GATE EE 2020

For a value to be a root, it must make the polynomial equal to zero.
(A) If we evaluate the polynomial at $x=0$, we get $P(0) = d$. By simply choosing $d=0$, we can guarantee that $x=0$ is a root, no matter the values of $a, b,$ and $c$.

(D) The nature of the roots (whether they are real or complex) depends on the relationships between all coefficients. The coefficient $c$ alone cannot force all roots to be real; for any given $c$, one can always choose the other coefficients to create complex roots.

For completeness: (B) is false, as a polynomial like $(x-1)^3 = 0$ has three identical roots. (C) is also false because for a polynomial with real coefficients, complex roots must come in conjugate pairs. A cubic must therefore have at least one real root.

To analyze the expression in option C, we can use the product-to-sum identity, which transforms the integrand into a sum of two sine functions: $\sin(p\theta)\cos(q\theta) = \frac{1}{2}[\sin((p+q)\theta) + \sin((p-q)\theta)]$. The integral of any term of the form $\sin(k\theta)$ is $-\frac{1}{k}\cos(k\theta)$. We must then evaluate the resulting cosine expressions at the limits of integration, $-\pi$ and $\pi$. For any constant $k$, this evaluation results in a term proportional to $\cos(k\pi) - \cos(-k\pi)$. Since the cosine function is even, meaning $\cos(x) = \cos(-x)$, this difference is always zero, causing the entire integral to be zero.

Consider a signal $x(t)$ that is bounded in amplitude and non-zero only over a finite time interval, say from $t_1$ to $t_2$. Its two-sided Laplace transform is given by the integral $X(s) = \int_{t_1}^{t_2} x(t)e^{-st} dt$. Because both the signal's value and the integration interval are finite, this integral is guaranteed to converge for every possible complex value of $s$. Consequently, the Region of Convergence (ROC) for such a signal is the entire s-plane. By definition, poles are locations where the transform $X(s)$ goes to infinity, and the ROC can never include any poles. If the ROC is the entire s-plane, there is no place for poles to exist.

Let's first analyze the original signal, $x[n] = \left(\frac{1}{2}\right)^n u[n]$. This is a standard right-sided sequence. Its Z-transform, $X(z)$, has a single pole at $z=1/2$, and its region of convergence (ROC) is the area outside this pole, so we have $|z| > 1/2$.

The signal in the question is $x[n-k]$, which is a time-delayed version of the original signal. The time-shifting property of the Z-transform tells us that the new transform is $z^{-k}X(z)$.

Multiplying by $z^{-k}$ may add or remove poles at $z=0$, but it does not change the location of any other poles from the original transform $X(z)$. Since the ROC is determined by the system's poles, it remains unaffected by a simple time shift. Therefore, the ROC for the shifted signal is the same as the original, which is $|z| > 1/2$.

To evaluate the integral, we first identify the singularities of the integrand by factoring the denominator:
$f(z) = \frac{z^{2+}1}{z^{2-}2z} = \frac{z^{2+}1}{z(z-2)}$
This shows that the function has simple poles at $z=0$ and $z=2$. The contour of integration $C$ is the unit circle, $|z|=1$. Of these two poles, only $z=0$ lies inside the contour.

By Cauchy's Residue Theorem, the integral is equal to $2\pi i$ times the residue at the enclosed pole. We can find the residue at the simple pole $z=0$ using the limit formula:
$\text{Res}(f, 0) = \lim_{z \to 0} (z-0) \frac{z^{2+}1}{z(z-2)} = \lim_{z \to 0} \frac{z^{2+}1}{z-2} = \frac{0^{2+}1}{0-2} = -\frac{1}{2}$
Therefore, the value of the integral is $2\pi i \times (-\frac{1}{2}) = -\pi i$.

Let's analyze how scaling a signal by a constant $k$ affects its average and RMS values, given the relationship $y(t) = kx(t)$.

The average (or mean) is a linear operation. This means a constant scalar can be factored out directly. Therefore, the average of the scaled signal $y(t)$ is simply $k$ times the average of the original signal $x(t)$. This gives us the relationship $y_A = kx_A$.

The RMS (root mean square) value is calculated from the signal's power. Power is proportional to the square of the signal, so we consider $y^2(t) = (kx(t))^2 = k^2x^2(t)$. The power of $y(t)$ is thus $k^2$ times the power of $x(t)$, which means $y_R^2 = k^2x_R^2$. Taking the square root gives $y_R = \sqrt{k^2x_R^2} = |k|x_R$. Assuming $k$ is a positive constant, as is common in such contexts, then $|k|=k$, which leads to $y_R = kx_R$.

The behavior of the generator is governed by the phasor equation $E_A = V_A + jI_A X_s$. The relationship between the magnitudes $|E_A|$ and $|V_A|$ depends entirely on how the voltage drop term, $jI_A X_s$, adds to the terminal voltage phasor $V_A$.

For a lagging load (Statement Q), the current $I_A$ lags $V_A$. The phasor $jI_A X_s$ will always add to $V_A$ in a constructive manner, making the resultant magnitude $|E_A|$ necessarily greater than $|V_A|$. Thus, statement Q is true.

For a leading load (Statement P), the current $I_A$ leads $V_A$. The phasor $jI_A X_s$ can oppose the $V_A$ phasor. While for a slightly leading load it's possible that $|V_A| < |E_A|$, for a sufficiently large leading power factor, we can have $|V_A| > |E_A|$. Since the statement claims this is always true, statement P is false.

A distance relay operates based on the impedance it measures to a fault. This relay is set to protect 80% of the line, so its tripping threshold corresponds to the impedance of that portion of the line.

The total line reactance is $0.2$ pu. The impedance seen by the relay for a fault at its 80% reach is:
$Z_{reach} = 0.80 \times 0.2 \text{ pu} = 0.16 \text{ pu}$

The threshold current for operation is the fault current that would flow for a solid fault occurring at this maximum reach. Using the given generator voltage of $1.0$ pu, we can find this current with Ohm's law:
$I_{fault} = \frac{V}{Z_{reach}} = \frac{1.0 \text{ pu}}{0.16 \text{ pu}} = 6.25 \text{ pu}$

In a load flow study, a PV bus represents a generator where the real power injection ($P$) and voltage magnitude ($V$) are specified as known inputs. The crucial question is how to determine the value of $P$ for each generator. This value is an operational setpoint, not just a physical rating. The process of economic load dispatch is used to optimally allocate the total required power generation among all available units to meet the system's load at the minimum possible cost. The solution of this economic dispatch calculation provides the exact real power output for each generator, which are then used as the specified $P$ values at the PV buses for the load flow analysis.

To find the poles of the system, we first need its transfer function, $H(s)$. We can obtain this by taking the Laplace transform of the differential equation, assuming zero initial conditions.

The equation $\frac{d^2y(t)}{dt^2}+4y(t)=6r(t)$ transforms into the s-domain as $s^2Y(s) + 4Y(s) = 6R(s)$.

Rearranging this to find the transfer function $H(s) = \frac{Y(s)}{R(s)}$, we get:
$H(s) = \frac{6}{s^{2+}4}$

The poles of a system are the roots of the denominator of its transfer function. We find them by solving the characteristic equation:
$s^{2+}4 = 0$

This gives $s^2 = -4$, so the poles are at $s = \pm\sqrt{-4}$, which simplifies to $s = \pm 2j$.

The presence of a very large inductor smooths the DC-side current, making it essentially constant. The full-bridge rectifier steers this current, causing the AC source to see a square wave of current drawn from it, which is in phase with the source voltage.

A fundamental principle of signal analysis is that a symmetrical square wave can be decomposed into a sum of sinusoids using a Fourier series. This series consists of a fundamental component at the wave's own frequency ($f$) and all its odd harmonics ($3f, 5f, 7f,$ etc.).

The amplitude of each harmonic is inversely proportional to its harmonic number ($n$). The Fourier series for the source current is given by $I_{source}(t) = \sum_{n=1,3,5,...}^{\infty } \frac{4I_{dc}}{n\pi} \sin(n\omega t)$.

The two "most dominant" components are those with the largest amplitudes, which are the fundamental ($n=1$) and the third harmonic ($n=3$). With a source frequency of $f = 50$ Hz, these correspond to frequencies of $1 \times 50 = 50$ Hz and $3 \times 50 = 150$ Hz.

To control the load power over its full range, we must find the firing angles ($\alpha$) corresponding to maximum and minimum power.

Maximum power (2 kW) is achieved at the minimum firing angle, $\alpha = 0^\circ$. Here, the thyristors conduct for the maximum possible duration, and the power delivered is $P_{max} = I_{rms}^2 R = (V_{rms}/|Z|)^2 \times R = (200/10)^2 \times (10 \cos 60^\circ) = 2 \text{ kW}$.

Minimum power (0 kW) occurs when the thyristors fail to turn on. The load is capacitive with a phase angle of $-60^\circ$, meaning the current leads the voltage by $60^\circ$. The steady-state current is proportional to $\sin(\omega t + 60^\circ)$, which becomes zero and then negative at $\omega t = 120^\circ$. If the firing pulse is applied at $\alpha \ge 120^\circ$, the thyristor will not conduct.

Therefore, to control the power from 2 kW down to 0, the firing angle $\alpha$ must be varied from $0^\circ$ to $120^\circ$.

To determine which option is equivalent, we can derive the system's transfer function, $H(z) = Y(z)/X(z)$, from the original diagram and compare it.

In the given graph, let's call the signal at the intermediate node $W(z)$. We can then write two equations: one for the feedback loop, $W(z) = X(z) + az^{-1}W(z)$, and one for the output, $Y(z) = W(z) + bz^{-1}W(z)$. By combining these, we find the overall transfer function is $H(z) = \frac{1+bz^{-1}}{1-az^{-1}}$.

Now, let's analyze the graph in option C. The output $Y(z)$ is formed by directly summing three signal paths: $Y(z) = X(z) + bz^{-1}X(z) + az^{-1}Y(z)$. If we rearrange this equation to solve for $Y(z)/X(z)$, we get $Y(z)(1-az^{-1}) = X(z)(1+bz^{-1})$. This yields the exact same transfer function, confirming that this representation is equivalent to the original.

The voltage gain ($A_v$) of a common-source amplifier is found by multiplying the negative of the transconductance ($g_m$) by the drain resistance ($R_D$). The fundamental formula is $A_v = -g_m R_D$. The negative sign indicates the characteristic 180° phase inversion between the input and output signals.

Using the given values, $g_m = 520 \mu A/V$ and $R_D = 4.7 k\Omega$. Note that the 10V power supply value is not needed for this calculation.

Substituting the values into the gain equation:
$A_v = -(520 \times 10^{-6} A/V) \times (4.7 \times 10^3 \Omega)$
$A_v = -2.444$

This sequence detector is designed to output a 1 the moment the target sequence $(1,0,1)$ is completed, and 0 otherwise. To find the output, we slide a 3-bit window across the input data and check for a match at each step. The first output is generated only after the third input bit is received.

Let's trace this process with the input $(1,1,0,1,0,0,1,1,0,1,0,1,1,0)$:
The first window is $(1,1,0)$, which is not a match (output 0). The second window is $(1,0,1)$, which is a match (output 1). Continuing this process, the sequence $(1,0,1)$ is detected again ending at the 10th and 12th input bits. For all other 3-bit windows, the output is 0.

Assembling the outputs generated from each consecutive window results in the final output sequence: $(0,1,0,0,0,0,0,1,0,1,0,0)$.

This problem presents a first-order linear differential equation. We start by arranging it into the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, which gives us $\frac{dy}{dx} + y = 2x$. The integrating factor is $I(x) = e^{\int 1 dx} = e^x$. Multiplying the standard form by $e^x$ allows us to simplify the left side to the derivative of a product: $\frac{d}{dx}(ye^x) = 2xe^x$.

Integrating both sides with respect to $x$ yields $ye^x = \int 2xe^x dx$. We solve the integral using integration by parts, which gives $ye^x = 2xe^x - 2e^x + C$. The general solution is found by dividing by $e^x$: $y = 2x - 2 + Ce^{-x}$.

Now, we apply the initial condition $y(0)=1$ to find the constant $C$. Plugging in the values, we get $1 = 2(0) - 2 + Ce^0$, which simplifies to $1 = -2 + C$, so $C=3$. The particular solution to our initial value problem is $y = 2x - 2 + 3e^{-x}$.

Finally, we substitute $x = \ln 2$ into this solution: $y = 2\ln 2 - 2 + 3e^{-\ln 2} = 2\ln 2 - 2 + 3/2$. Using a calculator, this is approximately $1.386 - 2 + 1.5 = 0.886$.

To find the speed regulation, we must first determine the motor's speed at no-load and full-load. Both speeds depend on the synchronous speed, $N_s$, which is the speed of the stator's rotating magnetic field.

$N_s = \frac{120f}{P} = \frac{120 \times 50}{4} = 1500$ rpm.

Using the given slips, we can find the no-load speed ($N_{NL}$) and full-load speed ($N_{FL}$):
$N_{NL} = N_s(1-s_{NL}) = 1500(1 - 0.01) = 1485$ rpm.
$N_{FL} = N_s(1-s_{FL}) = 1500(1 - 0.05) = 1425$ rpm.

Speed regulation is defined as the change in speed from no-load to full-load, expressed as a percentage of the full-load speed.
$\% \text{ Speed Regulation} = \frac{N_{NL} - N_{FL}}{N_{FL}} \times 100 = \frac{1485 - 1425}{1425} \times 100 = 4.21\%$.

According to Kirchhoff's Current Law (KCL), the current entering the junction, measured by ammeter A1, must equal the sum of the currents leaving it, measured by A2 and A3. Since these are AC currents, we must perform a phasor addition.

The total current $I_1$ is the sum of $I_2$ and $I_3$:
$I_1 = I_2 + I_3 = 1\angle 10^{\circ} + 1\angle 70^{\circ}$
To add these phasors, we convert them to rectangular coordinates, sum the real and imaginary parts, and then convert the result back to polar form. This calculation yields:
$I_1 \approx 1.732\angle 40^{\circ} \text{ A}$
An ideal ammeter measures the magnitude (or RMS value) of the current. Therefore, the reading on ammeter A1 is the magnitude of the resultant phasor, which is $1.732$ A.

To find the Thevenin voltage ($V_{TH}$), we can use the superposition theorem, analyzing the contribution of each source one at a time.

First, consider the 4 V voltage source and deactivate the 5 A current source by treating it as an open circuit. With the current path broken, no current flows through the 2 Ω and 3 Ω resistors. Therefore, there is no voltage drop across them, and the voltage at the terminals is simply the source voltage. So, the contribution from the voltage source is $V_{TH1} = 4 \text{ V}$.

Next, consider the 5 A current source and deactivate the 4 V voltage source by replacing it with a short circuit. Since the output terminals are open, the entire 5 A current flows through the 2 Ω resistor. The voltage at the node to the right of the 2 Ω resistor is $V_{TH2} = 5 \text{ A} \times 2 \, \Omega = 10 \text{ V}$.

The total Thevenin voltage is the sum of the individual contributions: $V_{TH} = V_{TH1} + V_{TH2} = 4 \text{ V} + 10 \text{ V} = 14 \text{ V}$.

Let's analyze the current flow in the circuit. By applying Kirchhoff's Current Law at the node connecting the load, the IGBT, and the diode, the load current $I_{load}$ splits between the other two components. This gives us the relationship: $I_{load} = I_{IGBT} + I_D$.

To find the current stress on the IGBT, we solve for its current: $I_{IGBT} = I_{load} - I_D$. Since the load is highly inductive, we can assume the load current $I_{load}$ remains nearly constant during the very fast switching event.

With $I_{load}$ constant, the IGBT current $I_{IGBT}$ will be at its maximum when the diode current $I_D$ is at its minimum. Looking at the provided graph of $I_D$ versus time, the current reaches its most negative value at point 3. At this moment, the IGBT must conduct the full load current plus the peak reverse recovery current of the diode.

The no-load losses in a transformer are core losses, which can be separated into hysteresis and eddy current losses. A crucial insight for this problem is that the ratio of applied voltage to frequency ($V/f$) is constant across all three scenarios ($200/50 = 160/40 = 100/25 = 4$). This condition implies a constant maximum flux density in the core.

When flux density is constant, the total core loss ($P_{loss}$) is modeled by the equation $P_{loss} = Af + Bf^2$, where $A$ and $B$ are constants representing hysteresis and eddy current effects, respectively. We can use the two given data points to create a system of equations:
$450 = 50A + (50)^2 B$
$320 = 40A + (40)^2 B$

To solve this system efficiently, we can divide each equation by its respective frequency:
$9 = A + 50B$
$8 = A + 40B$

Subtracting the second equation from the first gives $1 = 10B$, so $B = 0.1$. Substituting this value back, we find $A = 4$. Now we can calculate the losses for the 100 V, 25 Hz case:
$P_{loss} = (4)(25) + (0.1)(25)^2 = 100 + 62.5 = 162.5$ W.

The amplitude of the $n^{th}$ harmonic ($V_n$) for the given quasi-square wave is described by the formula $V_n = \frac{4V_s}{n\pi}\cos(n\sigma)$. Our first goal is to eliminate the 3rd harmonic, so we set its amplitude $V_3$ to zero. This condition, $\cos(3\sigma) = 0$, is satisfied when $3\sigma = \pi/2$, which gives a control angle of $\sigma = \pi/6$.

Now, we use this angle to find the magnitudes of the fundamental ($V_1$) and 5th harmonic ($V_5$) components. The ratio of their magnitudes is:
$|\frac{V_5}{V_1}| = |\frac{\frac{4V_s}{5\pi}\cos(5\sigma)}{\frac{4V_s}{\pi}\cos(\sigma)}| = |\frac{\cos(5\sigma)}{5\cos(\sigma)}|$

Substituting $\sigma = \pi/6$, we get:
$|\frac{V_5}{V_1}| = |\frac{\cos(5\pi/6)}{5\cos(\pi/6)}| = |\frac{-\sqrt{3}/2}{5(\sqrt{3}/2)}| = |-\frac{1}{5}| = 0.2$

Expressed as a percentage, the magnitude of the 5th harmonic is $0.2 \times 100\% = 20\%$ of the fundamental.

The defining characteristic of droop control is a linear relationship between the generator's power output and the system frequency. This means the slope of the frequency-power graph is constant.

First, let's calculate this slope from the initial load change. Power increased by $10$ MW (from $100$ MW to $110$ MW), causing a frequency drop of $0.25$ Hz (from $50$ Hz to $49.75$ Hz). The magnitude of the slope is:
$\text{Slope} = \frac{\Delta f}{\Delta P} = \frac{50 - 49.75 \text{ Hz}}{110 - 100 \text{ MW}} = \frac{0.25}{10}$
Now we use this constant slope for the second load change. The frequency drops further from $49.75$ Hz to $49.25$ Hz, a difference of $0.5$ Hz. Let the new power output be $P$.
$\frac{0.25}{10} = \frac{49.75 - 49.25}{P - 110} = \frac{0.5}{P - 110}$
Solving for the new power level $P$:
$P - 110 = \frac{0.5 \times 10}{0.25} = 20 \text{ MW}$
$P = 110 + 20 = 130 \text{ MW}$

We can solve this using the Nyquist stability criterion, which is given by the formula $N = P - Z$.

Here, $P$ represents the number of open-loop poles in the right-half of the s-plane. Looking at the denominator of $G(s)$, the term $(s-b)$ corresponds to a pole at $s=b$, so we have $P=1$.

The problem states that the Nyquist plot of $(1+G(s))$ encircles its origin once clockwise. This is equivalent to the plot of $G(s)$ encircling the critical point $(-1,0)$ once clockwise. A clockwise encirclement means $N = -1$.

Finally, $Z$ is the number of zeros of the characteristic equation $1+G(s)=0$ in the right-half plane. These are the closed-loop poles we are looking for. Substituting our values into the criterion gives $-1 = 1 - Z$. Solving this equation yields $Z = 2$.

The dashed box encloses the entire MOS device, which is an electrically isolated system. The device starts in an uncharged condition, meaning the total net charge within the box is initially zero.

Applying a voltage between the gate and body only causes a redistribution of charges within this closed system. For any positive charge that builds up in one region, a corresponding negative charge must appear elsewhere inside the box. Since no charge enters or leaves the box, the principle of charge conservation dictates that the total charge must remain zero. The given oxide charge $+Q$ is just one component of this internal distribution, but the overall sum remains unchanged.

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

First, we find the critical points by setting the derivative of $y = 3x^2 + 3x + 1$ to zero. The derivative is $\frac{dy}{dx} = 6x + 3$. Setting this to zero gives $6x + 3 = 0$, which yields the critical point $x = -1/2$. This point lies within our interval $[-2, 0]$.

Next, we evaluate the function at this critical point and at the two endpoints of the interval:

• At the endpoint $x = -2$: $y = 3(-2)^2 + 3(-2) + 1 = 7$.
• At the critical point $x = -1/2$: $y = 3(-1/2)^2 + 3(-1/2) + 1 = \frac{1}{4}$.
• At the endpoint $x = 0$: $y = 3(0)^2 + 3(0) + 1 = 1$.

Comparing the three resulting values $\{7, \frac{1}{4}, 1\}$, the largest is 7 (the maximum) and the smallest is $\frac{1}{4}$ (the minimum).

A vector field is conservative if, and only if, its curl is zero. For the given field $F$, this condition is $\nabla \times F = 0$. Let's break down the field into its components: $F_x = 5y - k_1z$, $F_y = 3z + k_2x$, and $F_z = k_3y - 4x$.

We calculate the curl and require each of its components to be zero:
The x-component: $\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = k_3 - 3 = 0 \implies k_3 = 3$.
The y-component: $\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} = -k_1 - (-4) = 4 - k_1 = 0 \implies k_1 = 4$.
The z-component: $\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = k_2 - 5 = 0 \implies k_2 = 5$.

Therefore, the constants must be $k_1=4$, $k_2=5$, and $k_3=3$.

To find the motor's speed under load, we'll compare the no-load and loaded conditions. The key is the relationship between speed ($N$), back EMF ($E_b$), and flux ($\phi$), given by $N \propto E_b / \phi$.

First, let's find the back EMF in each case using the formula $E_b = V - I_a R_a - V_{brush}$. The total brush voltage drop is $2 \times 1V = 2V$.
Under no-load conditions: $E_{b0} = 250V - (5A)(0.2\Omega) - 2V = 247V$.

Under loaded conditions, the shunt current is $I_{sh} = 250V / 100\Omega = 2.5A$. The armature current is $I_{aL} = I_{Line} - I_{sh} = 50A - 2.5A = 47.5A$.
So, the loaded back EMF is: $E_{bL} = 250V - (47.5A)(0.2\Omega) - 2V = 238.5V$.

Now, we can use the speed ratio. Let $N_0$ and $N_L$ be the no-load and loaded speeds. The flux is reduced by 5%, so $\phi_L = 0.95\phi_0$.
$\frac{N_L}{N_0} = \frac{E_{bL}}{E_{b0}} \times \frac{\phi_0}{\phi_L}$
Plugging in the values:
$N_L = 1200 \text{ rpm} \times \frac{238.5V}{247V} \times \frac{\phi_0}{0.95\phi_0} \approx 1218.4 \text{ rpm}$
This value is closest to 1220 rpm.

To determine the bus admittance matrix ($Y_{bus}$), we must express the injected bus currents ($I_i, I_j$) in terms of the bus voltages ($V_i, V_j$).

First, let's find the current, $I$, flowing through the transmission line itself. The ideal transformers modify the bus voltages, creating effective voltages of $t_iV_i$ and $t_jV_j$ at the line's terminals. The current through the line's admittance $Y$ is therefore $I = Y(t_iV_i - t_jV_j)$.

Next, we relate this line current back to the bus currents. The current injected at bus $i$, $I_i$, is the line current reflected through its transformer, giving $I_i = t_i I$. Similarly, the current injected at bus $j$ is $I_j = -t_j I$, where the negative sign indicates the direction of current injection relative to the line current.

By substituting the expression for $I$ into these equations, we get:
$I_i = t_i [Y(t_iV_i - t_jV_j)] = t_i^2 Y V_i - t_i t_j Y V_j$
$I_j = -t_j [Y(t_iV_i - t_jV_j)] = -t_i t_j Y V_i + t_j^2 Y V_j$

Arranging these two equations into the matrix form

gives us the final $Y_{bus}$ matrix.

This circuit is a standard model for a source like a solar cell. To transfer maximum power to a load, we must minimize the internal power losses.

The series resistor, $R_1$, represents an internal resistance that causes a voltage drop and dissipates power. To minimize this loss, $R_1$ should be as small as possible.

The shunt resistor, $R_2$, represents a leakage path that diverts current away from the output. To minimize this lost current, $R_2$ should be as large as possible.

Therefore, for maximum power transfer, the ideal design requires a very small series resistance ($R_1$) and a very large shunt resistance ($R_2$).

We can solve this using the Shockley diode equation, $I = I_0(e^{V_D / (n V_T)} - 1)$, where $I_0$ is the reverse saturation current. First, let's find the value of the term $n V_T$, which is the product of the ideality factor and the thermal voltage: $n V_T = (15/13) \times 26 \text{ mV} = 30 \text{ mV}$.

At the initial voltage of $V_{D1} = -0.03 \text{ V} = -30 \text{ mV}$, the current is $I_1 = I_0(e^{-30\text{mV}/30\text{mV}} - 1) = I_0(e^{-1} - 1)$.

We need to find the new voltage, $V_{D2}$, where the current is $1.5 I_1$. We can set up a second equation: $1.5 I_1 = I_0(e^{V_{D2} / 30\text{mV}} - 1)$. By substituting the first equation into the second, the $I_0$ terms cancel out: $1.5(e^{-1} - 1) = e^{V_{D2} / 30\text{mV}} - 1$.

Solving for the exponential term gives $e^{V_{D2} / 30\text{mV}} = 1.5(e^{-1} - 1) + 1 \approx 0.04$. To find $V_{D2}$, we take the natural logarithm of both sides, resulting in $V_{D2} = (30 \text{ mV}) \times \ln(0.04) \approx -0.0965 \text{ V}$.

The power delivered to the load is the product of its voltage and current, $P_L = V_L I_L$. To maximize this power, we must consider the operating limits of the power supply. The supply can deliver a maximum current of 4 A and can maintain a maximum voltage of 10 V. The single point where both of these maximums are achieved simultaneously is at the transition, or "corner," of its operating characteristic. At this point, the load voltage is $V_L = 10$ V and the load current is $I_L = 4$ A. Using Ohm's Law, the load resistance that creates this condition is $R_L = \frac{V_L}{I_L} = \frac{10 \text{ V}}{4 \text{ A}} = 2.5 \text{ } \Omega$.