V = 12 V and R = 24 Ω
Current, I = ?
From Ohm’s law, \(I=\frac{V}{R}=\frac{12}{24}=0.5A\)

The relation between drift velocity of electrons and current in a wire of cross- sectional area A is
\({ v }_{ d }=\frac { I }{ neA } =\frac { 0.2 }{ 8.4\times { 10 }^{ 28 }\times 1.6\times { 10 }^{ -19 }\times 0.5\times { 10 }^{ -6 } }\)
v_d = 0.03 x 10^-3 m s^-1

E = 570 N C^-1, e = 1.6 x 10^-19 C,
m = 9.11 x 10^-31 kg and a = ?
F = ma = eE
\(a=\frac { eE }{ m } =\frac { 570\times 1.6 \times { 10 }^{ -19 } }{ 9.11\times { 10 }^{ -31 } } \)
\(=\frac { 912\times { 10 }^{- 19 }\times { 10 }^{ 31 } }{ 9.11 } \)
= 1.001 x 10¹⁴ ms^-2

(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) 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. 

(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.

(i) Due to this internal resistance, the power delivered to the circuit is not equal to power rating mentioned in the battery.
(ii) For a battery of emf \({ \xi }_{ 1 }\) with an internal resistance r, the power delivered to the circuit of resistance R is given by
\(P=I\xi =I(V+Ir)\)
Here V is the voltage drop across the resistance R and it is equal to IR.
Therefore, P = I (IR +Ir)
P = I² R + I² r
(iii) Here Pr is the power delivered to the internal resistance and PR is the power delivered to the electrical device (here it is the resistance R). For a good battery, the internal resistance r is very small, then for P < r < P