Seebeck effect is used in
(i) Thermoelectric generators that are used in power plants to convert heat in to electricity.
(ii) In automobiles as automotive thermoelectric generators for increasing fuel efficiency.
(iii) Thermocouples and Thermopile are used to measure the temperature difference between two objects

(i) Electrical power P = VI
(ii) Erectrical power \(P=V\left(\frac{V}{R}\right)=\frac{V^{2}}{R} \quad \therefore P=\frac{V^{2}}{R}\)
(iii) P = Iv = I(IR) = I²R
(iv) P = I²R

Devices that follow Ohm's law are called Ohmic conductor. Devices that do not follow Ohm's law are called non - Ohmic conductors.

According to Joule’s heating law H = I² Rt
The current passed through the electrical heater \(\\ \\ \\ \\ =\frac { 220V }{ 10\Omega } =22A\)
The heat produced in one second by the electrical heater H = I² R
The heat produced in one second H = (22)² x 10 = 4840 J = 4.84 k J. In fact the power rating of this electrical heater is 4.84 k W.
The amount of energy to increase the temperature of 1kg water from 30°C to 60°C is
Q = ms ΔT (Refer XI physics vol 2, unit 8)
Here m = 1 kg,
s = 4200 J kg^-1K^–1,
ΔT = 30,
so Q = 1 x 4200 x 30 = 126 kJ
The time required to produce this heat energy \(t=\frac { Q }{ { I }^{ 2 }R } =\frac { 126\times { 10 }^{ 3 } }{ 4840 } \approx 26.03s\)

\(\frac { P }{ Q } =\frac { R }{ S } \), when the network is balanced
\(\frac { 500 }{ 800 } =\frac { x+400 }{ 1000 } \)
\(x+400=\frac { 5 }{ 8 } \times 1000\)
x + 400 = 625 
x = 625 – 400
x = 225 Ω

\(\frac { P }{ Q } =\frac { R }{ S } \)
\(S=\frac { Q }{ P } \times R\)
\(S=\frac { 1000 }{ 100 } \times 40S=400\Omega \)

We can denote the current that flows from 9V battery as I₁ and it splits up nto I₂ and (I₁ – I₂) at the junction E according Kirchoff’s current rule (KCR).
Now consider the loop EFCBE and apply KVR, we get
1I₂ + 3I₁ + 2I₁ = 9
5I₁ + I₂ = 9          (1)
Applying KVR to the loop EADFE, we get
3 (I₁ – I₂ ) – 1I₂ = 6
3I₁ – 4I₂ = 6           (2)
Solving equation (1) and (2), we get
I₁ = 1.83 A and I₂ = -0.13 A
It implies that the current in the 1 ohm resistor flows from F to E.

Thus applying Kirchoff’s second law to the closed loop EACE
I₁R₁ + I₂R₂ + I₃R₃ = ξ
and for the closed loop ABCA
I₄R₄ + I₅R₅ - I₂R₂ = 0

i) Equivalent emf ξ_eq = 5 V
ii) Equivalent internal resistance,
\({ R }_{ eq }=\frac { r }{ n } =\frac { 0.5 }{ 4 } =0.125\Omega \)
iii) total current, \(I=\frac { \xi }{ { R }_{ 5 }+\frac { r }{ n } } \)
\(I=\frac { 5 }{ 10+0.125 } =\frac { 5 }{ 10.125 } \)
I ≈ 0.5 A 
iv) Potential difference across each cell
V = IR = 0.5 x 10 = 5 V
v) Current from each cell, \(I'=\frac { I }{ n } \)
\(I'=\frac { 0.5 }{ 4 } =0.125A\)

Equivalent emf of the combination
ξ_eq = nξ = 4 x 9 = 36 V
ii) Equivalent internal resistance r_eq = nr = 4 x 0.1 = 0.4 Ω
iii) Total current \(I=\frac { n\xi }{ R+nr } \)
\(=\frac { 4\times 9 }{ 10+(4\times 0.1) } \)
\(=\frac { 4\times 9 }{ 10+0.4 } =\frac { 36 }{ 10.4 } \)
I = 3.46 A
iv) Potential difference across external resistance V = IR = 3.46 x 10 = 34.6 V. The remaining 1.4 V is dropped across the internal resistance of cells.
v) Potential difference across each cell \(\frac { V }{ n } =\frac { 34.6 }{ 4 } =8.65V\)

To check which bulb will be fused, the voltage drop across each bulb has to be calculated.
The resistance of a bulb,
\(R=\frac { V^{ 2 } }{ P } =\frac { { (Ratedvoltage) }^{ 2 } }{ Ratedpower } \)
For 20W - 220V bulb,
\({ R }_{ 1 }=\frac { { \left( 220 \right) }^{ 2 } }{ 20 } \Omega =2420\Omega \)
For 100W - 220V bulb,
\({ R }_{ 2 }=\frac { { \left( 220 \right) }^{ 2 } }{ 20 } \Omega =484\Omega \)
Both the bulbs are connected in series. So same current will pass through both the bulbs. The current that passes through the circuit, \(I=\frac { V }{ { R }_{ tot } } \)
R_tot (R₁ + R₂)
R_tot = (484 + 2420) \(\Omega\) = 2904 \(\Omega\)
\(I=\frac { 440V }{ 2904\Omega } \approx 0.151A\)
The voltage drop across the 20W bulb is
\(V_1=IR_1=\frac { 440V }{ 2904 }\times2420 \approx 366.6V\)
The voltage drop across the 100W bulb is
\({ V }_{ 2 }=I{ R }_{ 2 }=\frac { 440 }{ 2904 } 484\approx 73.3A\)
The 20 W bulb will get fused because the voltage across it is more than the voltage rating.

(a) The power delivered by the battery P = VI. Since the bulbs are connected in parallel, the voltage drop across each bulb is the same. If the voltage is kept fixed, then the power is directly proportional to current (P ∝ I). So 60 W bulb draws twice as much as current as 30 W and it will glow brighter than 30 W bulb.
(b) To calculate the resistance of the bulbs, we use the relation \(P=\frac { { v }^{ 2 } }{ R } \) In both the bulbs, the voltage drop is the same, so the power is inversely proportional to the resistance or resistance is inversely proportional to the power \(\left( R∝ \frac { 1 }{ P } \right) \). It implies that, the 30W has twice as much as resistance as 60 W bulb.
(c) When these two bulbs are connected in series, the current passing through each bulb is the same. It is equivalent to two resistors connected in series. The bulb which has higher resistance has higher voltage drop. So 30W bulb will glow brighter than 60W bulb. So the higher power rating does not always imply more brightness and it depends whether bulbs are connected in series or parallel.

Case (a)
To find the equivalent resistance between the points a and b, we assume that current is entering the junction at a. Since all the resistances in the outside loop are the same (1Ω), the current in the branches ac and ad must be equal.  Hence the points C and D are at the same potential and no current through 5 Ω. It implies that the 5 Ω has no role in determining the equivalent resistance and it can be removed. So the circuit is simplified as shown in the figure.

The equivalent resistance of the circuit between a and b is R_eq = 1Ω

In all the sections, the resistors are connected in parallel.
Section I
\(\frac { 1 }{ { R }_{ { P }_{ 1 } } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ { R }_{ { P }_{ 1 } } } =\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =\frac { 2 }{ 2 } \quad { R }_{ { p }_{ 1 } }=1\Omega \)

Section II
\(\frac { 1 }{ { R }_{ { P }_{ 1 } } } =\frac { 1 }{ 4 } +\frac { 1 }{ 4 } =\frac { 2 }{ 4 } ,\quad \frac { 1 }{ { R }_{ { P }_{ 2 } } } =\frac { 1 }{ 2 } ,{ R }_{ { p }_{ 2 } }=2\Omega \)

Section III
\(\frac { 1 }{ { R }_{ { P }_{ 3 } } } =\frac { 1 }{ 6 } +\frac { 1 }{ 6 } =\frac { 2 }{ 6 } \)
\(\frac { 1 }{ { R }_{ { P }_{ 3 } } } =\frac { 1 }{ 3 } ,{ R }_{ { p }_{ 3 } }=3\Omega \)
Equivalent resistance is given by
R = Rp₁ + Rp₂ + Rp₃
R = 1 Ω + 2 Ω + 3 Ω = 6 Ω
The circuit became,

Equivalent resistance between A and B is

R_s = R₁ + R₂ = 15 Ω           (1)
\({ R }_{ p }=\frac { { R }_{ 1 }{ R }_{ 2 } }{ { R }_{ 1 }{ +R }_{ 2 } } =\frac { 56 }{ 15 } \Omega \quad \) (2)
From equation (1) substituting for R₁ + R₂ in equation (2)
\(\frac { { R }_{ 1 }{ R }_{ 2 } }{ 15 } =\frac { 56 }{ 15 } \Omega \)
∴ R₁R₂ = 56
\({ R }_{ 2 }=\frac { 56 }{ 15 } \Omega \)         (3)
Substituting for R₂ in equation (1) from equation (3)
\({ R }_{ 1 }+\frac { 56 }{ { R }_{ 1 } } =15\)
Then, \(\frac { { R }_{ 1 }^{ 2 }+56 }{ { R }_{ 1 } } =15\)
R₁² + 56 = 15 R₁
R₁² - 15 R₁ + 56 = 0
The above equation can be solved using factorisation.
R₁ = 8 Ω (or) R₁ = 7 Ω
If (R₁ = 8 Ω)
Substituting in equation (1)
8 + R₂ = 15
R₂ = 15 – 8 = 7 Ω ,
R₂ = 7 Ω i.e , (when R₁ = 8 Ω ; R₂ = 7 Ω)
If R₁= 7 Ω
Substituting in equation (1)
7 + R₂ = 15
R₂ = 8 Ω , i.e , (when R₁ = 7 Ω ; R₂ = 8 Ω )

Since the resistances are connected in parallel, therefore, the equivalent resistance in the circuit is
\(\frac { 1 }{ { R }_{ p } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } =\frac { 1 }{ 4 } +\frac { 1 }{ 6 } \)
\(\frac { 1 }{ { R }_{ p } } =\frac { 5 }{ 12 } \Omega \quad or\quad { R }_{ p }=\frac { 12 }{ 5 } \Omega \)
The resistors are connected in parallel, the potential difference (voltage) across them is the same.
\({ I }_{ 1 }=\frac { V }{ { R }_{ 1 } } =\frac { 24V }{ 4\Omega } =6A\)
\({ I }_{ 2 }=\frac { V }{ { R }_{ 2 } } =\frac { 24 }{ 6 } =4A\)
The current I is the sum of the currents in the two branches. Then,
I = I₁ + I₂ = 6 A + 4 A = 10 A

Since the resistors are connected in series, the effective resistance in the circuit
= 4 Ω + 6 Ω = 10 Ω
The Current I in the circuit =\(\frac { V }{ { R }_{ eq } } =\frac { 24 }{ 10 } =2.4A\)
Voltage across 4Ω resistor
V₁= IR₁ = 2.4A x 4Ω = 9.6V
Voltage across 6 Ω resistor
V₂ = IR₂ = 2.4A x 6Ω = 14.4V

In the first case, the resistance of the block
\({ R }_{ AB }=\rho \frac { length }{ Area } =\rho \frac { C }{ AB } \)
The current \({ I }_{ AB }=\frac { V }{ { R }_{ AB } } =\frac { V }{ \rho } .\frac { AB }{ C } \quad (1)\)
In the second case, the resistance of the block \({ R }_{ BC }=\rho \frac { A }{ BC } \)
The current \({ I }_{ BC }=\frac { V }{ { R }_{ BC } } =\frac { V }{ \rho } .\frac { BC }{ C } \quad (2)\)
To express I_BC interms of I_AB, we multiply and divide equation (2) by AC, we get
\({ I }_{ BC }=\frac { V }{ \rho } .\frac { BC }{ A } \frac { AC }{ AC } =\left( \frac { V }{ \rho } .\frac { AB }{ C } \right) .\frac { { C }^{ 2 } }{ { A }^{ 2 } } =\frac { { C }^{ 2 } }{ { A }^{ 2 } }.{ I }_{ AB }\)
Since C > A, the current I_BC > I_AB

R₁ = 20 Ω, R₂ = ?
Let the original length of the wire (l₁) be l.
New length, l₂ = 8l₁ (i.,e) l₂ = 8l
Original resistance, R₁ = \(\rho \frac { { l }_{ 1 } }{ { A }_{ 1 } } \)
New resistance R₂ = \(\rho \frac { { l }_{ 2 } }{ { A }_{ 2 } } =\frac { \rho (8l) }{ { A }_{ 2 } } \)
Though the wire is stretched, its volume is unchanged.
Initial volume = Final volume
A₁l₁ =  A₂l₂ , A₁l = A₂(8l)
\(\frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { 8l }{ l } =8\)
By dividing equation R₂ by equation R₁, we get
\(\frac { { R }_{ 2 } }{ { R }_{ 1 } } =\frac { \rho (8l) }{ { A }_{ 2 } } \times \frac { { A }_{ 1 } }{ \rho l } \)
\(\frac { { R }_{ 2 } }{ { R }_{ 1 } } =\frac { { A }_{ 1 } }{ { A }_{ 2 } } \times 8\)
Substituting the value of \(\frac { { A }_{ 1 } }{ { A }_{ 2 } } \), we get
\(\frac { { R }_{ 2 } }{ { R }_{ 1 } } =8\times 8=64\)
R₂ = 64 x 20 = 1280 Ω
Hence, stretching the length of the wire has increased its resistance.