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.

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

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

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.

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 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

Resistivity ∝ temperature for conductor. so, copper → decreases
Resistivity ∝\(\frac{1}{\text {temperature for semiconductor}}\)
so, germanium → increases.

R_t = R_o [1 + α(T₂ - T₁)
2 = 1[1 + 0.00125(T₂ - 293)
I = 5/4 x 10^-3 (T₂ - 293)
T₂ = 1093 K
T₂ = 820 ^oC

\(\mathrm{V}_1  =220 \mathrm{~V}, \quad \mathrm{P}_1=60 \mathrm{~W}  \)
\(\mathrm{~V}_{\mathrm{U}}  =110 \mathrm{~V}, \mathrm{P}_{\mathrm{U}}=60 \mathrm{~W}  \)
\(P  =\frac{V^2}{R} \Rightarrow R=\frac{V^2}{P}  \)
\(\therefore R_l  =\frac{V_I^2}{P_l} \text { Similarly, } \quad \mathrm{R}_U=\frac{V_U^2}{P_U}  \)
\(R_l  =\frac{220 \times 220}{60} \quad \mathrm{R}_{\mathrm{U}}=\frac{110 \times 110}{60}  \)
\(R_I  =\frac{48400}{60} \quad R_U=\frac{12100}{60}  \)
\(\frac{R_U}{R_l}  =\frac{12100}{60} \times \frac{60}{48400}=\frac{1}{4}  \)
\(R_U  =\frac{R_l}{4}=\frac{R}{4}\)

Brown - 1
Black - 0
Yellow - 10⁴
(∴ R = 10 x 10⁴ Ω = 100 kΩ)

\(P=\frac{V^2}{R}=\frac{240 \times 240}{120}=480 \ W\)