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.

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

Atoms in metals have one or more electrons which are loosely bound to the nucleus. These electrons are called free electrons and can be easily detached from the atoms by applying small energy.

Electric power is the rate at which the electrical potential energy is delivered
\(P =\frac{d U}{d t} \)
\(P =\frac{VdQ}{d t}=\mathrm{V} \frac{d Q}{d t} \)
Since \(\frac{d Q}{d t}=I\), where I - electric current
∴ P = VI

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).

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

(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.

(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

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.

In all the sections, the resistors are connected in parallel.
Section I
\(\frac { 1 }{ { R }_{ { P }_{ 1 } } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ { R }_{ { P }_{ 1 } } } =\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =\frac { 2 }{ 2 } \quad { R }_{ { p }_{ 1 } }=1\Omega \)

Section II
\(\frac { 1 }{ { R }_{ { P }_{ 1 } } } =\frac { 1 }{ 4 } +\frac { 1 }{ 4 } =\frac { 2 }{ 4 } ,\quad \frac { 1 }{ { R }_{ { P }_{ 2 } } } =\frac { 1 }{ 2 } ,{ R }_{ { p }_{ 2 } }=2\Omega \)

Section III
\(\frac { 1 }{ { R }_{ { P }_{ 3 } } } =\frac { 1 }{ 6 } +\frac { 1 }{ 6 } =\frac { 2 }{ 6 } \)
\(\frac { 1 }{ { R }_{ { P }_{ 3 } } } =\frac { 1 }{ 3 } ,{ R }_{ { p }_{ 3 } }=3\Omega \)
Equivalent resistance is given by
R = Rp₁ + Rp₂ + Rp₃
R = 1 Ω + 2 Ω + 3 Ω = 6 Ω
The circuit became,

Equivalent resistance between A and B is

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 )

In the first case, the resistance of the block
\({ R }_{ AB }=\rho \frac { length }{ Area } =\rho \frac { C }{ AB } \)
The current \({ I }_{ AB }=\frac { V }{ { R }_{ AB } } =\frac { V }{ \rho } .\frac { AB }{ C } \quad (1)\)
In the second case, the resistance of the block \({ R }_{ BC }=\rho \frac { A }{ BC } \)
The current \({ I }_{ BC }=\frac { V }{ { R }_{ BC } } =\frac { V }{ \rho } .\frac { BC }{ C } \quad (2)\)
To express I_BC interms of I_AB, we multiply and divide equation (2) by AC, we get
\({ I }_{ BC }=\frac { V }{ \rho } .\frac { BC }{ A } \frac { AC }{ AC } =\left( \frac { V }{ \rho } .\frac { AB }{ C } \right) .\frac { { C }^{ 2 } }{ { A }^{ 2 } } =\frac { { C }^{ 2 } }{ { A }^{ 2 } }.{ I }_{ AB }\)
Since C > A, the current I_BC > I_AB

Total power = 15 x 40 + 5 x 100 + 5 x 80 + 1000
= 600 + 500 + 400 + 1000
= 2500 W
P = VI, V = 220 V
\(I=\frac{P}{V}=\frac{2500}{220}=11.363 \ A\)
≃ 12 A

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

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

(i) \(mobility=\cfrac { Drift\ velocity }{ electric\ field } \) (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.