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

(i) The emf of cell ε is measured by connecting a high resistance voltmeter across it without connecting the external resistance R.
(ii) Since the voltmeter draws very little current for deflection, the circuit may be considered as open.
(iii) Hence the voltmeter reading gives the emf of the cell.
(iv) Then, external resistance R is included in the circuit and current I is established in the circuit.
(v) The potential difference across R is equal to the potential difference across the cell V.

(vi) The potential drop across the resistor R is,
v = IR .........(1)
(vii) Due to internal resistance r of the cell, the voltmeter reads a value V, which is less than the emf of cell ε. It is because, certain amount of voltage (Ir) has dropped across the internal resistance r.
Then V = \(\varepsilon\) - Ir
Ir = \(\varepsilon\) - v  ......(2)
(viii) Dividing equation (2) by equation (1), we get
\(\frac{I r}{I R}=\frac{\varepsilon-V}{V} \)
\(r=\left|\frac{\varepsilon-V}{V}\right| R \)
(ix) Since \(\varepsilon\), V and R are known, internal resistance r can be determined.

Resistors in series:
(i) When two or more resistors are connected end to end, they are said to be in series. The resistors could be simple resistors or bulbs or heating elements or other devices. Figure (a) shows three resistors R₁, R₂ and R₃ connected in series.
(ii) The amount of charge passing through resistor R₁ must also pass through resistors R₂ and R₃ since the charges cannot accumulate anywhere in the circuit. Due to this reason, the current I passing through all the three resistors are the same.

(iii) According to Ohm's law, if same current pass through different resistors of different values, then the potential difference across each resistor must be different. Let V_1, V₂ and V₃ be the potential difference (voltage)across each of the resistors R₁, R₂ and R₃ respectively, then we can write V₁ = IR₁, V₂= RI₂ and V₃ = IR₃. But the total voltage V is equal to the sum of voltages across each resistor.
V = V₁ + V₂ + V₃ = IR₁+ IR₂ + IR₃ 
V = I (R₁ + R₂ + R₃)
V = IR_S 
where R_s is the equivalent resistance,
R_S = R₁ + R₂ + R₃
(iv) When several resistances are connected in series, the total or equivalent resistance is the sum of the individual resistances.
Note: The value of equivalent resistance in series connection will be greater than each individual resistance.
Resistors in parallel:
(i) Resistors are in parallel when they are connected across the same potential difference as shown in figure (a).
(ii) In this case, the total current I that leave the battery split into three separate components.
Let I₁, I₂ and I₃ be the current through the resistors R₁, R₂, and R₃ respectively. Due to the conservation of charge, total current in the circuit I is equal to sum of the currents through each of the three resistors.
I = I₁ + I₂ + I₃  .....(1)
(iii) Since the voltage across each resistor is the same, applying Ohm's law to each resistor, we have
\({ I }_{ 1 }=\cfrac { V }{ { R }_{ 1 } } ,{ I }_{ 2 }=\cfrac { V }{ { R }_{ 2 } } ,{ I }_{ 3 }=\cfrac { V }{ { R }_{ 3 } } \) 
Substituting these values in equation (1), we get,
\({ I }_{ 1 }=\cfrac { V }{ { R }_{ 1 } } +\cfrac { V }{ { R }_{ 2 } } +\cfrac { V }{ { R }_{ 3 } } =V\left[ \cfrac { 1 }{ { R }_{ 1 } } +\cfrac { 1 }{ { R }_{ 2 } } +\cfrac { 1 }{ { R }_{ 3 } } \right] \)
\(I=\cfrac { V }{ { R }_{ p } } \)
\(\cfrac { 1 }{ R_{ P } } =\cfrac { 1 }{ { R }_{ 1 } } +\cfrac { 1 }{ { R_{ 2 } } } +\cfrac { 1 }{ R_{ 3 } } \)  

Here R_P is the equivalent resistance of the parallel combination of the resistors. Thus, when a number of resistors are connected in parallel, the sum of the reciprocal of the values of resistance of the individual resistor is equal to the reciprocal of the effective resistance of the combination.
Note: The value of equivalent resistance in parallel connection will be lesser than each individual resistance.

(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.

(i) XY is a conductor of area cross section A. \(\vec { E } \)is the applied electric field. n is the number of electrons  per unit volume with same drift velocity  (V_d) .
(ii) Let electrons move through a distance dx in time interval dt.

(iii) The drift velocity of the electrons = v_d
(iv) If the electrons move through a distance dx within a small interval of time dt,
\({ v }_{ d }=\cfrac { dx }{ dt } ;dx={ v }_{ d }dt\)  ..(i)
(v)  Since A is the area of cross section of the conductor, the electrons available in the volume of length dx is
= volume x number of electrons per unit volume  = A dx x n ...(2)
(vi) Substituting for dx from equation (1) in (2)
= (A v_d dt) n
(vii) Total charge in volume element dQ =(charge) x (number of electrons in the volume element)
dQ = (e) (Av_ddt)n
Hence the current \(I=\cfrac { dQ }{ dt } =\cfrac { ne{ Av }_{ d }dt }{ dt } \)
\(I=ne{ Av }_d\)      ..........(3)
Current density (J):
(viii) The current density (J) is defined as the current per unit area of cross section of the conductor.
\(J=\cfrac { I }{ A } \)
(ix) The S.I unit of current density is \({ Am }^{ -2 }\)
\(J=\cfrac { neAv_{ d } }{ A } \) (∵I = nAeV_d)
\(J={ nev }_{ d }\)         .........(4)       
(x) The above expression holds only when the direction of the current is perpendicular to the area A.
In general, the current density is a vector quantity and it is given by,
\(\vec { J } =ne\vec { v_{ d } } \)
Substituting \(\vec { v_{ d } } \) from equation 
\(\vec { v_{ d } } =\cfrac { e\tau }{ m } \vec { E } \)
\(\vec { J } =\cfrac { n.{ e }^{ 2 }\tau }{ m } \vec { E } \)  ...(5)
\(\vec { J } =\sigma \vec { E } \)   ....(6)
(xi) But conventionally, we take the direction of (conventional) current density as the direction of electric field. So, the above equation becomes,
\(\vec { J } =\sigma \vec { E } \)   .....(7)
(xii) Where, \(\sigma =\cfrac { { ne }^{ 2 }\tau }{ m } \) is called conductivity. The equation (7) is called microscopic form of ohm's law.