By applying KVL in the first loop,

V₁ = I₁ + 10I₂ + 2 (I₁ + I₂)

⇒ V₁ = 3I₁ + 12I₂

By applying KVL in the second loop,

V₂ = 6I₂ + 2(I₁ + I₂)

⇒ V₂ = 2I₁ + 8I₂

$Z=\left[\begin{array}{cc}3& 12\\ 2& 8\end{array}\right]$

|z| = 24 – 24 = 0

⇒ y – Parameter matrix doesn’t exist for the given circuit.

$\mathrm{V}\left(\mathrm{t}\right)=220\sqrt{2}\sin 377t$

ω = 377 rad/sec.

inductive reactance X_L = jωL = j 0.096 × 377

= j36.192 Ω

Capacitive reactance $X_c=\frac{1}{j\omega C}=\frac{1}{j\times 377\times 73.29\times {10}^{-6}}=j36.192\mathrm{\Omega }$

$\left|X_L\right|=\left|X_c\right|$

So, the circuit is operating at resonant frequency. At resonance, the circuit acts as purely resistive and hence the power factor in unity. 

Drawing the Thevenin equivalent of network N₁ and N₂ with proper polarities of voltage v₁ and v₂ and directions of current i₁ and i₂ is:

From the given graph

$\left|v_1\right|=5v=\left|v_2\right|$

$\left|i_1\right|=5mA=\left|i_2\right|$

Substituting the value the given circuit radius is

Apply KVL

$I=\frac{10}{5k}=2mA$

The number of twigs on a tree is always one less than the number of nodes.

Number of twigs = N – 1

The given diagram can be simplified as

By applying nodal analysis at node V₁,

$\frac{V_1}{12}+\frac{V_1-V_2}{6}+\frac{V_1-36}{2}=0$

⇒ V₁ + 2V₁ – 2V₂ + 6V₁ – 216 = 0

⇒ 9V₁ – 2V₂ = 216      ----(1)

By applying nodal analysis at node V₂,

$\frac{V_2}{18}+\frac{V_2-36}{9}+\frac{V_2-V_1}{6}=0$

⇒ V₂ + 2V₂ – 72 + 3V₂ – 3V₁ = 0

⇒ 6V₂ – 3V₁ = 72      ----(2)

By solving the equations (1) and (2), we get

V₁ = 30 V, V₂ = 27 V

$i_0=\frac{36-V_1}{2}=\frac{36-30}{2}=3A$.

Thevenin equivalent resistance across the terminals A – B is

Thevenin resistance is the equivalent resistance after removing all the independent sources.

V_x = 0 V

⇒ R_th = 10 kΩ

The given network can be represented as

By considering the expression for the turns ratio in the terms of impedances.

${\left(\frac{1}{a}\right)}^2=\frac{16}{10\times {10}^3}$

⇒ a = 25

V_s = 40 cos (ωt – 15°) V

= 40 sin (ωt + 75°) V

i = 40 sin (ωt + 21.87°) mA

$z=\frac{V_s}{i}=1000\mathrm{\angle }{53.13}^{\circ }=600+j800\mathrm{\Omega }\to \left(1\right)$

From the given circuit,

$z=600+j\omega \left(3.2\right)+\frac{1}{j\omega \left(2.5\times {10}^{-6}\right)}$

$=600+j\left(3.2\omega -\frac{0.4\times {10}^{-6}}{\omega }\right)\to \left(2\right)$

From equations (1) and (2)

$3.2\omega -\frac{1.4\times {10}^6}{\omega }=800$

⇒ ω² – 250ω - 125000 = 0

⇒ ω = 500 and ω = -250.

Negative value of ω is not possible.

At t = 0^-, the circuit is as shown below

$i_0\left(0^{-}\right)=10\left(\frac{2}{2+2}\right)=5mA$

i₀ (0⁺) = i₀ (0^-) = 5 mA

At t = ∞, the circuit is as shown below

i_A = 0,

$i_0\left(\mathrm{\infty }\right)=10\left(\frac{2}{2+2}\right)=5mA$

$i_0\left(t\right)=i_0\left(\mathrm{\infty }\right)+\left[i_0\left(0^{+}\right)-i_0\left(\mathrm{\infty }\right)\right]e^{-t/\tau }$

= 5 mA + [5 - 5] e^-t/τ

= 5 mA

At any time t > 0, i₀ = 5 mA 

Line voltage (VL) = 400 V

Phase voltage $\left(V_{ph}\right)=\frac{400}{\sqrt{3}}V$

When the switch is open

$I_{an}=\frac{400}{\sqrt{3}\left(40+j80\right)}$

$=\frac{10}{\sqrt{3}\left(1+2j\right)}$

$\left|I_{an}\right|=\frac{10}{\sqrt{15}}A$

rms value of voltage across the switch is

$V_s=\frac{10}{\sqrt{15}}\times 80=206.56V$

By applying superposition theorem,

$\begin{array}{l}{I}_1=\frac{10\sqrt{2}\mathrm{s}\mathrm{i}\mathrm{n}\left(1000t\right)}{1+j\left(1000\right)\left(1\times {10}^{-3}\right)}=\frac{10\sqrt{2}\mathrm{s}\mathrm{i}\mathrm{n}\left(1000t\right)}{1+j}\\ =\frac{10\sqrt{2}\mathrm{s}\mathrm{i}\mathrm{n}\left(1000t\right)}{\sqrt{2}\mathrm{\angle }{45}^{\circ }}\end{array}$  

= 10 sin (1000 t – 45°)

$I_2=\frac{10\sqrt{2}\mathrm{c}\mathrm{o}\mathrm{s}\left(3000t\right)}{1+j\left(3000\right)\left(1\times {10}^{-3}\right)}=\frac{10\sqrt{2}\mathrm{c}\mathrm{o}\mathrm{s}\left(3000t\right)}{1+3j}$

$=\frac{10\sqrt{2}\mathrm{c}\mathrm{o}\mathrm{s}\left(3000t\right)}{\sqrt{10}\mathrm{\angle }{71.56}^{\circ }}=\sqrt{20}\cos \left(3000t-{71.56}^{\circ }\right)$

I = I₁ + I₂

$I=10sin\left(1000t-{45}^{\circ }\right)+\sqrt{20}\mathrm{c}\mathrm{o}\mathrm{s}\left(3000t-{71.56}^{\circ }\right)$

$I_{rms}=\sqrt{{\left(\frac{10}{\sqrt{2}}\right)}^2+{\left(\frac{\sqrt{20}}{\sqrt{2}}\right)}^2}$

$=\sqrt{50+10}=\sqrt{60}$

$Powerdissipated\left(P\right)=I_{rms}^{2}R$

$={\left(\sqrt{60}\right)}^2\left(1\right)=60Watts$