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.

The resistivity of materials is
(i) inversely proportional to the number density (n) of the electrons
(ii) inversely proportional to the average time between the collisions \(\left( \tau \right) \)

(i) Electrical power P = VI
(ii) Erectrical power \(P=V\left(\frac{V}{R}\right)=\frac{V^{2}}{R} \quad \therefore P=\frac{V^{2}}{R}\)
(iii) P = Iv = I(IR) = I²R
(iv) P = I²R

Applying Kirchoff’s rule to the point P in the circuit,
The arrows pointing towards P are positive and away from P are negative.
Therefore, 0.2A - 0.4A + 0.6A - 0.5A + 0.7A - I = 0
0.7 A - I = 0
1.5A - 0.9A – I = 0
0.6A - I = 0
I = 0.6 A

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

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

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 )

I = 0.2 A, R = 10 Ω, E = 2.1 V
\(I=\frac{ɛ}{R+r}\)
\(0.2=\frac{2.1}{10+r}\)
0.2 x (10 + r) = 2.1
2 + 0.2 r = 2.1
0.2 r = 2.1 - 2 = 0.1
Internal resistance, \(r=\frac{0.1}{0.2}=\frac{1}{2}\)
r = 0.5 Ω

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

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

(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.

\(\therefore\)  Student A picked up the correct resistor of 25 k\(\Omega \) .