In series the current, I is same through both wires Rate of production of heat \(P={ I }^{ 2 }R\Rightarrow P\alpha R\) or \(H={ I }^{ 2 }Rt\Rightarrow H\alpha R\)
H - amount of heat produced; R - resistance of the wire.
Nichrome wire is heated up more since resistivity is more and hence resistance of nichrome wire is much more than that of copper wire of same length & radius.

\({ V }_{ d }=\cfrac { eE }{ m } \tau =\cfrac { eV }{ ml } .\tau \left[ E=\cfrac { V }{ l } \right] \)
Keeping V constant, if 1 is doubled, v_d is halved. Again keeping area constant if length is doubled, if R will be doubled [R \(\alpha\) 1,if a is constant].

It consists of a tungsten filament (melting point 3380^oC) kept inside a glass bulb with inert gas maintained at low pressure and heated to incandescence by the current. Electric discharge lamps, electric welding and electric arc also utilize the heating effect of current.

It is a safety device and connected in series in a circuit to protect the electric devices from the heat developed by the passage of excessive current. It is a short length of a wire made of a low melting point material. It melts and breaks the circuit if current exceeds a certain value.

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

The instantaneous current I is defined as the limit of the average current \(\Delta t\rightarrow 0\) 
\(I=\underset { \Delta t-0 }{ lim } \cfrac { \Delta Q }{ \Delta t } =\cfrac { dQ }{ dt } \)

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).

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

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

Microscopic form of ohm's law is 
\(\vec{J}=\sigma\vec {E}\) 
\(J=\frac{ne^2\tau}{m}\vec{E}\)
\(\frac{e \tau}{m} \rightarrow Drift \ velocity \ v_d\)
where \(\vec{J} \) - current density
\(\sigma \)  - conductivity
\(\vec{E} \) - Electric field

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

(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.

(a) An electric cell converts chemical energy into electrical energy to produce electricity. It contains two electrodes immersed in an electrolyte as shown Several electric cells connected together form a battery.

(b) When a cell or battery is connected to a circuit, electrons flow from the negative terminal to the positive terminal through the circuit. By using chemical reactions, a battery produces potential differences across its terminals. This potential difference provides the energy to move the electrons through the circuit.

The drift velocity is the average velocity acquired by the electrons inside the conductor when it is subjected to an electric field. The average time between successive collisions is called the mean free time denoted by \(\tau \). The acceleration \(\vec { a } \) experienced by the electron in an electric field \(\vec { E } \)  is given by
\(\vec { a } =\cfrac { -e\vec { E } }{ m } \left( since\vec { F } =-e\vec { E } \right) \)
The drift velocity is given by
\({ \vec { V } }_{ d }=\vec { a } \tau \)
\({ \vec { V } }_{ d }=\cfrac { e\tau }{ m } \vec { E } \)
\({ \vec { V } }_{ d }=-\mu \vec { E } \)
Here \(\mu =\cfrac { e\tau }{ m } \) is the mobility of the electron and it is defined as the magnitude of the drift velocity per unit electric field \(\mu =\cfrac { \left| { \vec { v } }_{ d } \right| }{ \left| \vec { E } \right| } \)

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 )

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

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

Let R be the external resistance in series with the cell of emf E and external resistance 'r',
The current in the circuit is \(I=\cfrac { E }{ R+r } \)
Case I: 1 = 0.5 A, R = 12\(\Omega \) then
\(0.5=\cfrac { E }{ 12r } \)
E = 0.5 (12 + r)
E = 6.0 + 0.5 r
Case II: I = 0.25 A. R = 25 then
\(0.25=\cfrac { E }{ 25+r } \)
E = (0.25) (25 + r)
E  = 6.25 + 0.25 r
From (1) (2), we get
6.0 + 0.5 r = 6.25 + 0.25 r
\(r=1\Omega \)
in equation (1) E = 6.0 + 0.5 x (1)
E = 6.5 v

Potential difference V = 3 V
Current, I = 5 A
By Ohms law, \(R=\cfrac { V }{ I } =\cfrac { 3 }{ 5 } =0.6\Omega \)

(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.

(i) Suppose n cells, each of emf volts and internal resistance r ohms are connected in series with an external resistance R as shown in Figure

(ii) The total emf of the battery = nr
The total resistance in the circuit = nr + R
By Ohm's law, the current in the circuit is
\(I=\cfrac { total\ emf }{ total\ resistance } =\cfrac { n\xi }{ nr+5 } \)
Case (a) If r << R, then
\(I=\cfrac { n\xi }{ R } ={ nl }_{ 1 }\)
where II is the current due to a single cell
\(\left( { I }_{ 1 }=\cfrac { \xi }{ R } \right) \)
(iii) 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, \(I=\cfrac { n\xi }{ nr } =\cfrac { \xi }{ r } \)
(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. Cells in parallel
(i) In parallel connection all the positive terminals of the cells are connected to one point and all the negative terminals to a second point. These two points form the positive and negative terminals of the battery.
(ii) Let n cells be connected in parallel between the points A and B and a resistance R is connected between the points A and B as shown in Figure. Let be the emf and r the internal resistance of each cell.

(iii) The equivalent internal resistance of the battery is \(\cfrac { 1 }{ { { r }_{ eq } } } =\cfrac { 1 }{ r } +\cfrac { 1 }{ r } +...\cfrac { 1 }{ r } (netrms)=\cfrac { n }{ r } \)
So \(\cfrac { 1 }{ { r }_{ eq } } =\cfrac { r }{ n } \) and the total resistance in the circuit = \(R+\cfrac { r }{ n } \) The total emf is the potential difference between the points A and B, which is equal to \(\xi \) The current in the circuit is given by
\(I=\cfrac { \xi }{ \frac { r }{ n } +R } \)
\(I=\cfrac { n\xi }{ r+nR } \)
Case (a) If >> R,\(I=\cfrac { n\xi }{ r } ={ nl }_{ 1 }\)
Case (b) If < \(I=\cfrac { \xi }{ R } \)
where II is the current due to a single cell and is equal to \(\cfrac { \xi }{ r } \) when R is negligible. Thus, the current through the external resistance due to the whole battery is n times the current due to a single cell.
  

Wheatstone's bridge:
i) An important application of Kirchhoff's rule is Wheatstone's bridge. It is used to compare Resistances and also helps in determining the unknown resistance in electrical network. The bridge consists of four resistances P, Q, R and S connected as shown in Figure.
ii) A galvanometer G is connected between the points B and D. The battery is connected between the points A and C. The current through the galvanometer is I_G and its resistance is G.
Applying Kirchhoff's current rule to junction B
I₁ - I_G - I₃ = 0    ..(1)
Applying Kirchhoff's current rule to junction D,
I₂ + I_G - I₄ = 0      ...(2)

Applying Kirchhoff's voltage rule to loop ABDA,
I₁P + I_GG - I₂R = 0      ...(3)
Applying Kirchhoff's voltage rule to loop ABCDA,
I₁P + I₃Q - I₄S - I₂R = 0   ...(4)
(iii) When the points B and D are at the same potential, the bridge is said to be balanced. As there is no potential difference between B and D, no current flows through galvanometer (I_G = 0). Substituting I_G = 0 in equation (1), (2) and (3), we get
I₁ = I₃  ..(5)
I₂ = I₄  ..(6)
I₁P = I₂R    ..(7)
Substituting the equation (7) in equation (4),
I₃Q = I₄S  .....(8)
Dividing equation (7) by equation (8), we get
\(\cfrac { P }{ Q } =\cfrac { R }{ S } \)   .....(9)
(iv) This is the bridge balance condition. Only under this condition, galvanometer shows null deflection.

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.