^{\prime}}{\mathrm{R}}=\frac{2}{4}=0.5 \mathrm{~A}\)
Current through \(\mathrm{R}=6 \Omega \ is, \mathrm{I}=\frac{\mathrm{V}^1}{\mathrm{R}}=\frac{2}{6}=0.33 \mathrm{~A}\)
Current through \(\mathrm{R}=12 \Omega \ is, \mathrm{I}=\frac{\mathrm{V}^1}{\mathrm{R}}=\frac{2}{12}=0.17 \mathrm{~A}\)

Applying Kirchhoff's current rule at P we get I₁ + I₂ = 2 ....(1)
Applying Kirchhoff's voltage rule for the closed loop PQSA we get,
5I₁ + 10I_g - 15I₂ = 0
Dividing the equation by 5 we get,
I₁ + 2I_g - 3I₂ = 0   ...(2)
Applying Kirchhoff's voltage rule for the closed loop QRSQ, we get,
\(\left(I_{1}-I_{g}\right) 10-\left(I_{2}+I_{g}\right) 20-10 I_{g}=0 \)
\(\therefore 10 I_{1}-10 I_{g}-20 I_{2}-20 I_{g}-10 I_{g}=0 \)
\(10 I_{1}-20 I_{2}-40 I_{g}=0 \)   ....(3)
Dividing Eqn 3 by 10, we get,
\(I_1-2 I_2-4 I_s=0\)   ....(4)
Solving equtions (4) and (2)
\(\begin{array}{l} I_1-4 I_g-2 I_2=0 \\ I_1+2 I_g-3 I_2=0 \\ (-)(-)(+) \\ \hline-6 I_g+I_2=0 \end{array}\)    .......(5) 
From ∴ I₂ = 6I_g    ....(6)
From equation (1), I₂ = 2 - I₁    ....(7)
From equation 6 & 7, we get,
2 - I₁ = 6I_g
∴ -I₁ = 6I_g - 2
∴ I₁ = 2 -  6I_g .....(8)
Substituting equations (6) and (8) in equation (2), we get
\(2-6 I_{g}+2 I_{g}-3\left(6 I_{g}\right)=0 \)
\(2-6 I_{g}+2 I_{g}-18 I_{g}=0 \)
\(2-22 I_{g}=0\)
\(\therefore-22 I_{g}=-2 \)
\(\therefore I_{g}=\frac{-2}{-22} \)
\(I_{g}=\frac{1}{11} \mathbf{A} \)

Length of the potentiometer wire = 4 m,
External resistance R = 2980 \(\Omega\), EMF(E) = 4V
Potentiometer resistance R_p = 20 \(\Omega\)
∴ Resistance per unit length \(=\frac{20}{4}=5 \Omega\)
∴ Total resistance of potentiometer wire R = R + R_p = 2980 + 20 = 3000 \(\Omega\)
Current, \(I=\frac{E}{R+R_{p}}=\frac{E}{R_T}=\frac{4}{3000}\)
\(I=\frac{4}{3} \times 10^{-3} A\)
Potential along the wire, V = IR
\(=\frac{4}{3} \times 10^{-3} \times 5 =\frac{20}{3} \times 10^{-3} \)
\(V=6.66 \times 10^{-3} \mathrm{~V} \mathrm{m}^{-1}=0.666 \times 10^{-2} \mathrm{~V}/M \)
∴ Potential gradient along the wire = 0.666 x 10^-2 V/m

Apply Kirchoff's 1^st law at junction B
I_I - I₂ - I₃ = 0
I₃ = I₁ - I₂          ....(1)
Applying Kirchoff's II^nd law at path ABEFA,
100I₃ + 100 I₁ = 15   ....(2)
Substitute equation (1) in equation (2),
100 (I₁ - I₂) + 100 I₁ = 15
100 I₁ - 100 I₂ + 100 I₁ = 15
200 I₁ - 100 I₂ = 15  ....(3)
Apply Kirchhoffs 2^nd law ar path BCDEB,
100 I₂ - 100 I₃ = -9    ....(4)
Substitute equation (1) in above equation (4),
100 I₂ - 100(I₁ - I₂) 100 = -9
100 I₂ - 100 I₁ + 100 I₂ = -9
200 I₂ - 100 I₁ = - 9 ....(5)
Solve equation (3) and (5), we get
Eqn (3)       ⇒ 200 I₁ - 100 I₂ = 15
2 x Eqn (5) ⇒ -200 I₁ + 400I₂ = -18
                        ________________
                                      300 I₂ = -3
I₂ = -3 /300 = -0.01 A
I₂ = -0.01 A
Substitute I₂ value in equation (5)
200 I₂ -100 I₁ = - 9
200 x (- 0.01) - 100 I₁ = - 9
- 2 - 100 I₁ = -9
-100 I₁ = -9 + 2
-100 I₁ = -7
I₁ = 7/100
I₁ = 0.07A
Substitute I₁, I₂ value in equation (1)
I₃ = I₁ - I₂
= 0.07 + 0.01
I₃ = 0.08 A
I₁ = 0.07 A, I₂ = - 0.01 A and I₃ = 0.08 A

Case I : Current from a cell is I₁ = 0.9 A
Resistance R₁ = 2\(\Omega \)
Case II : Current flowing I₂ = 0.3 A
Resistance R₂ = 7\(\Omega \)
E = I(R + r) = IR + Ir
\(\therefore r=\frac{E}{I}-R\)
By applying for two cases we get,
\(\frac{E}{0.9}-2=\frac{E}{0.3}-7 \)
\(\frac{E}{0.9}-\frac{E}{0.3}=7+2=-5 \)
\(-\frac{2 E}{0.9}=-5 \)
\(E =\frac{5 \times 0.9}{2}=\frac{4.5}{2}=2.25 V \)
∴ Internal resistance, \(r=\frac{E}{I}-R\)
Where I = 0.3 A, R = 7\(\Omega\), E = 2.25 V
\(\therefore r=\frac{2.25}{0.3}-7=\frac{22.5}{3}-7\)
\(r=7.5-7=0.5 \Omega \)
∴ Internal resistance = 0.5 \(\Omega\)

Requirement of resistors = \(150\Omega \)
Available resistances are \(R_{ 1 }=220\Omega \quad { R }_{ 2 }=79\Omega \ and \ { R }_{ 3 }=92\Omega \)

\(\frac{1}{R_{p}} =\frac{1}{R_{1}}+\frac{1}{R_{2}} \)
\(\therefore \frac{1}{R_{p}} =\frac{1}{220}+\frac{1}{79} \)
\(\frac{1}{R_p}=\frac{79+220}{17380}=\frac{299}{17380} \)
\(\therefore R_{p} =\frac{17380}{299}=58.12709 \simeq 58 \Omega \)
R_{Available Resistance} = R_r+ R₃
= 58 + 92 \(=150\Omega \)
Parallel combination of 220 \(\Omega \) and 79 \(\Omega \) in series with 92\(\Omega \).

Resistivity of the first material \({ \rho }_{ 1 }=4\times { 10 }^{ -3 }\Omega m\)
Length of the first material, l₁ = 25 x 10² m   
Area of cross section of the first material, A₁ = 3 x 3 x 10^-6 
= 9 x 10^-6 m²
Resistivity of second material, \({ \rho }_{ 2}=5\times { 10 }^{ -3 }\Omega m\) 
Length of second material, \(l_{ 2 }=70\times { 10 }^{ -2 } m\) 
Area of cross section second material, A₂ = 3 x 3 x 10^-6
= 9 x 10^-6 m²
Resistivity of first material, \(\rho_{1}= R_{1} \times \frac {A_{1}}{l_{1}}\)  
Resistance first material, \(R_{1} = \frac {\rho_{1}l_1}{A_{1}}\) 
\(\therefore R_1=\frac{4 \times 10^{-3} \times 25\times 10^{-2}}{9 \times10^{-6}}\)
\(R_1=\frac{100}{9}\times10^{-5+6}=\frac{1000}{9}\Omega\)
Resistivity of second material, \(\rho_{2}=\frac{R_{2} A_{2}}{l_{2}} \)
∴ Resistance of second material, \(R_{2}=\frac{\rho_{2} l_{2}}{A_{2}} \)
\(\therefore R_{2} =\frac{5 \times 10^{-3} \times 70 \times 10^{-2}}{9 \times 10^{-6}} \)
\(R_2=\frac{350 \times 10^{-5}}{9 \times 10^{-6}}=\frac{350}{9} \times 10 \)
\(R_2=\frac{3500}{9} \Omega \)
Total resistance \(R_{t} =\frac{1000}{9}+\frac{3500}{9} \)
\(=\frac{4500}{9}=500 \Omega \)
\(\therefore\) Total resistance = 500 \(\Omega\)

At T_o = 20^oC, resistance R₀ = 10 \(\Omega \)
\(\alpha\) = 0.004/^oC, At T_o = 100^oC  R₁₀₀ = ? (at boiling point of water)
R_T = R₀[1 + \(\left.(\alpha( T-T_{0}\right))\)]
 R₁₀₀ = 10 [1 + (0.004 x (100 - 20))]
= 10 [1 + 0.32] = 10 x 1.32 = 13.2 \(\Omega \)
∴ Resistance at boiling point of water R₁₀₀ = 13.2 \(\Omega \)
(i.e) R_r = 13.2 \(\Omega \)
Comment : As the temperature increases, the resistance of the wire also increases.

Area of cross section  A = 10^-6 m² Current I = 2A
Number of electrons per cubic metre, n = 8 \(\times\)10²⁸
Current I = nAev_d
∴ Drift Velocity \(v_{d}=\frac{I}{n A e}\)
\(v_{d} =\frac{2}{8 \times 10^{28} \times 10^{-6} \times 1.6 \times 10^{-19}} \)
\(=\frac{2}{8 \times 10^{3} \times 1.6} \)
\(=\frac{20}{128} \times 10^{-3}=15.6 \times 10^{-5} \mathrm{~ms}^{-1} \)
Current density J = l / A.
\(J=\frac{2}{10^{-6}}\)
J = 2 x 10⁶ Am^-2

Emergency transferred, E = 10⁹ J
Potential difference, V = 5\(\times\)10⁷V
Time t = 0.2s.
(c) Power \(=\frac{\text { Energy }}{\text { Time }}=\frac{1 \times 10^{9}}{0.2}\) = 5 x 10⁹ W
P = 5 GW
(b) Current \(I =\frac{\text { Power }}{\text { Potiential difference }} \) \(=\frac{5 \times 10^{9}}{5 \times 10^{7}} \)
I = 10² = 100 A
∴ Q = 20C
(a) Charge Q = It   (Here, I = 100 A)
Q = 100 x 0.2 = 20 C
(a) Q = 20 C, (b) I = 100 A, (c) P = 5 GW

Resistance of the conductor, \(\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}\)
From Graph - I,
Resistance of the conductor A is, \(\mathrm{R}_{\mathrm{A}}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{2}{4}=0.5 \Omega\)
Resistance of the conductor B is, \(\mathrm{R}_{\mathrm{B}}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{4}{3}=1.33 \Omega\)
Resistance of the conductor C is, \(\mathrm{R}_{\mathrm{c}}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{5}{2}=2.5 \Omega\)
From Graph - II,
Resistance of the conductor D is, \(\mathrm{R}_{\mathrm{D}}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{4}{2}=2 \Omega\)
Resistance of the conductor E is, \(\mathrm{R}_{\mathrm{E}}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{3}{4}=0.75 \Omega\)
Resistance of the conductor F is, \(\mathrm{R}_{\mathrm{F}}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{2}{5}=0.4 \Omega\)
The conductor F has least resistance, \(R_F=\mathbf{0. 4} \Omega\)
The conductor C has maximum resistance, \(\mathbf{R}_{\mathrm{c}}=2.5 \Omega\)

Peltier discovered that, when an electric current is passed through a circuit of a thermocouple heat is evolved at one junction and absorbed at the other junction. This is known as Peltier effect.

Thomson showed that, if two points in a conductor are at different temperatures, the density of electrons at these points will differ and as a result the potential difference is created between these points.

Seebeck discovered that in a closed circuit consisting of two dissimilar metals, when the junctions are maintained at different temperature an emf is developed.

It states that the heat developed in an electrical circuit due to the flow of current varies directly as
(i) the square of the current
(ii) the resistance of the circuit and
(iii) the time of flow.

The internal resistance of a cell is the resistance offered to the flow of current (by the electrolyte) inside the cell.

The emf of the cell is directly proportional to the balancing length.

It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit.

It states that the algebraic sum of the currents at any junction of a circuit is zero.

Electric power is the rate at which the electrical potential energy is delivered
\(P =\frac{d U}{d t} \)
\(P =\frac{VdQ}{d t}=\mathrm{V} \frac{d Q}{d t} \)
Since \(\frac{d Q}{d t}=I\), where I - electric current
∴ P = VI

R = 10 Ω, I = 5 A, t = 5 minutes = 5 x 60 s
H = I² R t
= 5² x 10 x 5 x 60
= 25 x 10 x 300
= 25 x 3000
= 75000 J (or) 75 kJ

Q = 10 Ω
\(\frac { P }{ Q } =\frac { { l }_{ 1 } }{ 100-{ l }_{ 1 } } =\frac { { l }_{ 1 } }{ { l }_{ 2 } } \)
\(P=Q \times \frac{l_1}{100-l_1}\)
\(P=\frac { 10\times 55 }{ 100-55 } \)
\(P=\frac { 550 }{ 45 } =12.2\Omega \)

Q = 15 Ω, l₁: l₂ = 3 : 2
\(\frac { { l }_{ 1 } }{ { l }_{ 2 } } =\frac { 3 }{ 2 } \)
\(\frac { P }{ Q } =\frac { { l }_{ 1 } }{ { l }_{ 2 } } \)
\(P=Q\frac { { l }_{ 1 } }{ { l }_{ 2 } } \)
\(P=15 \times \frac { 3 }{ 2 } =22.5\Omega \)

Applying Kirchoff’s rule to the point P in the circuit,
The arrows pointing towards P are positive and away from P are negative.
Therefore, 0.2A - 0.4A + 0.6A - 0.5A + 0.7A - I = 0
0.7 A - I = 0
1.5A - 0.9A – I = 0
0.6A - I = 0
I = 0.6 A

The given values I = 3.93 A, ξ = 12 V,
R = 3 Ω
(a) The terminal voltage of the battery is equal to voltage drop across the resistor
V = IR = 3.93 x 3 = 11.79 V
The internal resistance of the battery,
\(r=\left[ \frac { \xi -V }{ V } \right] R=\left[ \frac { 12-11.79 }{ 11.79 } \right] \times 3=0.05\Omega \)
(b) The power delivered by the battery P = Iξ = 3.93 x 12 = 47.1 W
The power delivered to the resistor = I² R = 46.3 W
The remaining power = (47.1 – 46.3) = 0.8 W is delivered to the internal resistance and cannot be used to do useful work. (it is equal to I² r).

T₀ = 20^oC, T = 40^oC, R_o = 45 Ω , R = 85 Ω
\(\alpha =\frac { 1 }{ { R }_{ 0 } } \frac { \Delta R }{ \Delta T } \)
\(\alpha=\frac{1}{45}\left(\frac{85-45}{40-20}\right)=\frac{1}{45}(2)\)
\(\alpha=0.044 \text { per }^{\circ} C\) 

R₀ = 3 Ω, T = 100^oC, T₀ = 20^oC
α = 0.004/^oC, R_T = ?
R_T = R₀(1 + α(T - T₀))
R₁₀₀ = 3(1 + 0.004 x 80)
R₁₀₀ = 3.96 Ω

V = 12 V and R = 24 Ω
Current, I = ?
From Ohm’s law, \(I=\frac{V}{R}=\frac{12}{24}=0.5A\)

I = 32 A , t = 1 s
Charge of an electron, e = 1.6 x 10^-19 C
The number of electrons flowing per second, n = ?
\(I=\frac { q }{ t } =\frac { ne }{ t } \)
\(n=\frac { It }{ e } \)
\(n=\frac { 32\times 1 }{ 1.6\times { 10 }^{ -19 }C } \)
n = 20 x 10¹⁹ = 2 x 10²⁰ electrons

The relation between drift velocity of electrons and current in a wire of cross- sectional area A is
\({ v }_{ d }=\frac { I }{ neA } =\frac { 0.2 }{ 8.4\times { 10 }^{ 28 }\times 1.6\times { 10 }^{ -19 }\times 0.5\times { 10 }^{ -6 } }\)
v_d = 0.03 x 10^-3 m s^-1

E = 570 N C^-1, e = 1.6 x 10^-19 C,
m = 9.11 x 10^-31 kg and a = ?
F = ma = eE
\(a=\frac { eE }{ m } =\frac { 570\times 1.6 \times { 10 }^{ -19 } }{ 9.11\times { 10 }^{ -31 } } \)
\(=\frac { 912\times { 10 }^{- 19 }\times { 10 }^{ 31 } }{ 9.11 } \)
= 1.001 x 10¹⁴ ms^-2

Current density is J in a conductor is defined as the current flowing per unit area of cross-section of the conductor
\(\therefore J=\cfrac { I }{ A } \)
Where I - current
A - area of cross section

Electric power:
(i) The electric power P is the rate at which the electrical potential energy is delivered.
(ii) The electric power P is the rate at which the work is done.
\( P= \frac{d U}{d t} (or)= \frac{d W}{d t} (or)=VI(or)\frac{V^2}{R}\)
Unit: watt (W)
Electric energy:
(i) The electric energy is the product of power (P) and duration of the time (t) when electric energy is delivered.
(ii) E = Pt
Unit: watt-hour (Wh)

A superconductor is any material that can conduct electricity with no resistance. In most cases, materials such as metallic elements or compounds offer some resistance at room temperature, but offer less resistance at a temperature known as its critical temperature.

Temperature coefficient of resistance is defined as the ratio of increase in resistivity per degree rise in temperature to its resistivity at T_o.
\(\alpha=\frac{\rho_{T}-\rho_{0}}{\rho_{0}\left(T-T_{0}\right)}\)

Electrical resistivity of a material is defined as the resistance offered to current flow by a conductor of unit length having unit area of cross section.

Macroscopic form of Ohm's law is V = IR
Where V - Potential difference
I - current 
R - resistance of a conductor.

Microscopic form of ohm's law is 
\(\vec{J}=\sigma\vec {E}\) 
\(J=\frac{ne^2\tau}{m}\vec{E}\)
\(\frac{e \tau}{m} \rightarrow Drift \ velocity \ v_d\)
where \(\vec{J} \) - current density
\(\sigma \)  - conductivity
\(\vec{E} \) - Electric field

Current is defined as the ratio of the net (i.e. amount of) charge (Q) passing through any cross section of a conductor to time.
\(I=\frac{Q}{t}\)
Since current is the ratio of two scalar quantities, it is a scalar. In addition current I is defined as the scalar product of the current density and area vector at which the charges cross.
I = \(\vec{J}.\vec{A}\)

The current (rate of flow of charge) in the wire is 
\(I=\frac { Q }{ t } =\frac { 120 }{ 60 } =2A\)