(i) 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 the Peltier effect.
(ii) In the Cu-Fe thermocouple the junctions A and B are maintained at the same temperature.
(iii) Let a current from a battery flow through the thermocouple. At junction A, where the current flows from Cu to Fe, heat is absorbed and junction A becomes cold.
(iv) At junction B, where the current flows from Fe to Cu heat is liberated and it becomes hot.

Peltier effect: Cu - Fe thermocouple
(v) When the direction of current is reversed, junction A gets heated and junction B gets cooled as shown in Figure (b).
Hence Peltier effect is reversible.

If a current I flows through a conductor kept across a potential difference V for a time t, the work done or the electric potential energy spent is W = VIt
In the absence of any other external effect, this energy is spent on heating the conductor. The amount of heat(H) produced is H = VIt
For a resistance R,
H = I² Rt
This relation was experimentally verified by Joule and is known as Joule's law of heating. 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.

(i) Suppose n cells, each of emf \(\xi \) volts and internal resistance r ohms are connected in series with an external resistance R.
(ii) The total emf of the battery = \(n\xi \) The total resistance in the circuit = nr + R By Ohm's law, the current in the circuit is 
\(I=\cfrac { totalemf }{ taoal\ resistance } =\cfrac { n\xi }{ nr+R } \)
\(I=\cfrac { n\xi }{ R } =n{ l }_{ 1 }\)
(iii) where II is the current due to a single cell
\(\left( { I }_{ 1 }=\cfrac { \xi }{ R } \right) \)
Thus, if r is negligible when compared to R the current supplied by the battery is n times that supplied by a single cell.
Case (b) If r >> R,\(I=\cfrac { n\xi }{ nr } =\cfrac { \xi }{ r } \)
(iv) It is the current due to a single cell. That is, current due to the whole battery is the same as that due to a single cell and hence there is no advantage in connecting several cells.
(v) Thus series connection of cells is advantageous only when the effective internal resistance of the cells is negligibly small compared with R.

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.

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

The conductivity of a semiconductor increases with rise in temperature. When temperature increases, a large number of covalent bondsbreak. This produces a larger number of charge carriers. So current increases.

The bridge wire is soldered at the ends of the copper strips. Due to imperfect contact, some resistance, might be introduced at the contact. These are called end resistances. This error can be eliminated, if another set of readings are taken with P and Q interchanged and the average value of P is found.

Several cells can be connected to form a battery. In a series connection, the negative terminal of one cell is connected to the positive terminal of the second cell, the negative terminal of second cell is connected to the positive terminal of the third cell, and so on. The free positive terminal of the first cell and the free negative terminal of the last cell become the terminals of the battery.

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

(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

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).

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

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

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

\(\mathrm{V}=230 \mathrm{~V}  \)
\(P=\frac{V^2}{R} \text { since } \mathrm{V} \text { is same } \mathrm{P} \propto \frac{1}{R}  \)
\(\frac{1}{R_2}=\frac{1}{\frac{R_1}{2}}+\frac{1}{\frac{R_1}{2}}=\frac{2}{R_1}+\frac{2}{R_1}=\frac{2+2}{R_1}  \)
\(\frac{1}{R_2}=\frac{4}{R_1}  \)
\(\therefore R_2=\frac{R_1}{4}  \)
\(R_1=4 R_2  \)
\(\therefore \frac{P_2}{P_1}=\frac{R_1}{R_2}=\frac{4 R_2}{R_2}=4\)

Resistance, \(R=\frac{V}{I}=\frac{4}{2}=2 \ ohm\)

(i) \(mobility=\cfrac { Drift\ velocity }{ electric\ field } \) (or) \(\mu =\cfrac { { V }_{ d } }{ E } \)
(ii) \({ \mu }_{ d }=\cfrac { eE }{ mL } .\tau \) (or) \(\cfrac { { v }_{ d } }{ E } =\left( \cfrac { e }{ mL } \right) .\tau \)
\(\mu =\cfrac { e }{ mL } .\tau \)

(i) Drift velocity \({ u }_{ d }=\cfrac { ev }{ ml } .\tau \)
When I am doubled, drift velocity, become \(\cfrac { 1 }{ 2 } \) times the original v_d
(ii) Resistor \(R=\rho .\cfrac { l }{ A } \)
Resistor becomes doubled i.e. 2 times the original resistors.
(iii) Resistivity is not affected.