GATE ECE 2015 Set 1

To find the value of $k$ that allows for infinitely many solutions, we'll analyze the system using an augmented matrix and Gaussian elimination.

First, we set up the augmented matrix and perform row operations to reach row-echelon form:
$\left[\begin{array}{ccc|c} 1 & -2 & 3 & -1 \\ 1 & -3 & 4 & 1 \\ -2 & 4 & -6 & k \end{array}\right] \xrightarrow[\substack{R_2 \to R_2 - R_1 \\ R_3 \to R_3 + 2R_1}]{} \left[\begin{array}{ccc|c} 1 & -2 & 3 & -1 \\ 0 & -1 & 1 & 2 \\ 0 & 0 & 0 & k-2 \end{array}\right]$
A system has infinitely many solutions if the rank of the coefficient matrix, $\text{rank}(A)$, equals the rank of the augmented matrix, $\text{rank}([A|B])$, and this rank is less than the number of variables.

From the resulting matrix, we can see that $\text{rank}(A)=2$. For $\text{rank}([A|B])$ to also be 2, the last row must not represent a contradiction like $0=c$ where $c \neq 0$. This means the final entry in the last row must be zero.
Therefore, we must set $k-2=0$, which yields $k=2$.

The Mean Value Theorem states that for some $x$ in $(-1, 1)$, the derivative $f'(x)$ must equal the average rate of change over the interval $[-1, 1]$.

First, let's find the average rate of change, which is the slope of the line connecting the endpoints:
$\text{Slope} = \frac{f(1) - f(-1)}{1 - (-1)} = \frac{(1 - 1^2 + 1^3) - (1 - (-1)^2 + (-1)^3)}{2} = \frac{1 - (-1)}{2} = 1$

Next, we find the derivative of the function:
$f'(x) = -2x + 3x^2$

Now, we set the derivative equal to the slope and solve for $x$:
$3x^2 - 2x = 1 \implies 3x^2 - 2x - 1 = 0$

Factoring the quadratic equation gives $(3x+1)(x-1) = 0$, which yields solutions $x = -1/3$ and $x = 1$. Since the theorem requires the value to be in the open interval $(-1, 1)$, we choose $x = -1/3$.

The general addition rule for probability states that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. The simplified form $P(A \cup B) = P(A) + P(B)$ is only correct for mutually exclusive events, where the probability of their intersection is zero.

However, we are given that events A and B are independent. By definition, this means $P(A \cap B) = P(A)P(B)$. Since we are also told that $P(A) \neq 0$ and $P(B) \neq 0$, their product $P(A \cap B)$ must be a positive value. Thus, the subtraction term is non-zero, making the statement false.

To determine if a complex function is analytic, we must check if it satisfies the Cauchy-Riemann equations. Let's analyze the function $f(z) = \bar{z}$.

Writing this in terms of real and imaginary parts, we have $f(z) = x - iy$. This gives us the real part $u(x, y) = x$ and the imaginary part $v(x, y) = -y$.

The Cauchy-Riemann equations are $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$.
Let's check the first equation. We calculate the partial derivatives:
$\frac{\partial u}{\partial x} = 1$ and $\frac{\partial v}{\partial y} = -1$.

Since $1 \neq -1$, the first Cauchy-Riemann equation is not satisfied. Therefore, the function $\bar{z}$ is not analytic, making the statement that it is analytic untrue.

By definition, if a vector $\vec{v}$ is an eigenvector of a matrix $A$, then the product $A\vec{v}$ must be a scalar multiple of $\vec{v}$. This relationship is written as $A\vec{v} = \lambda\vec{v}$, where $\lambda$ is the eigenvalue.

Let's apply this to the given problem:
$\begin{bmatrix} 4 & 1 & 2 \\ p & 2 & 1 \\ 14 & -4 & 10 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = \lambda \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$
Performing the matrix multiplication on the left side yields a new vector:
$\begin{bmatrix} 12 \\ p+7 \\ 36 \end{bmatrix} = \begin{bmatrix} \lambda \\ 2\lambda \\ 3\lambda \end{bmatrix}$
From the first and third rows, which are free of the unknown $p$, we can determine the eigenvalue. The first row gives $12 = \lambda$, and the third row gives $36 = 3\lambda$, which also implies $\lambda = 12$.

Now, using the second row and our found eigenvalue $\lambda = 12$, we can solve for $p$:
$p+7 = 2\lambda \implies p+7 = 2(12) = 24$
Therefore, $p = 17$.

In a series RLC circuit at resonance, the inductive and capacitive reactances cancel each other out. The total impedance of the circuit is therefore purely resistive, $Z = R = 4 \, \Omega$.

This allows us to find the amplitude of the current flowing through the circuit:
$I = \frac{V}{R} = \frac{10 \, \text{V}}{4 \, \Omega} = 2.5 \, \text{A}$.

To find the voltage across the capacitor, we first calculate the resonant angular frequency, $\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(0.1 \times 10^{-3})(1 \times 10^{-6})}} = 10^5$ rad/s.

Using this frequency, we find the capacitive reactance: $X_C = \frac{1}{\omega C} = \frac{1}{10^5 \times (1 \times 10^{-6})} = 10 \, \Omega$.

Finally, the amplitude of the voltage across the capacitor is $V_C = I \cdot X_C = (2.5 \, \text{A})(10 \, \Omega) = 25 \, \text{V}$.

The key to simplifying this complex network is to recognize its symmetry. The circuit is symmetric both horizontally and vertically. This symmetry causes several nodes to be at the same electrical potential. For instance, the midpoint of the leftmost vertical resistor pair is at the same potential as the leftmost central node.

Because no current flows between points of equal potential, we can mentally remove the three resistors that connect these equipotential nodes. After this simplification, the circuit reduces to four parallel branches connected between terminals 'a' and 'b'. Two branches have a resistance of $R+R=2R$, and the other two (originating from the 'X' structures) each have an equivalent resistance of $R$.

The total equivalent resistance $R_{ab}$ is the parallel combination of these four branches:
$\frac{1}{R_{ab}} = \frac{1}{2R} + \frac{1}{2R} + \frac{1}{R} + \frac{1}{R} = \frac{1}{R} + \frac{2}{R} = \frac{3}{R}$
Therefore, the resistance is $R_{ab} = \frac{R}{3} = \frac{300 \, \Omega}{3} = 100 \, \Omega$.

Let's begin by applying Kirchhoff's Current Law (KCL) at the junction where the 5A current splits into the three parallel branches. The total current entering this junction is 5A.

The voltage across all parallel branches is $V_1$. The current in the middle $4\Omega$ resistor is $I$, so by Ohm's law, $V_1 = 4I$. This means the current in the other $4\Omega$ resistor is also $V_1/4 = I$. The third branch has a current of $2I$.

Summing these branch currents must equal the total incoming current: $5 \text{ A} = I + I + 2I = 4I$. Solving this gives us $I = 5/4 \text{ A}$.

Now we can directly calculate $V_1 = 4I = 4 \times (5/4) = 5 \text{ V}$.

To find $V_2$, we sum the voltage drops across the components in the main loop. $V_2$ equals the voltage drop across the series $4\Omega$ resistor plus the voltage $V_1$. The drop across the series resistor is $4\Omega \times 5\text{A} = 20 \text{ V}$. Therefore, $V_2 = 20 \text{ V} + V_1 = 20 \text{ V} + 5 \text{ V} = 25 \text{ V}$.

A negative differential resistance (NDR) region is the defining characteristic of a Tunnel (or Esaki) diode. This effect is based on the quantum mechanical tunneling of electrons. For tunneling to occur with a significant probability, the depletion region barrier must be extremely narrow (typically less than $10 \text{ nm}$). This is achieved by heavily doping both the P-region and the N-region (to concentrations above $10^{19} \text{ cm}^{-3}$). As the forward voltage increases, electrons tunnel from the N-side's conduction band to the P-side's valence band. Beyond a peak voltage ($V_p$), the energy bands misalign, reducing the available states for tunneling and causing the current to decrease, thus creating the NDR region.

The resistivity, $\rho$, of a material is the inverse of its electrical conductivity, $\sigma$. For a semiconductor, the conductivity depends on both electrons and holes. However, this sample is heavily doped with donors, making it an N-type semiconductor. In this case, the concentration of electrons ($n \approx N_D = 10^{16} \text{ cm}^{-3}$) is many orders of magnitude greater than the concentration of holes.

Consequently, the conductivity is dominated by the electrons, and we can use the simplified formula: $\sigma \approx n q \mu_n$, where $q$ is the elementary charge and $\mu_n$ is the electron mobility.

The resistivity can then be calculated as:
$\rho = \frac{1}{\sigma} \approx \frac{1}{n q \mu_n}$

Substituting the given values:
$\rho \approx \frac{1}{(10^{16} \text{ cm}^{-3})(1.6 \times 10^{-19} \text{ C})(1200 \text{ cm}^2/\text{V-s})} \approx 0.52 \text{ } \Omega\text{-cm}$

Let's analyze the circuit's behavior for each half of the input sine wave.

During the positive half-cycle ($v_{\in} > 0$), the input voltage attempts to drive current downwards. This forward biases both ideal diodes. The lower diode, being forward-biased and ideal, acts as a perfect short circuit. Since this short is directly in parallel with the output resistor, the voltage across the resistor, $v_{out}$, must be zero.

During the negative half-cycle ($v_{\in} < 0$), the polarity reverses. This reverse biases the lower diode, making it an open circuit. The circuit now effectively passes the input signal to the output resistor. For an ideal circuit, the output voltage simply follows the input voltage, so $v_{out} = v_{\in}$.

This combination of clipping the positive half and passing the negative half produces the waveform shown in option C.

To analyze this circuit, first determine if the Zener diode is active. We can do this by calculating the voltage across its terminals as if it were an open circuit. The circuit then becomes a simple voltage divider.

The voltage across the output resistor is calculated as:
$V_{o} = 10 \text{ V} \times \frac{1 \text{ k}\Omega}{1 \text{ k}\Omega + 1 \text{ k}\Omega} = 5 \text{ V}$

Since this calculated voltage ($5 \text{ V}$) is less than the Zener breakdown voltage ($V_Z = 6 \text{ V}$), the diode is not in its breakdown region. It behaves as an open circuit, so our initial voltage divider calculation correctly gives the final output voltage.

The total energy supplied by the source is the time integral of its power, $E = \int_0^{\infty } P(t) dt$, where the power is $P(t) = V_{source} \cdot i(t)$. First, we find the current $i(t)$ that flows while the capacitor is charging. The capacitor's voltage starts at $v_c(0) = 0$ V and charges towards its final value of $v_c(\infty ) = 3$ V.

The voltage across the capacitor follows the standard charging equation $v_c(t) = 3(1-e^{-t/\tau})$, with a time constant of $\tau = RC = 120 \, \Omega \times 0.1 \, \mu\text{F} = 12 \, \mu\text{s}$. The current is then found by $i(t) = C \frac{dv_c}{dt}$, which yields $i(t) = \frac{3C}{\tau}e^{-t/\tau} = \frac{1}{40}e^{-t/\tau}$ A.

Finally, we integrate the source power to find the total energy delivered:
$E = \int_0^\infty (3 \, \text{V}) \cdot i(t) \, dt = \int_0^\infty 3 \left(\frac{1}{40} e^{-t/\tau}\right) dt = \frac{3}{40}\$[-\tau e^{-t/\tau}]_0^\infty$
Evaluating the definite integral gives $E = \frac{3}{40}\tau = \frac{3}{40}(12 \times 10^{-6}) = 0.9 \times 10^{-6} \text{ J}$, which is $0.9 \, \mu\text{J}$.

In the 8085 microprocessor's architecture, the Accumulator, designated as register A, is the central register for arithmetic calculations. Following an addition, the resulting sum is stored in this register A.

Simultaneously, the processor updates various status flags to reflect the outcome of the operation. These flags, which include the Carry, Zero, Sign, and Overflow flags, are stored in a dedicated register known as the Flag Register, or F. Thus, the result resides in A and the overflow status in F.

The total memory capacity is 16 Kb, which is $16,384$ bits. In powers of two, this is $2^{14}$ bits.

A square memory array has an equal number of rows ($R$) and columns ($C$), so the total number of memory cells is $R \times C = R^2$.

We can set the total cells equal to the memory size: $R^2 = 2^{14}$. Solving for the number of rows gives $R = \sqrt{2^{14}} = 2^7$.

The row decoder must select one of these $2^7$ distinct rows. The number of address lines needed to select from $N$ items is $\log_2(N)$. Therefore, the decoder requires $\log_2(2^7) = 7$ address lines.

Step size $=0.0625 \mathrm{V}$ Decimal equivalent $=15$ Analog output $=15 \times 0.0625=0.9375 Volts$

$x(t)*\delta (-t-t_0)=x(-t)*\delta (t+t_0)=x(-t-t_0)$

First, let's find the expression for the new signal, $g(t)$. From the graph, the main triangular segment of $x(t)$ for the interval $-1 < t < 1$ is a line with the equation $x(t) = -3t$. Since $g(t) = x(\frac{t-1}{2})$, we substitute this new argument into the equation for $x(t)$, which gives us $g(t) = -3\left(\frac{t-1}{2}\right)$.

Next, we determine the period of $g(t)$. The original signal $x(t)$ is periodic with period $T_x = 3$. The transformation $t \to \frac{t}{2}$ is a time-scaling that expands the signal by a factor of 2. Therefore, the period of $g(t)$ is $T_g = 2 \times T_x = 6$. The time shift of $-1$ does not affect the period.

Now we can calculate the average power, which is the mean of the squared signal over one period. The original non-zero segment from $t \in (-1, 1)$ is mapped to $-1 < \frac{t-1}{2} < 1$, which means for $g(t)$ this segment exists over the interval $t \in (-1, 3)$. Since the signal is zero elsewhere in the period, we only need to integrate over this interval.

$P_{avg} = \frac{1}{T_g} \int_{\text{one period}} g(t)^2 dt = \frac{1}{6} \int_{-1}^{3} \left(-3\frac{t-1}{2}\right)^2 dt$
$P_{avg} = \frac{1}{6} \int_{-1}^{3} \frac{9}{4}(t-1)^2 dt = \frac{9}{24} \left[ \frac{(t-1)^3}{3} \right]_{-1}^{3} = \frac{3}{8} \left( \frac{2^3}{3} - \frac{(-2)^3}{3} \right) = \frac{3}{8} \left( \frac{16}{3} \right) = 2$.

Negative feedback is applied to an open-loop system with transfer function $G(s)$ to create a closed-loop system with transfer function $T(s) = \frac{G(s)}{1+G(s)H(s)}$. The denominator, $1+G(s)H(s)$, is known as the return difference. For most frequencies of interest, the magnitude of the loop gain $|G(s)H(s)|$ is large, making $|1+G(s)H(s)| > 1$. This reduces overall gain, reduces sensitivity to parameter variations ($S_G^T = \frac{1}{1+G(s)H(s)}$), and improves disturbance rejection by the same factor. However, feedback increases the system's responsiveness, which widens its bandwidth. For a first-order system, the closed-loop bandwidth is $\omega_{cl} = (1+K)\omega_{ol}$, where $K$ is the loop gain and $\omega_{ol}$ is the open-loop bandwidth. Thus, negative feedback increases, not reduces, bandwidth.

To find the gain $K$ where the root locus crosses the imaginary axis, we need to determine the condition for marginal stability. We can do this using the Routh-Hurwitz criterion. First, we find the characteristic equation of the closed-loop system: $1 + G(s) = 0$, which yields $s^3 + 4s^2 + 3s + K = 0$.

Next, we construct the Routh array from the coefficients of this equation:
$\begin{array}{c|cc} s^3 & 1 & 3 \\ s^2 & 4 & K \\ s^1 & \frac{12-K}{4} & 0 \\ s^0 & K & \end{array}$
The system becomes marginally stable, and the roots cross the imaginary axis, when an entire row of the array becomes zero. This occurs when the first term in the $s^1$ row is zero. By setting $\frac{12-K}{4} = 0$, we find that the required gain is $K = 12$.

To trace the polar plot, let's analyze the magnitude and phase of the system's frequency response, $G(j\omega) = \frac{10(1+j\omega)}{10+j\omega}$.

The phase angle of the system is given by $\phi(\omega) = \angle(1+j\omega) - \angle(10+j\omega)$, which simplifies to $\phi(\omega) = \tan^{-1}(\omega) - \tan^{-1}(0.1\omega)$.

At the starting frequency, $\omega=0$, the phase is $\phi(0) = \tan^{-1}(0) - \tan^{-1}(0) = 0^\circ$. At the ending frequency, $\omega \to \infty$, the phase approaches $90^\circ - 90^\circ = 0^\circ$.

For any frequency between these extremes ($0 < \omega < \infty$), the argument of the first tangent, $\omega$, is greater than the argument of the second, $0.1\omega$. Consequently, $\tan^{-1}(\omega) > \tan^{-1}(0.1\omega)$, which means the phase angle $\phi(\omega)$ is always positive.

Since the phase angle is always greater than or equal to zero and never exceeds $90^\circ$, the entire polar plot resides in the first quadrant.

To prevent slope overload, the maximum rate of change of the input signal must not exceed the maximum rate of change that the delta modulator can track.

For a sinusoidal signal, the maximum slope is its amplitude multiplied by its angular frequency, $A_m \cdot 2\pi f_m$. The modulator's maximum slope is its step size per sampling period, which is $\Delta / T_s$ or $\Delta \cdot f_s$.

Setting these two slopes equal gives the overload threshold:
$A_m \cdot 2\pi f_m = \Delta \cdot f_s$
Now, we can solve for the maximum amplitude, $A_m$, by substituting the given values:
$A_m \cdot 2\pi (2 \times 10^3) = (0.1) \cdot (20,000)$
$A_m = \frac{2000}{4000\pi} = \frac{1}{2\pi} \text{ V}$

The provided expression for $s(t)$ is the standard canonical representation for a band-pass signal. In this form, a baseband signal, described by its in-phase component $m(t)$ and quadrature component $\hat{m}(t)$, is modulated onto carriers at a high frequency $f_c$. This modulation process effectively shifts the spectrum of the original low-pass signal up to a frequency band centered around $f_c$. By definition, a signal whose energy is concentrated in a finite band of frequencies away from DC is a band-pass signal. This particular structure is also recognizable as a single-sideband (SSB) signal, which is a classic example of a band-pass signal.

The magnetic field intensity ($H$) around an infinitely long, straight current-carrying wire is described by Ampere's Law. At a radial distance $r$ outside the conductor, the magnitude of the azimuthal component of the magnetic field, $H_{\phi}$, is given by the formula $H_{\phi} = \frac{I}{2 \pi r}$, where $I$ is the current.

This equation shows that the field strength $H_{\phi}$ is inversely proportional to the distance $r$. This relationship, written as $H_{\phi} \propto \frac{1}{r}$, corresponds to a hyperbolic curve. The field is strongest near the conductor and weakens as you move farther away.

To determine the wavelength, we can compare the given electric field equation to the standard form of a plane wave, $E(z,t) = E_0 \cos(\omega t - \beta z)$. In our case, the equation is $\vec{E}(z,t)=\hat{a_{y}}2\cos(10^{8}t-\frac{z}{\sqrt{2}})$.

By matching the terms, we can identify the phase constant, $\beta$, which is the coefficient of the spatial variable $z$. From the equation, we see that $\beta = \frac{1}{\sqrt{2}}$ rad/m.

The phase constant and wavelength, $\lambda$, are related by the fundamental equation $\beta = \frac{2\pi}{\lambda}$.

To find the wavelength, we rearrange this formula to $\lambda = \frac{2\pi}{\beta}$.

Substituting the value of $\beta$ we found, we get $\lambda = \frac{2\pi}{1/\sqrt{2}} = 2\sqrt{2}\pi \approx 8.88$ m.

To solve this second-order differential equation, we begin with its characteristic equation: $m^2 + 2m + 1 = 0$.
Factoring this quadratic yields $(m+1)^2 = 0$, which gives us a repeated real root of $m = -1$.
The general form of the solution for a repeated root is $y(t) = (C_1 + C_2 t)e^{-t}$.
To find the constants $C_1$ and $C_2$, we apply the initial conditions. Using $y(0)=1$, we get $1 = (C_1 + 0)e^0$, which simplifies to $C_1 = 1$.
For the second condition, we first find the derivative: $y'(t) = C_2e^{-t} - (C_1 + C_2t)e^{-t}$.
Applying $y'(0)=1$, we get $1 = C_2e^0 - (C_1+0)e^0$, so $1 = C_2 - C_1$. Since $C_1=1$, we find that $C_2 = 2$.
Substituting the values of the constants, the specific solution is $y(t) = (1 + 2t)e^{-t}$.

To answer this, we need to check two conditions for the vector field $\vec{P}$: whether it is solenoidal (divergence is zero) and whether it is irrotational (curl is zero).

First, let's find the divergence of $\vec{P}$:
$\nabla \cdot \vec{P} = \frac{\partial}{\partial x}(x^3y) + \frac{\partial}{\partial y}(-x^2y^2) + \frac{\partial}{\partial z}(-x^2yz)$
$= 3x^2y - 2x^2y - x^2y = 0$
Since the divergence is zero, the vector field is solenoidal.

Next, we find the curl of $\vec{P}$ using the determinant formula:
$\nabla \times \vec{P} = \begin{vmatrix} \vec{a}_{x} & \vec{a}_{y} & \vec{a}_{z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^{3}y & -x^{2}y^{2} & -x^{2}yz \end{vmatrix} = -x^2z\,\vec{a}_{x} + 2xyz\,\vec{a}_{y} - (x^{3+}2xy^2)\vec{a}_{z}$
Because the result is not the zero vector, the field is not irrotational.

To identify the correct graph, we can analyze the function's turning points using calculus. First, find the derivative of $f(x)$ using the product rule: $f'(x) = e^{-x}(2x+1) - e^{-x}(x^{2+}x+1)$, which simplifies to $f'(x) = xe^{-x}(1-x)$. The critical points are where the slope is zero, so we set $f'(x) = 0$, giving us $x=0$ and $x=1$.

To classify these points, we use the second derivative test. The second derivative is $f''(x) = e^{-x}(x^2 - 3x + 1)$. Evaluating at our critical points, we find $f''(0) = 1$, which is positive, indicating a local minimum at $x=0$. We also find $f''(1) = -e^{-1}$, which is negative, indicating a local maximum at $x=1$. The only graph that features a local minimum at $x=0$ and a local maximum at $x=1$ is B.

Let a vertex of the rectangle in the first quadrant be $(x, y)$. By symmetry, the rectangle's dimensions are $2x$ and $2y$, making its area $A = 4xy$. To simplify maximization, we can work with the square of the area, $A^2 = 16x^2y^2$.

From the ellipse equation, $x^{2+}4y^2=1$, we can write $4y^2 = 1-x^2$. Substituting this into the squared-area expression gives us a function of $x$:
$A^2 = 4x^2(4y^2) = 4x^2(1-x^2) = 4x^2 - 4x^4$.

To find the maximum, we take the derivative with respect to $x$ and set it to zero:
$\frac{d}{dx}(4x^2 - 4x^4) = 8x - 16x^3 = 0$.

Solving $8x(1-2x^2)=0$ for a positive $x$ gives $x^2 = 1/2$, so $x = 1/\sqrt{2}$. Substituting this back into the ellipse equation gives $y = 1/\sqrt{8}$. The maximum area is $A = 4xy = 4 \left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{\sqrt{8}} \right) = 4 \left( \frac{1}{\sqrt{16}} \right) = 1$.

The damping ratio, represented by the Greek letter xi ($\xi$), is inversely related to the circuit's quality factor ($Q$). The fundamental relationship is $\xi = \frac{1}{2Q}$.

For a series RLC circuit, the quality factor is specifically defined as $Q = \frac{1}{R} \sqrt{\frac{L}{C}}$.

To find the damping ratio for this circuit, we can substitute the expression for $Q$ into the first equation:
$\xi = \frac{1}{2 \left( \frac{1}{R} \sqrt{\frac{L}{C}} \right)}$

Simplifying this expression gives us the final formula:
$\xi = \frac{R}{2} \sqrt{\frac{C}{L}}$

To solve for the capacitor voltage, we'll analyze the circuit's transient response. We need to find the initial voltage, the final (steady-state) voltage, and the time constant.

Given zero initial conditions, the capacitor voltage starts at $v_c(0) = 0$ V. As time approaches infinity, the capacitor acts like an open circuit. The final voltage across it is determined by the voltage divider formed by the resistors: $v_c(\infty ) = 10 \text{ V} \times \frac{2\Omega}{3\Omega+2\Omega} = 4$ V.

The time constant is $\tau = R_{eq}C$, where $R_{eq}$ is the Thévenin resistance seen by the capacitor. This is the parallel combination of the $3\Omega$ and $2\Omega$ resistors: $R_{eq} = \frac{3 \times 2}{3+2} = 1.2 \, \Omega$. This gives a time constant of $\tau = (1.2 \, \Omega)(\frac{5}{6} \, \text{F}) = 1$ second.

The general equation for the capacitor voltage is $v_c(t) = v_c(\infty ) + [v_c(0) - v_c(\infty )]e^{-t/\tau}$. Plugging in our values, we get $v_c(t) = 4 + (0-4)e^{-t/1} = 4(1 - e^{-t})$ Volts.

At $t = 1$ second, the voltage is $v_c(1) = 4(1 - e^{-1}) \approx 4(1 - 0.3679) \approx 2.528$ V.

To deliver maximum power to a resistive load from a complex source, the load resistance must equal the magnitude of the Thevenin impedance ($R_L = |Z_{Th}|$).

First, we find the Thevenin impedance $Z_{Th}$ by looking into the load terminals with the voltage source shorted. This places the $2\,\Omega$ resistor and $j2\,\Omega$ inductor in parallel:
$Z_{Th} = \frac{2 \times j2}{2+j2} = 1+j1 \, \Omega$.
For maximum power, we set $R_L = |Z_{Th}| = |1+j1| = \sqrt{1^{2+}1^2} = \sqrt{2} \approx 1.414 \, \Omega$.

Next, we find the Thevenin voltage $V_{Th}$ using the voltage divider rule across the open load terminals:
$V_{Th} = (4\angle0^\circ \text{ V}) \frac{j2}{2+j2} = 2\sqrt{2}\angle45^\circ \text{ V} \approx 2.828\angle45^\circ$ V.

The current through the load is $I = \frac{V_{Th}}{Z_{Th} + R_L}$. The magnitude of this current is $|I| = \frac{|2.828\angle45^\circ|}{|(1+j1) + 1.414|} \approx 1.08$ A.

The maximum power transferred to the load is then $P_{max} = |I|^2 R_L = (1.08)^2 (1.414) \approx 1.649$ W.

For an abrupt p-n junction, the junction capacitance ($C_J$) is inversely proportional to the square root of the total potential across the depletion region. This total potential is the sum of the built-in potential ($V_{bi}$) and the applied reverse bias ($V_R$).

We can express this relationship as $C_J \propto \frac{1}{\sqrt{V_{bi} + V_R}}$. To find the new capacitance ($C_{J2}$) from the initial capacitance ($C_{J1}$), we can set up a ratio:
$\frac{C_{J2}}{C_{J1}} = \sqrt{\frac{V_{bi} + V_{R1}}{V_{bi} + V_{R2}}}$
Substituting the given values: $V_{bi} = 0.75$ V, $C_{J1} = 5$ pF, $V_{R1} = 1.25$ V, and $V_{R2} = 7.25$ V.
$C_{J2} = 5 \text{ pF} \times \sqrt{\frac{0.75 + 1.25}{0.75 + 7.25}} = 5 \times \sqrt{\frac{2}{8}} = 5 \times \sqrt{\frac{1}{4}}$
This simplifies to $C_{J2} = 5 \times \frac{1}{2} = 2.5$ pF.

In saturation, a MOSFET's output resistance, $r_o$, accounts for the slight slope of the I-V curve caused by channel length modulation. This resistance is defined as the inverse of the output conductance, $g_o = \frac{\partial I_D}{\partial V_{DS}}$.

The drain current equation including this effect is $I_D = I_{D,sat}(1 + \lambda V_{DS})$, where $I_{D,sat}$ is the ideal saturation current and $\lambda$ is the channel length modulation coefficient.

Taking the derivative gives the conductance: $g_o = \frac{\partial I_D}{\partial V_{DS}} = \lambda I_{D,sat}$.
The output resistance is the reciprocal: $r_o = \frac{1}{\lambda I_{D,sat}}$.

We can approximate $I_{D,sat}$ with the given operating current $I_D = 1$ mA. Plugging in the values:
$r_o = \frac{1}{0.05 \text{ V}^{-1} \times 1 \times 10^{-3} \text{ A}} = 20 \times 10^3 \text{ } \Omega = 20 \text{ k}\Omega$.