Current produces thermal energy, thermal energy may also be suitably used to produce an electromotive force. This is known as thermoelectric effect.

The resistanct is defined as the ratio of potential difference across the given conductor to the current passing through the conductor.
\(R=\cfrac { V }{ I } \)

If a net charge Q passes through any cross section of a conductor in time t, then the current is defined as \(I=\cfrac { Q }{ t } \)

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.

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.

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

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

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

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

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 = \(\vec{J}.\vec{A}\)

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.

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.

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

\(R \propto \frac{1}{A}, R \propto \frac{1}{r^2}  \)
\(R_A \propto \frac{1}{r_A^2}, R_B \propto \frac{1}{r_B^2} \)
\(\frac{r_A}{r_B}=\left(\frac{R_B}{R_A}\right)^{1 / 2}=\left(\frac{R_B}{3 R_B}\right)^{1 / 2}=\frac{1}{3^{\frac{1}{2}}}=\frac{1}{\sqrt{3}}\)

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

(i) The ohm's law can be derived from the equation \(J=\sigma E\) Consider a segment of wire of length I and cross-sectional area A as shown in Figure.

(ii) When a potential difference V is applied across the wire, a net electric field is created in the wire which constitutes the current in the wire.
(iii) For simplicity, we assumed that the electric field is uniform in the entire length of the wire, the potential difference (voltage V) can be written as V = EI
(iv) As we know, the magnitude of current density
\(J=\sigma E=\sigma \cfrac { V }{ l } \)   
(v) But \(J=\cfrac { I }{ A } \), so we write the equation as,
\(\cfrac { I }{ A } =\sigma \cfrac { V }{ l } \)       
(vi) By rearranging the above equation we get,
\(V=I\left( \cfrac { I }{ \sigma A } \right) \)  
(vii) The quantity \(\cfrac { l }{ \sigma A } \)is called resistance of the conductor and it is denoted as R. Note that the resistance is directly proportional to the length of the conductor and inversely proportional to area of cross-section.
(viii) Therefore, the macroscopic form of ohm's law can be stated as V = IR.