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 = \(\frac{4}{17}+\frac{6}{17}+\frac{-4}{17}+\frac{6}{17}+\frac{-2}{17}=\frac{10}{17} A\)

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 charge on the capacitor is
Q = CV = 900 x 10^–12 F x 100 V = 9 x 10^–8 C
The energy stored by the capacitor is
= (1/2) CV² = (1/2) QV
= (1/2) x 9 x 10^–8C x 100 V = 4.5 x 10^–6 J
(b) In the steady situation, the two capacitors have their positive plates at the same potential, and their negative plates at the same potential. Let the common potential difference be V′. The charge on each capacitor is then Q′ = CV′. By charge conservation, Q′ = Q/2. This implies V′ = V/2. The total energy of the system
\(=2 \times \frac{1}{2} Q^{\prime} V^{\prime}=\frac{1}{4} Q V=2.25 \times 10^{-6} \mathrm{~J}\)
Thus in going from (a) to (b), though no charge is lost; the final energy is only half the initial energy.
There is a transient period before the system settles to the situation (b). During this period, a transient current flows from the first capacitor to the second. Energy is lost during this time in the form of heat and electromagnetic radiation

Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be the x-axis; the negative charge is taken to be on the right side of the origin

Let P be the required point on the x-axis where the potential is zero. If x is the x-coordinate of P, obviously x must be positive. (There is no possibility of potentials due to the two charges adding up to zero for x < 0.) If x lies between O and A, we have
\(\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{3 \times 10^{-8}}{x \times 10^{-2}}-\frac{2 \times 10^{-8}}{(15-x) \times 10^{-2}}\right]=0\)
where x is in cm. That is,
\(\frac{3}{x}-\frac{2}{15-x}=0\)
which gives x = 9 cm
If x lies on the extended line OA, the required condition is
\(\frac{3}{x}-\frac{2}{x-15}=0\)
which gives
x = 45 cm
Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the formula for potential used in the calculation required choosing potential to be zero at infinity.

Electric field, E = 3 \(\times\)10³ \(\hat{i}\)N/C, i.e. electric field is directed towards X-axis (due to involvement of \(\hat{i}\)).
(i) As the surface is in YZ-plane, so the area vector (normal to the square) is along X-axis.

Area, S = 10 \(\times\) 10 = 100 cm² = 10^-2m²
Area vector, S = 10^-2 \(\hat{i}\)m²
Using the formula of electric flux,
\(\phi\)= E.S = ES cos \(\theta\)
=ES   [\(\because\) angle between E and S is 0^\(\circ\)]
= 3 \(\times\)10³ \(\times\)10^-2 = 30 N-m²/C
(ii) Now, the area vector makes an angle of 60^\(\circ\)with X-axis.
E=3\(\times\)10³ \(\hat{i}\) N/C
S = 100 cm² = 10^-2 m², \(\theta\)= 60^\(\circ\)
Using the formula of electric flux,
\(\phi\)= E.S
\(\Rightarrow\) \(\phi\)=ES cos \(\theta\)= 3 \(\times\)10³ \(\times\)10^-2 cos 60^\(\circ\)
= 3 \(\times\)10 \(\times\)\(\frac{1}{2}\)=15 N-m²/C

Given, p= 4 X10^-9 C-m, E= 5 x10⁴,
\( \theta =30^{\circ} \)
\(\therefore \ \tau =p E^2 \sin \theta \)
\(=4 \times 10^{-4} \times 5 \times 10^4 \times \sin 30^{\circ} \)
\(=4 \times 10^{-4} \times 5 \times 10^4 \times \frac{1}{2} \quad\left[\because \sin 30^{\circ}=\frac{1}{2}\right] \)
\(=10 \times 10^{-4}=10^{-4} \mathrm{~N}-\mathrm{m}\)

Since the electric field has only an x component, for faces perpendicular to x direction, the angle between E and ΔS is ± \(\pi \)/2. Therefore, the flux φ = E.ΔS is separately zero for each face of the cube except the two shaded ones. Now the magnitude of the electric field at the left face is
E_L = \(\alpha\)x^1/2 = \(\alpha\)a^1/2
(x = a at the left face).
The magnitude of electric field at the right face is
E_R = \(\alpha\)x^1/2 = \(\alpha\)(2a)^1/2
(x = 2a at the right face
The corresponding fluxes are
φ_L= E_L . ΔS = ΔS E_L. \(\hat{n}\) _L = E_L ΔS cosθ = –E_L ΔS, since θ = 180°
= – E_La₂
φ_R = E_R . ΔS = E_R ΔS cosθ = E_R ΔS, since θ = 0°
= E_Ra²
Net flux through the cube
= φ_R + φ_L = E_Ra² – E_La² = a² (E_R – E_L) = \(\alpha\)a² [(2a)^1/2 – a^1/2]
= \(\alpha\)a^5/2 ( \(\sqrt{2}\)– 1)
= 800 (0.1)^5/2 ( \(\sqrt{2}\) – 1)
= 1.05 N m² C^–1
(b) We can use Gauss’s law to find the total charge q inside the cube. We have φ = q / ε₀ or q = φε₀. Therefore,
q = 1.05 x 8.854 x 10^–12 C = 9.27 x 10^–12 C.

(1) Given, C₁ = 2pF, C₂ = 3pF and C₃ = 4pF.
Equivalent capacitance for the parallel combination is given by C_eq .
Therefore, C_eq = C₁ + C₂ + C₃ = 2 + 3 + 4 = 9pF
Hence, the total capacitance of the combination is 9pF.
(2) Supply voltage, V = 100V
The three capacitors are having the same voltage, V = 100v
q = VC
where,
q = charge
C = capacitance of the capacitor
V = potential difference
for capacitance, c = 2pF
q = 100 x 2 = 200pC = 2 x 10^-10C
for capacitance, c = 3pF
q = 100 x 3 = 300pC = 3 x 10^-10C
for capacitance, c = 4pF
q = 100 x 4 = 400pC = 4 x 10^-10 C

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

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.

(a) \(U=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r}=9 \times 10^{9} \times \frac{7 \times(-2) \times 10^{-12}}{0.18}=-0.7 \mathrm{~J}\)
(b) W = U₂ – U₁ = 0 – U = 0 – (–0.7) = 0.7 J.
(c) The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find
\(q_{1} V\left(\mathbf{r}_{1}\right)+q_{2} V\left(\mathbf{r}_{2}\right)=A \frac{7 \mu \mathrm{C}}{0.09 \mathrm{~m}}+A \frac{-2 \mu \mathrm{C}}{0.09 \mathrm{~m}}\)
and the net electrostatic energy is
\(q_{1} V\left(\mathbf{r}_{1}\right)+q_{2} V\left(\mathbf{r}_{2}\right)+\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r_{12}}=A \frac{7 \mu C}{0.09 m}+A \frac{-2 \mu C}{0.09 m}-0.7 \mathrm{~J}\)
= 70 − 20 − 0.7 = 49.3 J

Let the value of unknown charge be q.
Electric field at 20 cm away, E = 1.5 \(\times\) 10³ N/C

From the formula, electric field,
\(\begin{aligned} E & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2} \end{aligned}\)
\(\begin{aligned} \Rightarrow 1.5 \times 10^3 & =\frac{9 \times 10^9 \times \mathrm{q}}{\left(20 \times 10^{-2}\right)^2} \\ \end{aligned}\)
\(\begin{aligned} \therefore \quad q =\frac{1.5 \times 10^3 \times 20 \times 20 \times 10^{-4}}{9 \times 10^9} \\ \end{aligned}\)
\(\begin{aligned} & =6.67 \times 10^{-9} \mathrm{C} \end{aligned}\)
As the electric field is radially inwards which shows that the nature of unknown charge q is negative
\(\begin{aligned} & \mathrm{E}_i=\lim _{q_0 \rightarrow 0} \frac{\mathrm{F}_i}{q_0}=\lim _{q_0 \rightarrow 0}\left[\frac{1}{q_0}\left(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_i q_0}{r_i^2} \hat{\mathrm{r}}_i\right)\right] \\ \end{aligned}\)
\(\begin{aligned} & \mathrm{E}_i=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_i}{r_i^2} \hat{\mathrm{r}}_i \end{aligned}\)
If E is electric field at point P due to the system of charges, then by principle of superposition of electric fields, E = E₁ + E₂ + E₃ + ... + E_n = \( \sum_{i=1}^{n}E_{i}\)
Using Eq. (i), we get \(\mathrm{E}=\sum_{i=1}^n \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_i}{r_i^2} \hat{\mathbf{r}}_i\)
or \(\mathrm{E}=\frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{r_i^2} \hat{\mathbf{r}}_i\)
\(\therefore\) E is a vector quantity.

Let us consider a charge q is placed at the centre of a cubic Gaussian surface. As per the question,
q = 2 \(\mu\)C = 2 \(\times\)10^-6 C
Length of edge = 9 cm

According to Gauss' theorem, the net electric flux (\(\phi\)) through the surface is given by
\(\phi=\frac{q}{\varepsilon_{0}}=\frac{2\times 10^{-6}}{8.854\times 10^{-12}} \)
= 2.26 \(\times\)10⁵ N-m² /C
Thus, the net electric flux through the surface is 2.26 \(\times\)10⁵ N-m²/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.