When the current through the element is very small, heating effects can be ignored and the temperature T₁ of the element is the same as room temperature. When the toaster is connected to the supply, its initial current will be slightly higher than its steady value of 2.68 A. But due to heating effect of the current, the temperature will rise. This will cause an increase in resistance and a slight decrease in current. In a few seconds, a steady state will be reached when temperature will rise no further, and both the resistance of the element and the current drawn will achieve steady values. The resistance R₂ at the steady temperature T₂ is
\(R_{2}=\frac{230 \mathrm{~V}}{2.68 \mathrm{~A}}=85.8 \Omega\)
Using the relation
R₂ = R₁ [1 + α (T₂ - T₁)]
with α = 1.70 x 10^-4 °C^-1, we get
\(T_{2}-T_{1}=\frac{(85.8-75.3)}{(75.3) \times 1.70 \times 10^{-4}}=820^{\circ} \mathrm{C}\)
that is, T₂ = (820 + 27.0) °C = 847 °C
Thus, the steady temperature of the heating element (when heating effect due to the current equals heat loss to the surroundings) is 847 °C.

Number density of free electrons in a copper conductor, n = 8.5 x 10²⁸ m^-3 Length of the copper wire, l = 3.0 m
Area of cross-section of the wire, A = 2.0 x 10^-6 m²
Current carried by the wire, I = 3.0 A, which is given by the relation,
I = nAeV_d
Where,
e = Electric charge = 1.6 x 10⁻¹⁹ C
V_d = Drift velocity = \(\frac{\text { Length of the wire (l) }}{\text { Time taken to cover l(t) }}\)
\(I=n A e \frac{l}{t}\)
\(t=n A e \frac{l}{I}\)
\(=\frac{3 \times 8.5 \times 10^{28} \times 2 \times 10^{-6} \times 1.6 \times 10^{-19}}{3.0}\)
\(=2.7 \times 10^{4} s\)
Therefore, the time taken by an electron to drift from one end of the wire to the other is 2.7 x 10⁴ s.

Temperature, T₁ = 27.5°C
Resistance of the silver wire at T₁, R₁ = 2.1 Ω
Temperature, T₂ = 100°C
Resistance of the silver wire at T₂, R₂ = 2.7 Ω
Temperature coefficient of silver = α
It is related with temperature and resistance as 
\(\alpha=\frac{R_{2}-R_{1}}{R_{2}\left(T_{2}-T_{1}\right)}\)
\(=\frac{2.7-2.1}{2.1(100-27.5)}=0.0039^{\circ} \mathrm{C}^{-1}\)
Therefore, the temperature coefficient of silver is 0.0039°C⁻¹.

Given, resistance of heating element at temperature 27°C,   
R₂₇ = 100 \(\Omega\)
Resistance of heating element at temperature t°C,
R_t = 117 \(\Omega\)
\(\alpha\)=1.70 \(\times\)10^-4 °C^-1, t = ?
By using the formula of temperature coefficient of resistance,
\(\alpha=\frac{R_2-R_1}{R_1\left(t_2-t_1\right)}\)   ...(i)
Here, R₁ = R_t, R₁ = R₂₇, t₂ = t and t₁ = 27°C
Such that, \(\alpha=\frac{R_t-R_{27}}{R_{27}(t-27)}\)
 Substituting given values in Eq. (i), we get
\(\begin{aligned} 1.70 \times 10^{-4} & =\frac{117-100}{100(t-27)} \\ \end{aligned}\)
\(\begin{aligned} \text { or } \quad t-27 & =\frac{17}{100 \times 1.70 \times 10^{-4}} \end{aligned}\)
or t = 1000 + 27 = 1027°C 

The network is not reducible to a simple series and parallel combinations of resistors. There is, however, a clear symmetry in the problem which we can exploit to obtain the equivalent resistance of the network.
The paths AA', AD and AB are obviously symmetrically placed in the network. Thus, the current in each must be the same, say, I. Further,  at the corners AA', B and D, the incoming current I must split equally into the two outgoing branches. In this manner, the current in all the 12 edges of the cube are easily written down in terms of I, using Kirchhoff’s first rule and the symmetry in the problem. Next take a closed loop, say, ABCC'EA, and apply Kirchhoff’s second rule:
–IR – (1/2)IR – IR + \(\epsilon\) = 0
where R is the resistance of each edge and e the emf of battery. Thus,
\(\epsilon = \frac{5}{2}IR\)
The equivalent resistance R_eq of the network is
\(R_{eq}= \frac{\epsilon}{3I}=\frac{5}{6}R\)
For R = 1 Ω, R_eq = (5/6) Ω and for e = 10 V, the total current (= 3I ) in
the network is
3I = 10 V/(5/6) Ω = 12 A, i.e., I = 4 A
The current flowing in each edge can now be read off from the figure.

Current flowing through various branches of the circuit is represented in the given figure.

I₁ = Current flowing through the outer circuit
I₂ = Current flowing through branch AB
I₃ = Current flowing through branch AD
I₂ - I₄ = Current flowing through branch BC
I₃ + I₄ = Current flowing through branch CD
I₄ = Current flowing through branch BD
For the closed circuit ABDA, potential is zero i.e.,
10I₂ + 5I₄ - 5I₃ = 0
2I₂ + I₄ - I₃ = 0
I₃ = 2I₂ + I₄ … (1)
For the closed circuit BCDB, potential is zero i.e.,
5(I₂ - I₄) - 10(I₃ + I₄) - 5I₄ = 0
5I₂ + 5I_{4 -} 10I₃ - 10I₄ - 5I₄ = 0
5I₂ - 10I₃ - 20I₄ = 0
I₂ = 2I₃ + 4I₄ … (2)
For the closed circuit ABCFEA, potential is zero i.e.,
-10 + 10 (I₁) + 10(I₂) + 5(I₂ - I₄) = 0
10 = 15I₂ + 10I₁ - 5I₄
3I₂ + 2I₁ - I₄ = 2 … (3)
From equations (1) and (2), we obtain
I₃ = 2(2I₃ + 4I₄) + I₄
I₃ = 4I₃ + 8I₄ + I4
- 3I₃ = 9I₄
- 3I₄ = + I₃ … (4)
Putting equation (4) in equation (1), we obtain
I₃ = 2I₂ + I₄
- 4I₄ = 2I₂
I₂ = - 2I₄ … (5)
It is evident from the given figure that,
I₁ = I₃ + I₂ … (6)
Putting equation (6) in equation (1), we obtain
3I₂ +2(I₃ + I₂) - I₄ = 2
5I₂ + 2I₃ - I₄ = 2 … (7)
Putting equations (4) and (5) in equation (7), we obtain
5(- 2 I₄) + 2(- 3 I₄) - I₄ = 2
- 10I₄ - 6I₄ - I₄ = 2
17I₄ = - 2
\(I_{4}=-\frac{2}{17} A\)
Equation (4) reduces to
I₃ = - 3(I₄)
\(=-3\left(-\frac{2}{17}\right)=\frac{6}{17} A\)
\(I_{2}=-2\left(I_{4}\right)\)
\(=-2\left(-\frac{2}{17}\right)=\frac{4}{17} A\)
\(I_{2}-I_{4}=\frac{4}{17}-\left(-\frac{2}{17}\right)=\frac{6}{17} A\)
\(I_{3}+I_{4}=\frac{6}{17}+\left(-\frac{2}{17}\right)=\frac{4}{17} A\)
\(I_{1}=I_{3}+I_{2}\)
\(=\frac{6}{17}+\frac{4}{17}=\frac{10}{17} A\)
Therefore, current in branch \(A B=\frac{4}{17} A\)
In branch BC \(=\frac{6}{17} A\)
In branch CD = \(-\frac{4}{17} A\)
In branch AD = \(\frac{6}{17} A\)
In branch BD = \(\left(-\frac{2}{17}\right) A\)
Total current = \(\frac{4}{17}+\frac{6}{17}+\frac{-4}{17}+\frac{6}{17}+\frac{-2}{17}=\frac{10}{17} A\)

Considering the mesh BADB, we have
100I₁ + 15I_g - 60I₂ = 0
or 20I₁ + 3I_g - 12I₂ = 0
Considering the mesh BCDB, we have
10 (I₁ - I_g) - 15I_g - 5 (I₂ + I_g) = 0
10I₁ - 30I_g - 5I₂ = 0
2I₁ - 6I_g - I₂ = 0
Considering the mesh ADCEA,
60I₂ + 5 (I₂ + I_g) = 10
65I₂ + 5I_g = 10
13I₂ + I_g = 2
Multiplying by 10
20I₁ - 60I_g - 10I₂ = 0
From we have
63I_g - 2I₂ = 0
I₂ = 31.5I_g
Substituting the value of I₂ into we get
13 (31.5I_g ) + I_g = 2
410.5 I_g = 2
I_g = 4.87 mA

Each branch of the network is assigned an unknown current to be determined by the application of Kirchhoff’s rules. To reduce the number of unknowns at the outset, the first rule of Kirchhoff is used at every junction to assign the unknown current in each branch. We then have three unknowns I₁, I₂ and I₃ which can be found by applying the second rule of Kirchhoff to three different closed loops. Kirchhoff’s second rule for the closed loop ADCA gives,
10 - 4(I₁ - I₂) + 2(I₂ + I₃ - I₁) - I₁ = 0 
that is, 7I₁ -  6I₂ - 2I₃ = 10
For the closed loop ABCA, we get
10 - 4I₂ - 2 (I₂ + I₃) - I₁ = 0
that is, I₁ + 6I₂ + 2I₃ = 10 
For the closed loop BCDEB, we get
5 - 2 (I₂ + I₃) - 2 (I₂ + I₃ - I₁) = 0
that is, 2I₁ - 4I₂ - 4I₃ = -5
Equations (3.61 a, b, c) are three simultaneous equations in three unknowns. These can be solved by the usual method to give
\(I_1=2.5 \mathrm{~A}, \quad I_2=\frac{5}{8} \mathrm{~A}, \quad I_3=1 \frac{7}{8} \quad \mathrm{~A}\)
The currents in the various branches of the network are 
\(\mathrm{AB}: \frac{5}{8} \mathrm{~A}, \quad \mathrm{CA}: 2 \frac{1}{2} \mathrm{~A}, \quad \mathrm{DEB}: 1 \frac{7}{8} \mathrm{~A}\)
\(\mathrm{AD}: 1 \frac{7}{8} \mathrm{~A}, \quad \mathrm{CD}: 0 \mathrm{~A}, \quad \mathrm{BC}: 2 \frac{1}{2} \mathrm{~A}\)
It is easily verified that Kirchhoff’s second rule applied to the remaining closed loops does not provide any additional independent equation, that is, the above values of currents satisfy the second rule for every closed loop of the network. For example, the total voltage drop over the closed loop BADEB
\(5 \mathrm{~V}+\left(\frac{5}{8} \times 4\right) \mathrm{V}-\left(\frac{15}{8} \times 4\right) \mathrm{V}\)
equal to zero, as required by Kirchhoff’s second rule.

(a) The direction of drift velocity of conduction electrons is opposite to the electric field direction, i.e., electrons drift in the direction of increasing potential. The drift speed v_d is given v_d = (I/neA)
Now, e = 1.6 x 10^-19 C, A = 1.0 x 10^-7 m², I = 1.5 A. The density of conduction electrons, n is equal to the number of atoms per cubic metre (assuming one conduction electron per Cu atom as is reasonable from its valence electron count of one). A cubic metre of copper has a mass of 9.0 x 10³ kg. Since 6.0 x 10²³ copper atoms have a mass of 63.5 g,
\(n=\frac{6.0 \times 10^{23}}{63.5} \times 9.0 \times 10^{6}\)
= 8.5 x 10²⁸ m^-3
which gives,
\(v_{d}=\frac{1.5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 1.0 \times 10^{-7}}\)
= 1.1 x 10^-3 m s^-1 = 1.1 mm s^-1
(b) (i) At a temperature T, the thermal speed  of a copper atom of mass M is obtained from \(\left[<(1 / 2) M v^{2}>=(3 / 2) k_{\mathrm{B}} T\right]\) and is thus typically of the order of \(\sqrt{k_{B} T / M}\), where k_B is the Boltzmann constant. For copper at 300 K, this is about 2 x 10² m/s. This figure indicates the random vibrational speeds of copper atoms in a conductor. Note that the drift speed of electrons is much smaller, about 10^-5 times the typical thermal speed at ordinary temperatures.
(ii) An electric field travelling along the conductor has a speed of an electromagnetic wave, namely equal to 3.0 x 10⁸ m s^-1  The drift speed is, in comparison, extremely small; smaller by a factor of 10^-11.

Given, E = 10 V, r = 3 \(\Omega\), I = 0.5 A
As, \(\begin{aligned} I & =\frac{E}{R+r} \end{aligned}\)
\(\begin{aligned} R & =\frac{E}{I}-r \end{aligned}\)
\(\begin{aligned} =\frac{10}{0.5}-3=17 \Omega \end{aligned}\)
and terminal voltage, V = IR = 0.5 \(\times\)17 = 8.5 V
 

Given, potential difference = 230 V
Initial current at 27°C = I_27°C = 3.2 A
Final current at t°C = I_t°C = 2.8 A
Room temperature = 27°C
Temperature coefficient of resistance, \(\alpha=1.70 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\)
Resistance at 27°C, R_27°C\(=\frac{V}{I_{27^{\circ} \mathrm{C}}}=\frac{230}{3.2}=\frac{2300}{32} \Omega\)
Resistance at t°C, R_t°C = \(\frac{V}{I_{t^{\circ} \mathrm{C}}}=\frac{230}{2.8}=\frac{2300}{28} \Omega\)
Temperature coefficient of resistance
\(\begin{aligned} \alpha & =\frac{R_t-R_{27}}{R_{27}(t-27)} \end{aligned}\)
\(\begin{aligned} \Rightarrow 1.70 \times 10^{-4} & =\frac{\frac{2300}{28}-\frac{2300}{32}}{\frac{2300}{32}(t-27)} \\ \end{aligned}\)
\(\begin{aligned} \text { or } \quad t-27 & =\frac{82.143-71.875}{71.875 \times 1.70 \times 10^{-4}}=840.347 \end{aligned}\)
or t = 840.3 + 27 = 867.3 °C
Thus, the steady temperature of heating element is 867.3 °C

Emf of the storage battery, E = 8.0 V
Internal resistance of the battery, r = 0.5 Ω
DC supply voltage, V = 120 V
Resistance of the resistor, R = 15.5 Ω
Effective voltage in the circuit = V¹
R is connected to the storage battery in series. Hence, it can be written as
V¹ = V - E
V¹ = 120 - 8 = 112 V
Current flowing in the circuit = I, which is given by the relation,
\(I=\frac{V^{1}}{R+r}\)
\(=\frac{112}{15.5+5}=\frac{112}{16}=7 A\)
Voltage across resistor R given by the product, IR = 7 x 15.5 = 108.5 V
DC supply voltage = Terminal voltage of battery + Voltage drop across R
Terminal voltage of battery = 120 - 108.5 = 11.5 V
A series resistor in a charging circuit limits the current drawn from the external source. The current will be extremely high in its absence. This is very dangerous.

R₀ = 5 Ω, R₁₀₀ = 5.23 Ω and R_t = 5.795 Ω
Now, \(t=\frac{R_{t}-R_{0}}{R_{100}-R_{0}} \times 100, \quad R_{t}=R_{0}(1+\alpha t)\)
\(=\frac{5.795-5}{5.23-5} \times 100\)
\(=\frac{0.795}{0.23} \times 100=345.65^{\circ} \mathrm{C}\)

(a) Electric field is established throughout the circuit, almost instantly (with the speed of light) causing at every point a local electron drift. Establishment of a current does not have to wait for electrons from one end of the conductor travelling to the other end. However, it does take a little while for the current to reach its steady value.
(b) Each ‘free’ electron does accelerate, increasing its drift speed until it collides with a positive ion of the metal. It loses its drift speed after collision but starts to accelerate and increases its drift speed again only to suffer a collision again and so on. On the average, therefore, electrons acquire only a drift speed.
(c) Simple, because the electron number density is enormous, ~10²⁹ m^-3.
(d) By no means. The drift velocity is superposed over the large random velocities of electrons.
(e) In the absence of electric field, the paths are straight lines; in the presence of electric field, the paths are, in general, curved.

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 \(2\times { 10 }^{ -7 }\Omega -m\)

Emf of the battery, E = 12 V
Internal resistance of the battery, r = 0.4 Ω
Maximum current drawn from the battery = I
According to Ohm’s law,
E = Ir
\(I=\frac{E}{r}\)
\(=\frac{12}{0.4}=30 A\)
The maximum current drawn from the given battery is 30 A.