Junction rule: At any junction; the sum of the currents entering the junction is equal to the sum of currents leaving the junction.
Alternatively, Δ i = 0
Justification: Conservation of charge 
Loop rule: The Algebraic sum of charges in the potential around any closed loop involving resistors and cells in the loop is zero.
Alternatively: Δ V = 0, where Δ V is the change in potential
Justification: Conservation of energy

Let the resistivity of the material be \(\rho \).
∴ Resistance of wire,\(\ R=\rho \frac { l }{ A } \)
or \(\rho =\frac { RA }{ l } \)
= \(\quad \frac { 5\times 6\times { 10 }^{ -7 } }{ 15 } \)
= \(2\times { 10 }^{ -7 }\Omega -m\)
Thus the resistivity of the material at the temperature of the experiment is \(2\times { 10 }^{ -7 }\Omega -m\)

Let l be the length of four segments will be l, 2l, 4l, 8l. Their corresponding resistance will be R, 2R, 4R and 8R. Given R = \(2 \ \Omega\)
Resistance of the original wire 
= R + 2 R + 4 R + 8 R
= 15 R 
= 15 x 2
= 30Ω

In a pure semiconductors (Ge or Si) called intrinsic semiconductor the electrically conductivity is related by the electrons thermally excited from the valence band to the conduction band. There are equal number densities of free electrons and holes in conduction band and valence band of intrinsic semiconductor.
When a pure semiconductor of Ge or Si is doped with impurity atoms of valence three (like B or A1), some additional energy levels are created just above the upper energy level of valence band. Due to it, band gap of \(p\)-type semiconductor becomes smaller than intrinsic semiconductor. Also in \(p\)-type semiconductor, the number density of holes is more than that of electrons.
A \(p\)-type semiconductor is obtained when a trivalent atoms (Bor A1) are doped in pure semiconductor of Ge or Si. Here each doped trivalent atom shares its three valence electrons with the three atoms of Ge or Si and form bonds. While the fourth bond remains unbounded. Due to it, a hole is created. Since the impurity atoms and atoms of semiconductor are electrically neutral, hence \(p\)-type semiconductor is also neutral.

Note that thermally generated electrons (ni ~10¹⁶m^–3) are negligibly small as compared to those produced by doping.
Therefore, ne \(\approx\) ND
Since n_e n_h = \(n_{i}^{2}\) , The number of holes
n_h = (2.25 x 10³² ) / (5 x10²²)
= ~ 4.5 x 10⁹ m^–3

1.2\(\Omega\)

As we know E = V + Ir and \(V=\frac{E}{2}, r=1 \Omega\)
\(\therefore E=\frac{E}{2}+I r \ \Rightarrow \ I=\frac{E}{2 r}=\frac{1.5}{2 \times 1}=0.75 \mathrm{~A}\)

The circuit diagram of a full-wave rectifier.

The working principle of full rectifier ;
The full-wave rectifier converts alternating input to variable DC output. It uses the principle of operation of the diode in forward and reverse bias condition. During the positive half cycle of the AC input wave, D1 is forward bias and D2 is reverse bias. We get a positive output at the voltage. During the negative half cycle of the AC input wave, D2 is forward bias and D1 is reverse bias. We get a positive output at the voltage.

The donor energy level E_D is just below the bottom of the conduction band. At room temperature this small energy gap is easily converted by the thermally excited electrons. The conduction band has more electrons as they have been contributed both by thermal excitation and donor impurities. Whereas the acceptor energy level  E_A lies slightly above the top of the valence band. At room temperature, many electrons of the valence band get excited to these acceptor energy levels, leaving behind equal number of holes in the valence band. These holes can conduct current. Thus, the valence band has more holes than the electrons in the conduction band.

In n-type extrinsic semiconductors, the number of free electrons in conduction band is much more than the number of holes in valence band. The donor energy level lies just below the conduction band. In p-type extrinsic semiconductor, the number of the holes in valence band is much more than the number of free electrons in conduction band. The acceptor energy level lies just above the valence band.
(ii) 
At absolute zero temperature (0 K) conduction band of semiconductor is completely empty, i.e., σ=0
Hence, the semiconductor behaves as an insulator. At room temperature, some valence electrons acquire enough thermal energy and jump to the conduction band where they are free to conduct electricity. Thus, the semiconductor acquires a small conductivity at room temperature.

No current flows through 4Ω resistor as capacitor offers infinite resistance in DC circuits.
Also, 2Ω and 3Ω are in parallel combination
∴ R_AB = \(\frac { 2\times 3 }{ 2+3 } =\frac { 6 }{ 5 } \)
Applying Kirchhoffs second rule in outer loop AB and cell.
Let I current flow through outer loop in clockwise direction.
-1.2I - 2.8I + 6 = 0
4I = 6
I = 3/2A
∴ Potential difference across AB
V_AB = IR_AB = 3/2 x 1.2 = 1.8V
∴ 3Ω and 2Ω are in parallel combination.
∴ Potential difference across 2Ω  resistor is given by
I = \(\frac { V }{ R } =\frac { 1.8 }{ 2 } \) = 0.9A
I = 0.9 A

Drift velocity: The average velocity with which the free electrons drift under the influence of an external field.
Relaxation time: Average time interval between two successive collisions of an electron with the ions/atoms of the conductor.
The drifts velocity will be inversely proportional to (or \(v_b\ \propto \frac{1}{l}\) ) and hence it will become one-third of 1 its initial value.

Let n electrons pass through wire in 2 minutes, when a current of 300 mA or 0.3A pass for 2 minutes
Charge passing = I = dt =
\(0.3\times 120=36\quad C\)
\(Also \ charge \ with \ n \ electrons\)
\( =n\times 1.6\times { 10 }^{ -19 }C\\ n\times 1.6\times { 10 }^{ -19 }=36\)
\(\\ or\quad n=\frac { 36 }{ 1.6\times { 10 }^{ -19 } } \)
\(\\ =\frac { 360 }{ 16 } \times { 10 }^{ 19 }\)
\(\\ n=22.5\times { 10 }^{ 19 }\)

Given
\({ R }_{ 1 }=1\Omega ,{ R }_{ 2 }=2\Omega ,{ R }_{ 3 }=3\Omega \)
(a) Total resistance of series combination
\({ R }_{ s }={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }\)
\({ R }_{ s }\) = 1 + 2 + 3 = \(6\Omega \)

(b) Since E = I (R + r)
I = \(\frac { E }{ { R }_{ s }+0 } =\frac { E }{ { R }_{ s } } =2A\)
\({ V }_{ 1 }={ IR }_{ 1 }=2\times 1=2V\)
\({ V }_{ 2 }={ IR }_{ 2 }=2\times 2=4V\)
\({ V }_{ 3 }={ IR }_{ 3 }=2\times 3=6V\)

Given, input frequency = 50 Hz
For a half-wave rectifier, the output frequency is equal to the input frequency.
\(\therefore\) Output frequency=50 Hz
For a full wave rectifier, the output frequency is twice the input frequency
\(\therefore\) Output frequency = 2 \(\times\) 50 = 100 Hz.

(a) In case of an intrinsic semiconductor the number density of free electrons is equal to number density of holes. There are no external impurity atoms in the intrinsic semiconductors and the conductivity is low in comparison to that of doped (extrinsic semiconductors). When the intrinsic semiconductor is doped with acceptor impurity atoms, a p-type semiconductor is formed. In case of p-type semiconductor, n_h >> n_e. Therefore, the conductivity of p-type semiconductor is higher than that of the intrinsic semiconductor.
The number of negatively charged immobile ions is equal to the number of oppositively charged holes. Therefore, p type semi-conductor is electrically neutral.
(b) The heavy doping of p and n sides of a p-njunction results into a reduced width of depletion region. As, for the same value of barrier potential, the width decreases and the electric field across the junction Increases.
\(\text { i.e. } \ V=E d \Rightarrow E=\frac{V}{d}\)
As, the electric field of the junction acquires very high value for a large variation of current, the potential difference in breakdown region remains almost same.
This property of zener diode is used in voltage regulation. Zener diode as a voltage regulator.
The unregulated de is given as an input through a series resistor R_s . The zener is always kept under reverse bias. As the input voltage increases, the current through resistance R_s and zener increases. This causes more voltage rise in resistance R_s keeping the voltage across Z the same. This is possible, as at zener voltage, the current increases without any change in potential. As the input drops, the current through resistance R_s and Z drops causing no change in voltage across zener. Thus, at both higher and lower potential inputs, a regulated output is obtained.

It states that current flowing through a conductor is proportional to the potential difference across its two ends provided the physical conditions of the conductor remain unchanged.
If V is the potential difference between two ends of a conductor and I is the current flowing through it, then
\(V\alpha I\)
\(V=RI\)
Where R is the constant of proportionality and is called the resistance of the conductor. Its value depends upon
(i) Shape of the conductor
(ii) Length of conductor
(iii) Nature of the material

If a graph is plotted between V and I, the graph will be a straight line passing through the origin.
(b) Resistance is the property of a material by virtue of which it opposes the flow of current through it and quantitatively it is given by,
\(R=\frac { V }{ I }\)
\( =\frac { Potential \ difference }{ Current } \)
definition of ohm
Hence a conductor has a resistance of one ohm if a current of one ampere flows through it when a potential difference of one volt is maintained across its two ends.

(a) Electric current: The flow of charge in a definite direction constitutes the electric current and the rate of flow of charge through any cross-section of a conductor is the measure of current i.e.,
\(=\frac { Total \ charge \ flowing \ (q) }{ Time \ taken \ (t) } \)
If dq charge is flowing through a section of a conductor in time dt, the electric current is given by
\(I=\frac { dq }{ dt } \)
Unit of electric current S.I unit of current is ampere. It is also called the practical unit of current. It is denoted by A,

\(1 \ ampere(A)=\frac { 1 \ coulomb }{ 1 \ second } = \ 1C{ s }^{ -1 }\)
Thus the current through a wire is said to be 1 ampere if one coulomb of charge is flowing per second through a section of the wire.
Current is a scalar quantity, the direction of flow of positive charges is considered as a conventional direction of the current.
In solids, positive charges being heavy do not move and currently is due to free electrons, therefore, the conventional direction of current is opposite to electronic current.
\(I=\frac { q }{ t } =\frac { ne }{ t } \)
\(n=\frac { It }{ e } =\frac { 1\times 1 }{ 1.6\times { 10 }^{ -19 } } \)
\(=6.25\times { 10 }^{ 18 }electrons \ per \ second.\)

​

(i) (c) :In insulator, energy band gap is > 3 eV
(ii) (b) : In conductor, separation between conduction and valence bands is zero and in insulator, it is greater than 1 eV. Hence in semiconductor the separation between conduction and valence band is 1 eV.
(iii) (a): According to band theory the forbidden gap in conductors E_g = 0, in insulators E_g > 3 eV and in semiconductors E_g < 3 eV.
(iv) (a): The four valence electrons of C, Si and Ge lie respectively in the second, third and fourth orbit.Hence energy required to take out an electron from these atoms (i.e. ionisation energy Eg) will be least for Ge, followed by Si and highest for C. Hence, the number of free electrons for conduction in Ge and Si are significant but negligibly small for C.
(v) (b)

(I) (b): \(q=10^{6} \times 1.6 \times 10^{-19} \mathrm{C}=1.6 \times 10^{-13} \mathrm{C}\)
t = 10^-3 s
\(I=\frac{q}{t}=\frac{1.6 \times 10^{-13}}{10^{-3}}=1.6 \times 10^{-10} \mathrm{~A}\)
(ii) (C): C S^-1
(iii) (C): The current flowing through a conductor of non-uniform cross-section remain same in the whole of the conductor.
(iv) (a): When a constant current is flowing through a conductor of non-uniform cross-section, electron density does not depend upon the area of cross section, while current density, drift velocity and electric field all vary inversely with area of cross-section.
(v) (a): Given, I = 40 A ;A = 10m²
\(\therefore\) Current density, \(J=\frac{I}{A} \text { or } J=\frac{40}{10}=4 \mathrm{~A} / \mathrm{m}^{2}\)

(c): In a semiconductor, there are no free electrons at 0 K. The number of free electrons increases with increase in temperature because with increase in temperature the electron get sufficient energy to cross forbidden band and reach conduction band. But total number of free electrons in a semiconductor is less than that in a conductor.

(c): Due to diffusion of holes from the p-region to the n-region and of electrons from the n-region to the p-region an electric field is set up across the junction barrier. Once the depletion layer is formed it is in equilibrium and becomes free of mobile charge carriers.

(b): On increasing temperature of wire the kinetic energy of free electrons increase and so they collide more rapidly with each other and hence their drift velocity decreases. Also when temperature increases, resistance increase and resistance is inversely proportional to conductivity of material.

(b): Current is a scalar quantity, it is justified by the following two observations
(i) If current carrying wire is bent at some point, then also current in the wire remains same, while a vector quantity always changes by changing its direction.
(ii) Current flowing in the circuit do not follow the laws of vector addition. It follows according to ordinary rule of algebra. This makes it clear that current is not a vector but a scalar quantity. Also current is defined as rate of flow of charge through the wire 1= dq/ dt.