GATE ECE 2014 Set 3

To locate the maximum value of a function, we start by finding its critical points. This is done by taking the derivative and setting it equal to zero.

The derivative of $f(x) = \ln(1+x) - x$ is $f'(x) = \frac{1}{1+x} - 1$.

Setting the derivative to zero to find critical points: $f'(x) = 0 \implies \frac{1}{1+x} - 1 = 0$. This equation simplifies to $\frac{1}{1+x} = 1$, which is only true when $x=0$.

To verify that this is a maximum, we apply the Second Derivative Test. The second derivative is $f''(x) = -\frac{1}{(1+x)^2}$.

At our critical point $x=0$, the second derivative is $f''(0) = -\frac{1}{(1+0)^2} = -1$. Since $f''(0) < 0$, the function is concave down at this point, confirming a local maximum. Thus, the maximum value occurs at $x=0$.

A first-order differential equation is defined as linear when it can be expressed in the form $\frac{dy}{dx} + P(x)y = Q(x)$. In this structure, both $P(x)$ and $Q(x)$ must be functions of the independent variable $x$ only.

The equation is further classified as "non-homogeneous" when the term $Q(x)$ is not zero.

Analyzing the equation $\frac{dy}{dx}+xy=e^{-x}$, we can identify that $P(x) = x$ and $Q(x) = e^{-x}$. Since $Q(x)$ is a non-zero function of $x$, this equation fits the definition of a linear non-homogeneous differential equation.

Here is a step-by-step explanation matching each application to its numerical method:

• P1: Numerical integration is the process of approximating a definite integral, $\int_a^b f(x)dx$. Simpson's 1/3-rule (M3) is a classic numerical method used for this exact purpose.
• P2: Solution to a transcendental equation involves finding the root of a non-algebraic equation (e.g., $e^x - x = 5$). The Newton-Raphson Method (M1) is an iterative root-finding algorithm for such equations.
• P3: Solution to a system of linear equations (e.g., $Ax=b$) is a fundamental problem in linear algebra. The Gauss Elimination Method (M4) is a standard direct method for solving these systems.
• P4: Solution to a differential equation of the form $\frac{dy}{dx} = f(x,y)$ is found numerically using step-by-step methods. The Runge-Kutta Method (M2) is one of the most widely used and accurate methods for this task.

This leads to the correct matching: P1-M3, P2-M1, P3-M4, P4-M2.

This problem describes a sequence of two independent events. First, for the fourth head to land on the tenth toss, we must have exactly 3 heads in the first 9 tosses. Second, the tenth toss itself must be a head.

The probability of getting 3 heads in 9 tosses is a binomial probability, calculated as ${}^9C_3 \left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^6 = {}^9C_3 \left(\frac{1}{2}\right)^9$.

The probability of the tenth toss being a head is simply $\frac{1}{2}$.

To find the total probability, we multiply the probabilities of these two events:
$P(\text{Total}) = \left[ {}^9C_3 \left(\frac{1}{2}\right)^9 \right]\$ \times \frac{1}{2} = {}^9C_3 \left(\frac{1}{2}\right)^{10}$
Calculating the values, we get:
$\frac{84}{1024} \approx 0.082$

To find the relationship between the derivatives, we must first compute the partial derivatives of $z = xy \ln(xy)$ with respect to both $x$ and $y$.

First, let's find $\frac{\partial z}{\partial x}$. We use the product rule, treating $y$ as a constant:
$\frac{\partial z}{\partial x} = y\ln(xy) + xy\left(\frac{1}{xy} \cdot y\right) = y\ln(xy) + y = y[\ln(xy)+1]$.

Next, we find $\frac{\partial z}{\partial y}$. The function is symmetric in $x$ and $y$, so we expect a similar result. Treating $x$ as a constant:
$\frac{\partial z}{\partial y} = x\ln(xy) + xy\left(\frac{1}{xy} \cdot x\right) = x\ln(xy) + x = x[\ln(xy)+1]$.

Now, we test these results. Let's multiply the first derivative by $x$ and the second by $y$:
$x\frac{\partial z}{\partial x} = x \cdot y[\ln(xy)+1] = xy(\ln(xy)+1)$.
$y\frac{\partial z}{\partial y} = y \cdot x[\ln(xy)+1] = xy(\ln(xy)+1)$.

Since both expressions are identical, we have confirmed that $x\frac{\partial z}{\partial x} = y\frac{\partial z}{\partial y}$.

To find the shape of the current $i(t)$, we can solve the given circuit equation. The Laplace transform is an effective tool for this, as it converts the integral equation into a simpler algebraic one. Applying the transform to the equation yields $\frac{V_S}{s} = RI(s) + \frac{1}{Cs}I(s)$, where $I(s)$ is the Laplace transform of the current.

We can now algebraically solve for $I(s)$. Factoring out $I(s)$ gives $\frac{V_S}{s} = I(s)\left(R + \frac{1}{Cs}\right)$. Rearranging, we find $I(s) = \frac{V_S}{Rs + \frac{1}{C}}$, which can be written in a standard form as $I(s) = \frac{V_S/R}{s + \frac{1}{RC}}$.

Finally, we take the inverse Laplace transform to find the current $i(t)$ in the time domain. The expression for $I(s)$ corresponds to the time function $i(t) = \frac{V_S}{R}e^{-t/RC}$. This equation describes a current that starts at a maximum value of $\frac{V_S}{R}$ at $t=0$ and exponentially decays toward zero, which matches the graph in option A.

To find the current $I$, we will use the superposition theorem by calculating the current from each power source independently.

First, let's find the current due to the 5V source alone, treating the 1A source as an open circuit. The three resistors are now in series, so the current is $I_1 = \frac{5\text{V}}{5\Omega + 5\Omega + 10\Omega} = \frac{5}{20}\text{A} = 0.25\text{A}$.

Next, we find the current due to the 1A source alone, treating the 5V source as a short circuit. We use the current divider rule to find the current $I_2$ flowing through the rightmost branch: $I_2 = 1\text{A} \times \frac{5\Omega}{5\Omega + (5\Omega + 10\Omega)} = 0.25\text{A}$.

Since both currents $I_1$ and $I_2$ flow downwards through the 10Ω resistor, we add them to find the total current: $I = I_1 + I_2 = 0.25\text{A} + 0.25\text{A} = 0.5\text{A}$.

In a MOSFET, the channel is the region in the semiconductor directly beneath the gate electrode where current flows from source to drain. The length of this channel, denoted as $L$, is a critical parameter that dictates the transistor's performance. This physical dimension is set by the length of the gate itself. The process of creating this precisely shaped gate from a layer of polysilicon is known as poly-silicon gate patterning. Therefore, it is this specific lithography and etching step that defines the all-important channel length.

In a semiconductor under illumination, electron-hole pairs are generated, creating excess carriers. The net recombination rate ($R$) is the rate at which these excess carriers recombine to return to thermal equilibrium. For a P-type semiconductor under low-level injection, the majority carrier (hole) concentration $p_0$ is much larger than the excess carrier concentration.

The recombination rate is fundamentally defined by how many excess minority carriers are present to recombine. For P-type silicon, the minority carriers are electrons. The recombination rate ($R$) is given by the excess minority carrier concentration ($\delta n$) divided by the minority carrier recombination lifetime ($\tau_n$):

$R = \frac{\delta n}{\tau_n}$

This equation shows that the recombination rate ($R$) is directly proportional to the excess minority carrier concentration ($\delta n$). While the lifetime $\tau_n$ is inversely proportional to the majority carrier concentration ($p_0$), the most direct relationship for the rate itself is with the quantity of what is recombining, which is $\delta n$.

The relationship between the diffusion constant ($D_p$) and the mobility ($\mu_p$) of charge carriers is described by the Einstein relation. This fundamental equation is given by:

$\frac{D_p}{\mu_p} = \frac{kT}{q} = V_T$

Here, $V_T$ represents the thermal voltage. To find the hole diffusion constant, we rearrange the formula to $D_p = \mu_p \times V_T$.

Using the provided values, we substitute $\mu_p = 500 \text{ cm}^2/\text{V-s}$ and the thermal voltage $V_T = 26 \text{ mV} = 0.026 \text{ V}$.

The calculation is:
$D_p = (500 \text{ cm}^2/\text{V-s}) \times (0.026 \text{ V}) = 13 \text{ cm}^2/\text{s}$.

A transconductance amplifier is a voltage-controlled current source (VCCS), where the output current is a function of the input voltage, expressed as $I_{out} = g_m \cdot V_{\in}$.

On the input side, to maximize the voltage transfer from the source to the amplifier's input, the amplifier's input resistance ($R_{\in}$) must be much higher than the source resistance ($R_s$). An ideal voltage-sensing device has infinite input resistance to avoid "loading" the source.

On the output side, the amplifier acts as a current source. This source is modeled in parallel with the amplifier's output resistance ($R_{out}$). To deliver the maximum possible current to the load ($R_L$), $R_{out}$ must be much larger than $R_L$ so that most of the generated current flows through the load.

Therefore, the desirable characteristics for a transconductance amplifier are a high input resistance and a high output resistance.

To find the value of $R_C$, we first need to determine the collector current, $I_C$. We can find this by analyzing the base-emitter input loop. The emitter voltage is $V_E=10V$ and the base-emitter drop is $|V_{BE}|=0.7V$, which means the voltage at the base is $V_B = 10V - 0.7V = 9.3V$. This voltage drives the base current $I_B$ through $R_B$ to ground, so $I_B = \frac{V_B}{R_B} = \frac{9.3V}{100k\Omega} = 0.093mA$.

The collector current is then simply the base current amplified by $\beta$: $I_C = \beta I_B = 50 \times 0.093mA = 4.65mA$.

This collector current flows through $R_C$, creating the output voltage $V_0$ across it. Applying Ohm's Law to the output, $V_0 = I_C R_C$. Since we want $V_0$ to be $5V$, we can solve for the required resistance:
$R_C = \frac{V_0}{I_C} = \frac{5V}{4.65mA} \approx 1.075 k\Omega$.

In this circuit, the capacitor acts as a filter, charging up to the peak input voltage and then slowly discharging. This keeps the output voltage approximately constant at the peak value, which is $V_p = 10 \text{ V}$.

The average current drawn by the load is determined by this DC voltage across the resistor. By conservation of charge, the average current supplied by the diode over a full cycle must equal this average load current. Therefore, we can estimate the average diode current as:

$I_{D, avg} \approx \frac{V_{out, DC}}{R} \approx \frac{V_p}{R} = \frac{10 \text{ V}}{100 \, \Omega} = 0.1 \text{ A}$

To find the maximum quantization error, we first need to calculate the size of each quantization interval, known as the step size ($\Delta$). This is found by dividing the total voltage range by the number of available digital levels. For a 4-bit converter, we have $2^4 = 16$ levels.

The step size is:
$\Delta = \frac{\text{Voltage Range}}{\text{Number of Levels}} = \frac{8 \text{ V}}{16} = 0.5 \text{ V}$

The quantization error is the difference between the true analog value and the quantized digital level. The maximum possible error occurs when the analog signal is exactly halfway between two quantization levels. This maximum error is always half of the step size.

Maximum Quantization Error = $\frac{\Delta}{2} = \frac{0.5 \text{ V}}{2} = 0.25 \text{ V}$

This circuit is a master-slave D flip-flop, a configuration that connects two gated D latches in series. The "master" latch and the "slave" latch are driven by complementary clock signals.

When the clock signal is at one level (e.g., high), the master latch is active and captures the input value from $D$. During this time, the slave latch is inactive. When the clock signal transitions to the opposite level (low), the master is disabled, and the slave becomes active, passing the value captured by the master to the final output. This two-step process effectively makes the flip-flop edge-triggered.

Let's determine the function of the circuit by analyzing each multiplexer stage.

First, consider the left multiplexer. Its output is controlled by the select line $S_1$. The output is $W$ when $S_1=0$ and $\bar{W}$ when $S_1=1$. This relationship can be written as an equation: $\bar{S_1}W + S_1\bar{W}$, which is the definition of the XOR function, $W \oplus S_1$.

Now, let this result, $W \oplus S_1$, be the input $X$ to the second multiplexer. The second MUX takes $X$ and its inverse $\bar{X}$ as its data inputs, selected by $S_2$. This configuration is identical in structure to the first MUX.

Therefore, the final output $F$ is given by $F = X \oplus S_2$.

Substituting the expression for $X$ back into our equation for $F$, we get the complete function: $F = (W \oplus S_1) \oplus S_2$. This is the expression for a three-input XOR gate.

The original signal, $x(t)$, contains two distinct frequencies: $f_1 = \frac{10\pi}{2\pi} = 5$ Hz and $f_2 = \frac{30\pi}{2\pi} = 15$ Hz. When a signal is sampled, its frequency spectrum is replicated at integer multiples of the sampling frequency, $f_s$. In this case, sampling at $f_s = 20$ Hz creates a periodic spectrum containing the original frequencies plus their aliases.

The spectrum of the sampled signal will thus have components at frequencies $|f_{\text{original}} \pm k \cdot f_s|$. The set of all positive frequencies present in this spectrum includes the original 5 Hz and 15 Hz, as well as their images like $|5-20|=15$ Hz, $5+20=25$ Hz, $15+20=35$ Hz, and so on.

The final step is reconstruction using an ideal low-pass filter with a cutoff frequency $f_c = 20$ Hz. This filter passes all frequency components below 20 Hz and rejects all components above it. Looking at the frequencies present in the sampled signal, both 5 Hz and 15 Hz are below the 20 Hz cutoff and will therefore be present in the reconstructed signal.

An all-pass system is defined by its magnitude response being unity for all frequencies. This property imposes a specific relationship between its poles and zeros: they must occur in conjugate reciprocal pairs.

Let's identify the pole and zero of the given transfer function, $H(z) = \frac{z^{-1}-b}{1-az^{-1}}$. The denominator, $1-az^{-1}$, gives a pole at $z=a$. The numerator, $z^{-1}-b$, gives a zero at $z=1/b$.

For the system to be all-pass, the zero must be the conjugate reciprocal of the pole. Since the pole is at $a$, the zero must be located at $1/a^*$. We can now equate the two expressions for the zero location:
$\frac{1}{b} = \frac{1}{a^*}$
Solving for $b$ gives us the relationship $b = a^*$.

To recover the message signal $m(t)$, we must capture all the information it contains. The message signal has frequency components up to a maximum frequency, $f_m = 5$ kHz.

According to the Nyquist-Shannon sampling theorem, the minimum rate required to perfectly represent a signal is twice its highest frequency. Even though we are sampling the modulated signal $y(t)$, the rate is dictated by the bandwidth of the original message we wish to recover.

Therefore, the minimum sampling rate is determined by the properties of $m(t)$.
$f_{s,min} = 2 \times f_m = 2 \times 5 \text{ kHz} = 10 \text{ kHz}$
This is because we can use bandpass sampling (a form of undersampling) on $y(t)$ at this rate to successfully alias the message signal's spectrum down to baseband for recovery.

Converting the block d iagram into signal flow graph on Forward paths, $P_{1}=G_{1} G_{2} ; P_{2}=G_{2} \cdot 1 ; P_{3}=1 \cdot 1=1$ So, the transfer function is $\frac{C(s)}{R(s)}=G_{1} G_{2}+G_{2}+1$

To find the system's output at steady state, we first analyze the system's response in the Laplace domain. The transform of the input signal, $x(t) = -3e^{2t}u(t)$, is $X(s) = \frac{-3}{s-2}$.

The output, $Y(s)$, is the product of the transfer function $H(s)$ and the input transform $X(s)$:
$Y(s) = H(s)X(s) = \left(\frac{s-2}{s+3}\right) \left(\frac{-3}{s-2}\right) = \frac{-3}{s+3}$
This result shows a pole-zero cancellation. The initial condition contributes a transient term to the response, but because the system's pole is at $s=-3$ (in the stable left-half plane), this transient term will decay to zero over time.

We can find the final value using the Final Value Theorem on the stable response:
$y(\infty ) = \lim_{s \rightarrow 0} sY(s) = \lim_{s \rightarrow 0} s\left(\frac{-3}{s+3}\right) = (0)\left(\frac{-3}{3}\right) = 0$

The actual propagation delay for a passband signal is captured by the group delay, $\tau_g$. This is a crucial concept that describes how long it takes for the "envelope" of the signal to travel from the transmitter to the receiver.

The formula for group delay is derived from the phase response: $\tau_g = -\frac{1}{2\pi} \frac{d\varphi(f)}{df}$.

To find the delay, we differentiate the given phase response, $\varphi(f) = -2\pi \alpha (f-f_{c})-2\pi \beta f_{c}$, with respect to frequency $f$. Since $\alpha$, $\beta$, and $f_c$ are constants, the derivative is simply $\frac{d\varphi(f)}{df} = -2\pi\alpha$.

Substituting this result back into the group delay formula gives:
$\tau_g = -\frac{1}{2\pi} (-2\pi\alpha) = \alpha$.

The instantaneous frequency deviation from the carrier, $\Delta f(t)$, is found by taking the time derivative of the argument of the cosine function (the total phase) and subtracting the carrier frequency component. Equivalently, we can just differentiate the phase modulation term and divide by $2\pi$.

The phase modulation term is $\phi(t) = \beta_1\sin(2\pi f_1 t) + \beta_2\sin(2\pi f_2 t)$.

Differentiating this term with respect to time and dividing by $2\pi$ gives the frequency deviation:
$\Delta f(t) = \frac{1}{2\pi} \frac{d\phi(t)}{dt} = \beta_1 f_1 \cos(2\pi f_1 t) + \beta_2 f_2 \cos(2\pi f_2 t)$.

The maximum value of this expression, $\Delta f_{\max}$, occurs when both cosine terms reach their peak value of 1. This gives the maximum possible deviation as the sum of the individual peak deviations: $\beta_1 f_1 + \beta_2 f_2$.

For an air-filled rectangular waveguide, the cut-off frequency $f_c$ of the $TE_{mn}$ mode is determined by its dimensions, $a$ and $b$. The governing equation is $f_c = \frac{c}{2} \sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}$, where $c$ is the speed of light.

Given the dimensions $a = 5 \text{ cm} = 0.05 \text{ m}$ and $b = 3 \text{ cm} = 0.03 \text{ m}$, we analyze the $TE_{21}$ mode, so the mode indices are $m=2$ and $n=1$.

Substituting these values into the formula:
$f_c = \frac{3 \times 10^8 \text{ m/s}}{2} \sqrt{\left(\frac{2}{0.05 \text{ m}}\right)^2 + \left(\frac{1}{0.03 \text{ m}}\right)^2}$

This simplifies to $f_c = 1.5 \times 10^8 \sqrt{(40)^2 + (33.33)^2} \approx 1.5 \times 10^8 \sqrt{2711.1}$.
The final calculation yields a cut-off frequency of approximately $7.81 \times 10^9 \text{ Hz}$, or $7810 \text{ MHz}$.

Signal distortion in this system is caused by wave reflections along the coaxial cable. A reflection occurs at any point where there is an impedance mismatch. If the receiver impedance doesn't match the cable ($Z_R \neq Z_0$), the signal reflects off the receiver. This reflected wave travels back to the transmitter. If the transmitter impedance is also mismatched ($Z_T \neq Z_0$), the wave reflects yet again, creating a series of echoes. These multiple reflections interfere with the primary signal, forming standing waves and corrupting the shape of the wideband signal, which is the definition of distortion. Thus, mismatches at both ends are required for this to occur.

To find the maximum value of a continuous function on a closed interval, we must check the function's values at the endpoints and at any critical points within the interval.

First, find the derivative of the function: $f'(x) = 6x^2 - 18x + 12$.
To find the critical points, we set the derivative to zero: $6x^2 - 18x + 12 = 0$.
Factoring out a 6 gives $x^2 - 3x + 2 = 0$, which solves to $(x-1)(x-2) = 0$.
The critical points are $x=1$ and $x=2$, which are both in the interval $[0, 3]$.

Now, evaluate the function $f(x)$ at the critical points and the endpoints ($x=0, x=3$):
$f(0) = -3$
$f(1) = 2 - 9 + 12 - 3 = 2$
$f(2) = 2(8) - 9(4) + 12(2) - 3 = 1$
$f(3) = 2(27) - 9(9) + 12(3) - 3 = 6$

Comparing these four values, the maximum value is $6$.

A key property of a real symmetric matrix is that its eigenvalues are guaranteed to be real. However, the statement's claim that they are also always positive is incorrect. We can show this with a simple counterexample.

Consider the matrix

. This matrix is both real and symmetric. The eigenvalues of a diagonal matrix are its diagonal elements, which for this matrix are $\lambda_1 = -2$ and $\lambda_2 = -1$. These eigenvalues are real, but they are negative. This disproves the assertion that the eigenvalues must always be positive.

Let's determine the average number of tosses by considering the two possible starting outcomes. The process stops once both a Head (H) and a Tail (T) have appeared.

We can partition all possible outcomes into two groups: sequences that start with a Head and sequences that start with a Tail.

1. Sequences starting with H: To meet the condition, these sequences must be HT, HHT, HHHT, and so on. We can calculate the contribution to the total expectation from this group by summing (number of tosses) × (probability of sequence) for each sequence. This gives the series:
   $E_H = 2 \cdot P(HT) + 3 \cdot P(HHT) + \dots = 2\left(\frac{1}{2}\right)^2 + 3\left(\frac{1}{2}\right)^3 + 4\left(\frac{1}{2}\right)^4 + \dots$
   This arithmetico-geometric series evaluates to $3/2$.

2. Sequences starting with T: By symmetry, the contribution from sequences starting with a Tail (TH, TTH, TTH, etc.) is identical: $E_T = 3/2$.

The total average number of tosses is the sum of the contributions from these two mutually exclusive groups, covering all possibilities:
Average Tosses = $E_H + E_T = \frac{3}{2} + \frac{3}{2} = 3$.

We can find this probability by integrating over the volume of the sample space, which is a unit cube. A straightforward approach is to condition on the value of $X_3$. By the law of total probability, we can average the probability $P(X_1+X_2 \leq x_3)$ over all possible values of $x_3 \in [0,1]$.

$P(X_1+X_2 \leq X_3) = \int_0^1 P(X_1+X_2 \leq x_3) \cdot f_{X_3}(x_3) \, dx_3$

Since $X_3$ is uniform on $[0,1]$, its density $f_{X_3}(x_3)$ is 1. For a fixed $x_3$, the probability $P(X_1+X_2 \leq x_3)$ corresponds to the area of a right triangle in the $X_1$-$X_2$ unit square with an area of $\frac{1}{2}x_3^2$.

Substituting this area into the integral gives:
$\int_0^1 \frac{x_3^2}{2} dx_3 = \left[ \frac{x_3^3}{6} \right]_0^1 = \frac{1}{6}$

The probability is $\frac{1}{6}$, which is approximately $0.167$.

To find the overall transfer function, we must analyze the full two-stage circuit, accounting for the loading effect the second network has on the first. A clear way to do this is with nodal analysis. Let's label the intermediate voltage between the two stages as $V_x(s)$.

Applying Kirchhoff's Current Law (KCL) at the output node, $V_3(s)$, gives a relationship between $V_x$ and $V_3$:
$\frac{V_x(s) - V_3(s)}{1/(Cs)} = \frac{V_3(s)}{R}$
Similarly, applying KCL at the intermediate node $V_x(s)$ provides a second equation involving $V_1(s)$, $V_x(s)$, and $V_3(s)$. Solving these two equations simultaneously for the ratio $\frac{V_3(s)}{V_1(s)}$ yields the transfer function:
$\frac{V_3(s)}{V_1(s)} = \frac{(RCs)^2}{1 + 3RCs + (RCs)^2}$
The given component values result in a time constant $RC = (10 \text{ k}\Omega)(100 \text{ }\mu\text{F}) = 1 \text{ s}$. Substituting $RC=1$ into our expression gives the final simplified result.
$\frac{V_3(s)}{V_1(s)} = \frac{s^2}{1 + 3s + s^2}$

To find the node voltage $V_2$, we can use nodal analysis. The voltage source between nodes $V_1$ and $V_2$ creates a supernode. The constraint equation for this supernode is given by the voltage source itself: $V_1 - V_2 = 10 \text{ V}$, which means $V_1 = V_2 + 10$.

Next, we write a KCL equation for the entire supernode. The sum of currents leaving the supernode must equal the sum of currents entering. This gives us:
$\frac{V_1}{-j3} + \frac{V_2}{6} + \frac{V_2}{j6} = 4$

Now, substitute $V_1 = V_2 + 10$ into the KCL equation:
$\frac{V_2 + 10}{-j3} + V_2 \left(\frac{1}{6} + \frac{1}{j6}\right) = 4$

To simplify, multiply the entire equation by $6j$:
$-2(V_2 + 10) + jV_2 + V_2 = 24j$
$(-1+j)V_2 - 20 = 24j \implies (-1+j)V_2 = 20 + j24$

Finally, we solve for $V_2$:
$V_2 = \frac{20 + j24}{-1 + j} = \frac{(20 + j24)(-1 - j)}{(-1)^2 + 1^2} = \frac{-20 - j20 - j24 + 24}{2} = \frac{4 - j44}{2} = 2 - j22 \text{ V}$

To find the angular frequency $\omega$ where the circuit is purely resistive, we first need to determine the equivalent impedance $Z_{bb'}$ as seen from the terminals $b-b'$. This is found by deactivating the independent voltage source (replacing it with a short circuit). The impedance then consists of the 1 F capacitor in series with the parallel combination of the 1 Ω resistor and the 0.5 H inductor.

The expression for the total impedance is:
$Z_{bb'} = Z_C + (R || Z_L) = \frac{1}{j\omega(1)} + \frac{1 \cdot (j\omega \cdot 0.5)}{1 + j\omega \cdot 0.5}$

To find when this impedance is purely resistive, we must find the value of $\omega$ that makes the imaginary part of $Z_{bb'}$ equal to zero. Let's separate the expression into its real and imaginary components:
$Z_{bb'} = -j\frac{1}{\omega} + \frac{j0.5\omega(1-j0.5\omega)}{1+(0.5\omega)^2} = \left(\frac{0.25\omega^2}{1+0.25\omega^2}\right) + j\left(\frac{0.5\omega}{1+0.25\omega^2} - \frac{1}{\omega}\right)$

Setting the imaginary part to zero gives the condition:
$\frac{0.5\omega}{1+0.25\omega^2} - \frac{1}{\omega} = 0 \implies 0.5\omega^2 = 1+0.25\omega^2$

Solving this equation for $\omega$ yields $0.25\omega^2 = 1$, or $\omega^2 = 4$. Since frequency must be positive, we find that $\omega = 2$ rad/s.

To determine the value of $R_{1}$, we must convert the given star (Y) network into its equivalent delta ($\Delta$) configuration.

The formula for the delta resistor opposite a specific leg of the star network is the sum of the two adjacent star resistors, plus their product divided by the third, opposite star resistor. For $R_{1}$, the adjacent resistors are $5\,\Omega$ and $3\,\Omega$, and the opposite resistor is $7.5\,\Omega$.

Applying the Y-to-$\Delta$ conversion formula:
$R_{1} = 5 + 3 + \frac{5 \times 3}{7.5}$
$R_{1} = 8 + \frac{15}{7.5} = 8 + 2 = 10\,\Omega$.

First, we find the built-in potential, $V_{bi}$, using the thermal voltage $V_T = kT/q$, the given donor ($N_D$) and acceptor ($N_A$) concentrations, and the intrinsic carrier concentration ($n_i$).
$V_{bi} = V_T \ln\left(\frac{N_A N_D}{n_i^2}\right) = (0.026 \text{ V}) \ln\left(\frac{(5 \times 10^{18})(1 \times 10^{16})}{(1.5 \times 10^{10})^2}\right) \approx 0.86 \text{ V}$
Next, we calculate the total depletion width, $W$, using the previously found built-in potential.
$W = \sqrt{\frac{2 \epsilon_{si}}{q} \left(\frac{1}{N_A} + \frac{1}{N_D}\right) V_{bi}}$
Since the acceptor concentration $N_A$ is much larger than the donor concentration $N_D$, the term $(1/N_A + 1/N_D)$ is dominated by the lighter doping side and simplifies to approximately $1/N_D$.
$W \approx \sqrt{\frac{2(1.04 \times 10^{-12})}{1.6 \times 10^{-19}} \left(\frac{1}{1 \times 10^{16}}\right) (0.86)} \approx 3.3 \times 10^{-5} \text{ cm}$