GATE EE 2015 Set 2

This statement is true due to the sum rule of differentiation, which holds for complex functions just as it does for real functions.

The differentiability of $g(z)$ and $h(z)$ at $z_0$ means that the limits defining their derivatives exist:
$g'(z_0) = \lim_{\Delta z \to 0} \frac{g(z_0 + \Delta z) - g(z_0)}{\Delta z}$
$h'(z_0) = \lim_{\Delta z \to 0} \frac{h(z_0 + \Delta z) - h(z_0)}{\Delta z}$

To check if $f(z)$ is differentiable, we examine its derivative's limit definition:
$f'(z_0) = \lim_{\Delta z \to 0} \frac{f(z_0 + \Delta z) - f(z_0)}{\Delta z} = \lim_{\Delta z \to 0} \frac{[g(z_0 + \Delta z) + h(z_0 + \Delta z)] - [g(z_0) + h(z_0)]}{\Delta z}$

Using the properties of limits, we can separate this into:
$f'(z_0) = \lim_{\Delta z \to 0} \frac{g(z_0 + \Delta z) - g(z_0)}{\Delta z} + \lim_{\Delta z \to 0} \frac{h(z_0 + \Delta z) - h(z_0)}{\Delta z}$

Since both individual limits exist, their sum exists. Therefore, $f(z)$ is differentiable at $z_0$ and its derivative is $f'(z_0) = g'(z_0) + h'(z_0)$.

Let's analyze the system of equations using the matrix form $Ax=b$, where $A$ is the coefficient matrix. The four statements P, Q, R, and S are classic, equivalent conditions for a matrix to be invertible.

We can establish their equivalence by linking each one to statement S: the determinant of the coefficient matrix is nonzero.

• S $\equiv$ P: If the determinant of $A$ is nonzero ($\det(A) \neq 0$), the matrix is invertible. This allows us to find a unique solution given by $x = A^{-1}b$.
• S $\equiv$ Q: A square matrix has linearly independent rows if and only if its determinant is nonzero. Since the rows of $A$ represent our equations, this means the equations are linearly independent.
• S $\equiv$ R: The determinant of a matrix is equal to the product of its eigenvalues. For this product to be nonzero, every single eigenvalue must be nonzero.

Since P, Q, and R are all logically equivalent to S, they are all equivalent to each other.

This question requires matching four fundamental theorems to their mathematical expressions. Let's examine each one.

Stokes's Theorem (P) connects the line integral of a vector field around a closed loop to the surface integral of the field's curl. This relationship, $\iint (\nabla \times \mathbf{A})\cdot d\mathbf{s} = \oint \mathbf{A} \cdot d\mathbf{l}$, corresponds to expression (4).

Gauss's Theorem (Q), often called Gauss's Law in electromagnetism, states that the electric flux through a closed surface equals the enclosed charge $Q$, matching formula (1): $\oiint \mathbf{D}\cdot d\mathbf{s} = Q$.

The Divergence Theorem (R) relates the volume integral of a vector field's divergence to the flux of that field through a closed surface, as shown in formula (3): $\iiint (\nabla \cdot \mathbf{A})dV = \oiint \mathbf{A}\cdot d\mathbf{s}$.

Finally, Cauchy's Integral Theorem (S) from complex analysis states that the integral of an analytic function $f(z)$ over a closed path is zero, which is expression (2): $\oint f(z)dz = 0$.

The key insight here is to see how the function $g(t)$ relates to $f(t)$. Let's find the derivative of $f(t)$ with respect to time.

First, write $f(t)$ using exponents: $f(t) = 2\sqrt{\frac{t}{\pi}} = \frac{2}{\sqrt{\pi}} t^{1/2}$.
Now, using the power rule for differentiation:
$\frac{df}{dt} = \frac{d}{dt}\left(\frac{2}{\sqrt{\pi}}t^{1/2}\right) = \frac{2}{\sqrt{\pi}} \cdot \frac{1}{2}t^{-1/2} = \frac{1}{\sqrt{\pi}}t^{-1/2} = \sqrt{\frac{1}{\pi t}}$.
This shows that $g(t)$ is the time derivative of $f(t)$.

We can now use the time-differentiation property of the Laplace transform: $L\left\{\frac{df}{dt}\right\} = sF(s) - f(0)$, where $F(s)$ is the Laplace transform of $f(t)$.
Since $f(0) = 2\sqrt{0/\pi} = 0$, the property simplifies to $L\{g(t)\} = sF(s)$.
Given that $F(s) = s^{-3/2}$, we calculate the transform of $g(t)$ as:
$L\{g(t)\} = s \cdot s^{-3/2} = s^{1 - 3/2} = s^{-1/2}$.

Here is a step-by-step explanation of the matching:

• P: Permanent Magnet Moving Coil (PMMC) instruments have a deflection direction that depends on the current's polarity. A reversing AC current would cause the pointer to oscillate around zero, making them suitable for DC Only (P-1).
• Q: Moving iron with a current transformer (CT): While a moving iron instrument works for both AC and DC, a current transformer operates on the principle of electromagnetic induction, which requires an alternating magnetic flux. Thus, a CT works only with AC, restricting the entire setup to AC Only (Q-2).
• R: Rectifier instruments use a rectifier circuit to convert AC into DC, which is then measured by a DC meter (like a PMMC). This combination allows the measurement of both AC and DC quantities (R-3).
• S: Electrodynamometer instruments produce a torque proportional to the product of currents in their fixed and moving coils. For single-phase measurements, this torque is proportional to the current squared ($T \propto I^2$), resulting in a positive deflection for both AC and DC (S-3).

In the two-wattmeter method, the sum of the readings, $W_1 + W_2$, equals the total power consumed. Therefore, we have our first equation: $W_1 + W_2 = 5$ kW.

Next, we determine the phase angle, $\phi$, from the given power factor: $\phi = \cos^{-1}(0.707) = 45^\circ$.

The relationship between the phase angle and the wattmeter readings is expressed as $\tan(\phi) = \frac{\sqrt{3}(W_1 - W_2)}{W_1 + W_2}$.

Substituting the known values gives $\tan(45^\circ) = 1 = \frac{\sqrt{3}(W_1 - W_2)}{5}$. This provides our second equation: $W_1 - W_2 = \frac{5}{\sqrt{3}} \approx 2.89$ kW.

Solving this system of two linear equations ($W_1 + W_2 = 5$ and $W_1 - W_2 = 2.89$) gives the individual readings: $W_1 = 3.94$ kW and $W_2 = 1.06$ kW.

The output voltage for this capacitive voltage divider is described by the equation $V_{out} = V_{bus} \left( \frac{C_1}{C_1 + C_2} \right)$. To find the maximum possible output voltage, we need to maximize the value of the fraction $\frac{C_1}{C_1+C_2}$. This is achieved by using the largest possible capacitance for $C_1$ and the smallest possible capacitance for $C_2$.

Given the $\pm 10\%$ tolerances, the extreme values are:
$C_{1,max} = 1 \, \mu\text{F} \times (1 + 0.10) = 1.1 \, \mu\text{F}$
$C_{2,min} = 9 \, \mu\text{F} \times (1 - 0.10) = 8.1 \, \mu\text{F}$

Substituting these "worst-case" values into the voltage divider formula gives the maximum output:
$V_{out, max} = 100 \, \text{kV} \times \frac{1.1}{1.1 + 8.1} = 100 \times \frac{1.1}{9.2} \approx 11.95 \, \text{kV}$

This circuit acts as a full-wave voltage doubler. The input signal has a peak amplitude of $V_m = 100$ V and an angular frequency of $\omega = 100\pi$ rad/s. The problem gives the condition $100\pi RC = 50$, which can be written as $\omega RC = 50$.

Because the product $\omega RC$ is much greater than 1, the time constant $\tau = RC$ is significantly longer than the period of the input voltage. This means the capacitors have very little time to discharge through the resistor in each cycle. As a result, in the steady state, each capacitor charges up to the peak input voltage, $V_m$. The total voltage across resistor R is the sum of the voltages from the two series capacitors, yielding an average voltage of $V_{R,avg} \approx 2V_m = 2(100 \text{ V}) = 200 \text{ V}$.

To solve this, we apply the boundary conditions for electromagnetic fields at the interface between two dielectrics. The boundary is the plane at $y=0$.

First, we separate the electric field in region 1, $\vec{E}_1 = 3\hat{a}_{x}+4\hat{a}_{y}+2\hat{a}_{z}$, into its tangential and normal components. The component normal to the boundary is $\vec{E}_{1n} = 4\hat{a}_{y}$, and the component tangential to the boundary is $\vec{E}_{1t} = 3\hat{a}_{x}+2\hat{a}_{z}$.

Across a dielectric boundary, the tangential component of the electric field is always continuous. Therefore, the tangential component in region 2 is the same as in region 1:
$\vec{E}_{2t} = \vec{E}_{1t} = 3\hat{a}_{x}+2\hat{a}_{z}$.

Assuming no free surface charge at the boundary, the normal component of the electric displacement field, $\vec{D} = \epsilon\vec{E}$, is continuous. This gives us the relation $\epsilon_1 E_{1n} = \epsilon_2 E_{2n}$, or $\epsilon_{r1}E_{1n} = \epsilon_{r2}E_{2n}$. Plugging in the values, we get $2 \times 4 = 5 \times E_{2n}$, which yields $E_{2n} = 1.6$. The normal field component in region 2 is thus $\vec{E}_{2n} = 1.6\hat{a}_{y}$.

Finally, we find the total electric field in region 2 by adding its tangential and normal components: $\vec{E}_2 = \vec{E}_{2t} + \vec{E}_{2n} = 3\hat{a}_{x} + 1.6\hat{a}_{y} + 2\hat{a}_{z}$.

The induced voltage in the circular turn is given by Faraday's Law, $V = -d\Phi/dt$, where $\Phi$ is the time-varying magnetic flux. The flux is defined as $\Phi(t) = \vec{B} \cdot \vec{A}(t)$, which for a rotating loop becomes $\Phi(t) = BA\cos(\omega t)$.

First, we calculate the magnetic field $B = \mu_0 H = (4\pi \times 10^{-7})(10^7) = 4\pi$ T. The area of the turn is $A = \pi r^2 = \pi(1)^2 = \pi$ m$^2$. The angular frequency $\omega$ is converted from 60 rpm to $\omega = 2\pi \times (60/60) = 2\pi$ rad/s.

Substituting these into the flux equation gives $\Phi(t) = (4\pi)(\pi) \cos(2\pi t) = 4\pi^2 \cos(2\pi t)$.

Now, we find the induced voltage by taking the time derivative:
V_{turn}(t) = -\frac{d}{dt} \[4\pi^2 \cos(2\pi t)]$ = -4\pi^2[-2\pi \sin(2\pi t)] = 8\pi^3 \sin(2\pi t)$.

The peak value of this sinusoidal voltage is its amplitude, $V_{peak} = 8\pi^3$, which calculates to approximately $248.05$ V.

This circuit is an active low-pass filter built with an inverting op-amp. The voltage gain is determined by the ratio of the feedback impedance to the input impedance, $H(s) = -Z_f/Z_{\in}$. The feedback impedance $Z_f$ is the parallel combination of a 1 kΩ resistor and a 0.1 µF capacitor, while the input impedance $Z_{\in}$ is a 1 kΩ resistor. This gives a transfer function of $H(s) = \frac{-1}{1+10^{-4}s}$.

For the sinusoidal input $V_i = 2\sin(2\pi \times 2000t)$, the angular frequency is $\omega = 2\pi(2000) = 4000\pi$ rad/s. To find the amplitude of the output, we calculate the magnitude of the gain at this specific frequency.
The gain magnitude is $|H(j\omega)| = \frac{1}{\sqrt{1+(\omega \cdot 10^{-4})^2}}$. At $\omega = 4000\pi$, this becomes $|H(j4000\pi)| = \frac{1}{\sqrt{1 + (0.4\pi)^2}} \approx 0.6226$.

Finally, the output amplitude is the input amplitude (2 V) multiplied by this gain: $V_{o, amp} = 2 \times 0.6226 \approx 1.245$ V.

Excellent! Let's craft a clearer, more insightful explanation.

This circuit uses a fixed-bias configuration. The base current, $I_B$, is set by the +10 V supply and the resistor $R_B$. Since these values do not change, $I_B$ remains constant.

As the transistor stays in the active mode, the collector current $I_C = \beta I_B$ must also be constant. We can express this constant $I_C$ using Ohm's Law across the collector resistor for both scenarios.

Initially, with $V_C = 2$ V, the collector current is $I_C = \frac{10 - V_C}{R_C} = \frac{10 - 2}{R_C} = \frac{8}{R_C}$.
When we replace $R_C$ with $R'_C$ to get $V_C = 4$ V, the current is $I_C = \frac{10 - 4}{R'_C} = \frac{6}{R'_C}$.

Since $I_C$ is the same in both cases, we equate these expressions:
$\frac{8}{R_C} = \frac{6}{R'_C}$
Rearranging to find the ratio gives:
$\frac{R'_C}{R_C} = \frac{6}{8} = 0.75$

To convert a Sum of Products (SOP) expression to a Product of Sums (POS) form, we switch our focus from where the function is true to where it is false. An SOP expression is a sum of minterms, which are the input combinations that result in an output of '1'. A POS expression is a product of maxterms, which are derived from the input combinations that result in an output of '0'.

The given function $F$ has three variables, so there are $2^3 = 8$ possible input combinations. The SOP expression lists 5 minterms, so the function's output is '1' for 5 inputs. This means the output must be '0' for the remaining $8 - 5 = 3$ combinations. By identifying the input combinations not represented in the SOP, we find the function is '0' for $010$, $100$, and $110$.

Each of these "zero" combinations is then converted into a sum term (a maxterm). For example, the input $A=0, B=1, C=0$ will make the sum term $(A+\bar{B}+C)$ equal to zero. Applying this rule to all three zero-producing inputs yields the terms of the POS expression.

The circuit in Figure (c) is designed to sum the outputs of filter F1 and filter F2. From their frequency response graphs, we can identify F1 as a low-pass filter (LPF) with a cutoff frequency of $f_1$, and F2 as a high-pass filter (HPF) with a cutoff frequency of $f_2$.

Let's analyze the behavior of this summed output. For frequencies below $f_1$, the LPF passes the signal while the HPF blocks it, leading to a high total output. Similarly, for frequencies above $f_2$, the HPF passes the signal while the LPF blocks it, again resulting in a high total output.

Given that $f_1 < f_2$, there is a frequency range between these two cutoffs. In this band ($f_1 < f < f_2$), both the LPF and HPF will attenuate the input signal. Consequently, their sum will be very low. A circuit that passes low and high frequencies but rejects an intermediate band is known as a band-stop filter.

A bipolar junction transistor (BJT) operates in saturation mode when it is fully "on," behaving like a closed switch. For a BJT to turn on, the base-emitter (BE) junction must be forward biased, typically requiring a base-emitter voltage of $V_{BE} \approx 0.7V$. This allows current to flow from the emitter into the base.

In saturation, a sufficiently large base current causes the collector current to reach its maximum possible value, limited by the external circuit. This forces the collector-emitter voltage ($V_{CE}$) to drop to a very small saturation value, $V_{CE(sat)} \approx 0.2V$. For an NPN transistor, the base-collector voltage is $V_{BC} = V_{BE} - V_{CE}$. With typical values, $V_{BC} \approx 0.7V - 0.2V = 0.5V$. Since $V_{BC}$ is positive, the collector-base (CB) junction is also forward biased. Therefore, in saturation, both junctions are forward biased.

Initially, the system is in steady state, delivering power across two lines. This requires a constant, non-zero rotor angle, $\delta_0$. At time $t_1$, one transmission line is tripped, which increases the total series reactance between the generator and the infinite bus. This reduces the system's power transfer capability, so the electrical power output $P_e$ instantaneously drops below the constant mechanical power input $P_m$. This power surplus causes the rotor to accelerate, increasing the angle $\delta$. As the transient is stable, the rotor angle oscillates with decreasing amplitude before settling at a new, higher equilibrium value.

The diagonal element of the bus admittance matrix, $Y_{33}$, represents the total admittance connected to bus 3. This value is the sum of the series admittances from lines connected to other buses and the total shunt admittance at bus 3 itself.

First, we find the shunt admittance. The series admittances connected to bus 3 are the negative of the off-diagonal elements: $y_{31} = -Y_{31} = -j4$ and $y_{32} = -Y_{32} = -j5$. The total shunt admittance at bus 3, $y_{3,sh}$, is then:
$y_{3,sh} = Y_{33} - (y_{31} + y_{32}) = -j8 - (-j4 - j5) = j1$ pu.

Next, we account for the compensation. Compensating 50% of the shunt capacitance means adding an admittance that cancels half of the capacitive shunt admittance. The change in admittance is thus $-0.5 \times y_{3,sh} = -0.5 \times j1 = -j0.5$ pu.

Finally, this change is added to the original diagonal element to find the new value:
$Y_{33, \text{new}} = Y_{33, \text{old}} + \text{change} = -j8 + (-j0.5) = -j8.5$ pu.

The circuit's behavior is described by the KVL equation: $V = iR + L\frac{di}{dt}$. Since the current through an inductor cannot change instantaneously and the circuit starts from rest, the current at the moment the switch is closed is zero, so $i(0^+) = 0$. Substituting this into the KVL equation at $t=0^+$ gives $V = (0)R + L\frac{di}{dt}(0^+)$, which simplifies to $\frac{di}{dt}(0^+) = \frac{V}{L}$.

To find the second derivative, we differentiate the entire KVL equation with respect to time. Since $V$ is a constant, its derivative is zero: $0 = R\frac{di}{dt} + L\frac{d^2i}{dt^2}$. This equation holds for all time $t>0$, including at $t=0^+$. Evaluating it at $t=0^+$ and rearranging gives $\frac{d^2i}{dt^2}(0^+) = -\frac{R}{L}\frac{di}{dt}(0^+)$. Finally, substituting the value we found for the first derivative gives the answer: $\frac{d^2i}{dt^2}(0^+) = -\frac{R}{L}\left(\frac{V}{L}\right) = -\frac{RV}{L^2}$.

The arrangement of the five 1Ω resistors forms a Wheatstone bridge. You can see this more clearly if you mentally redraw the circuit, placing the 5V source on the left and the 2Ω resistor on the right.

A Wheatstone bridge is considered "balanced" when the ratio of resistances in its upper and lower arms are equal. Here, the ratio for the left side is $1\,\Omega / 1\,\Omega = 1$, and the ratio for the right side is also $1\,\Omega / 1\,\Omega = 1$.

Since the bridge is balanced, the electrical potential at the top and bottom output terminals is identical. This results in a potential difference of zero volts across these two points.

The 2Ω resistor is connected between these two points. According to Ohm's Law ($I = V/R$), if the voltage ($V$) across a resistor is zero, the current ($I$) flowing through it must also be zero.

To solve this, we'll find the total input impedance ($Z_{\in}$) by reflecting all components from the load resistor back to the input terminals. An impedance is reflected across a transformer by dividing it by the square of the turns ratio ($n^2$).

Starting from the right, the total impedance can be written as:
$Z_{\in} = \frac{1}{b^2}\left(\frac{R}{a^2} + j\omega L\right) + \frac{1}{j\omega C} = \frac{R}{a^2b^2} + j\left(\frac{\omega L}{b^2} - \frac{1}{\omega C}\right)$
We are given that $Z_{\in}$ is a purely resistive value of $2.5 \Omega$, meaning $Z_{\in} = 2.5 + j0$. For the impedance to be resistive, its imaginary part must be zero. This condition allows us to solve for $b$.
$\frac{\omega L}{b^2} - \frac{1}{\omega C} = 0 \implies \frac{5000 \cdot 1 \cdot 10^{-3}}{b^2} = \frac{1}{5000 \cdot 10 \cdot 10^{-6}} \implies \frac{5}{b^2} = 20 \implies b=0.5$
Next, we equate the real part of $Z_{\in}$ to the specified resistance of $2.5 \Omega$.
$\frac{R}{a^2b^2} = 2.5 \implies \frac{2.5}{a^2b^2} = 2.5 \implies a^2b^2 = 1$
Substituting the value we found for $b$, we can now solve for $a$.
$a^2(0.5)^2 = 1 \implies a^2(0.25) = 1 \implies a^2=4 \implies a=2$

For the motor and load to operate at a steady-state speed, the torque produced by the motor must match the torque required by the load. Since rotational losses are neglected, the motor's developed torque, $T_m$, must equal the load torque, $T_L$.

The load's torque requirement is given by the relationship $T_L = 2.78 \omega_r$. We also know one point on the motor's performance curve: it produces a rated torque of $T_m = 500$ Nm at a speed of $180$ rad/s.

Let's find the speed at which the load would demand this exact torque. By setting $T_L = 500$ Nm, we solve the load equation for speed:
$500 = 2.78 \omega_r$
$\omega_r = \frac{500}{2.78} \approx 179.85$ rad/s.

This calculated speed is almost identical to the motor's speed ($180$ rad/s) when producing $500$ Nm. This confirms that the equilibrium point where the motor and load characteristics intersect is approximately $180$ rad/s.

The motor's torque is the total developed mechanical power divided by the rotor's angular speed. We can calculate this by first analyzing a single phase.

Given a slip $s=0.05$, the effective rotor resistance is $R'_r/s = 0.05/0.05 = 1 \Omega$. The per-phase rotor current magnitude is $|I'_r| = \frac{|V_g|}{|R'_r/s + jX'_r|} = \frac{210}{|1 + j0.22|} \approx 205.09$ A.

The power transferred across the air gap per phase is $P_g = |I'_r|^2 (R'_r/s) = (205.09)^2 \times 1 \approx 42062$ W. The corresponding mechanical power developed per phase is $P_{m,ph} = (1-s)P_g = (1-0.05) \times 42062 \approx 39959$ W.

For a three-phase motor, the total mechanical power is $P_{m,total} = 3 \times P_{m,ph} \approx 119877$ W. The synchronous speed for a 2-pole, 50 Hz motor is $N_s = 3000$ rpm, so the rotor speed is $N_r = N_s(1-s) = 2850$ rpm, or $\omega_m = 2850 \times \frac{2\pi}{60} \approx 298.45$ rad/s.

Finally, the total torque is $T = \frac{P_{m,total}}{\omega_m} = \frac{119877}{298.45} \approx 401.6$ Nm.

The generated voltage in a DC machine is inversely proportional to the number of parallel paths, $A$. For the initial wave winding, $A_{wave} = 2$. When reconnected as a lap winding for a 4-pole machine, the number of parallel paths becomes $A_{lap} = P = 4$. Since the speed and field are unchanged, the ratio of the new voltage to the old voltage is $\frac{V_{new}}{V_{old}} = \frac{A_{old}}{A_{new}} = \frac{2}{4} = \frac{1}{2}$. The new rated voltage is therefore $\frac{1}{2} \times 230 \text{ V} = 115 \text{ V}$.

The power rating of a machine is limited by the current capacity of its armature conductors. Let the current rating per conductor path be $I_{path}$. The total armature current is $I_a = A \times I_{path}$. By changing from a wave winding ($A=2$) to a lap winding ($A=4$), we double the number of parallel paths. This doubles the total current capacity of the armature ($I_{a, new} = 2 I_{a, old}$) because the current rating of the individual conductors remains the same.

The new power rating is $P_{new} = V_{new} \times I_{a, new}$. Since the voltage is halved ($V_{new} = V_{old}/2$) and the current capacity is doubled ($I_{a, new} = 2 I_{a, old}$), the power rating remains the same: $P_{new} = (\frac{V_{old}}{2}) \times (2 I_{a, old}) = V_{old} \times I_{a, old} = 5 \text{ kW}$.

The transfer function of a system is the Laplace transform of its response to a unit impulse. For the given response, $e^{-2t}(\sin 5t + \cos 5t)$, the transfer function $C(s)$ is the sum of the individual transforms:
$C(s) = \frac{5}{(s+2)^{2+}5^2} + \frac{s+2}{(s+2)^{2+}5^2}$
The DC gain represents the system's steady-state response to a constant input, which is found by evaluating the transfer function at $s=0$.
$\text{DC Gain} = C(0) = \frac{5}{(0+2)^{2+}5^2} + \frac{0+2}{(0+2)^{2+}5^2}$
This simplifies to $\frac{5}{4+25} + \frac{2}{4+25} = \frac{7}{29}$, which is approximately $0.241$.

The first plot for $G_1(s)$ traces the negative imaginary axis, which is the characteristic Nyquist plot for a pure integrator, $G_1(s) = \frac{1}{s}$. The second plot for $G_2(s)$ traces the positive imaginary axis, which corresponds to a pure differentiator, $G_2(s) = s$.

To find the Nyquist plot of the product, we multiply the two transfer functions:
$G(s) = G_1(s) \cdot G_2(s) = \frac{1}{s} \cdot s = 1$

The resulting transfer function is a constant value of 1. For any frequency $\omega$, the complex value is always $1+j0$. Therefore, the Nyquist plot is a single point located at $(1, 0)$ on the real axis of the complex plane.

To find the volume, we need to integrate the height of the surface, $f(x,y) = e^x$, over the triangular base area. The triangle is defined by the lines $x=0$, $y=1$, and $x=y$. We can set up a double integral where the outer variable, $y$, ranges from $0$ to $1$. For each value of $y$, the inner variable, $x$, ranges from the left boundary ($x=0$) to the right boundary ($x=y$).

This gives us the iterated integral:
$\text{Volume} = \int_{0}^{1} \int_{0}^{y} e^x \,dx\,dy$
First, we solve the inner integral with respect to $x$:
$\int_{0}^{1} \left[e^x\right]_0^y \,dy = \int_{0}^{1} (e^y - e^0) \,dy = \int_{0}^{1} (e^y - 1) \,dy$
Now, integrating with respect to $y$ yields the final volume:
$\left[e^y - y\right]_0^1 = (e^1 - 1) - (e^0 - 0) = e - 1 - 1 = e - 2 \approx 0.718$

To check if the coin tosses are independent, we must see if the probability of any joint outcome equals the product of the individual probabilities. First, let's find the individual probability of heads for each coin from the given data.
For coin R, the probability of heads is $P(H_R) = P(H_R H_S) + P(H_R T_S) = 0.28 + 0.30 = 0.58$.
For coin S, the probability of heads is $P(H_S) = P(H_R H_S) + P(T_R H_S) = 0.28 + 0.24 = 0.52$.

Now, let's test for independence using the event "both are heads". The actual probability is given as $P(H_R H_S) = 0.28$. If the tosses were independent, this probability would be the product of the individual probabilities: $P(H_R) \times P(H_S) = 0.58 \times 0.52 = 0.3016$.
Since $0.28 \neq 0.3016$, the events are not independent, meaning the coin tosses are dependent.

This differential equation is separable. We can rearrange it to $\frac{di}{dt} = 0.2i$ and then separate the variables to get $\frac{di}{i} = 0.2\,dt$. Integrating both sides gives $\ln(i) = 0.2t + K$, which can be written in exponential form as the general solution $i(t) = Ce^{0.2t}$.

Next, we use the given condition $i(4)=10$ to find the constant of integration, $C$. Plugging this into our general solution gives $10 = Ce^{0.2(4)} = Ce^{0.8}$. Solving for $C$, we find $C = \frac{10}{e^{0.8}} \approx 4.493$.

This gives us the particular solution $i(t) \approx 4.493e^{0.2t}$.

Finally, we can find $i(-5)$ by substituting $t=-5$ into this equation:
$i(-5) \approx 4.493e^{0.2(-5)} = 4.493e^{-1} \approx 1.653$.

To find the Fourier Transform, we apply its definition, $X(\omega) = \int_{-\infty }^{\infty } x(t)e^{-j\omega t}dt$. Since the signal $x(t)$ is only non-zero on the interval $[-1, 1]$, the integral simplifies to:
$X(\omega) = \int_{-1}^{1} e^{j10t} e^{-j\omega t} dt$
By combining the exponents, the integral becomes $\int_{-1}^{1} e^{j(10-\omega)t} dt$.
Evaluating this definite integral gives us:
$\left[ \frac{e^{j(10-\omega)t}}{j(10-\omega)} \right]_{-1}^{1} = \frac{e^{j(10-\omega)} - e^{-j(10-\omega)}}{j(10-\omega)}$
Using Euler's identity, $e^{j\theta} - e^{-j\theta} = 2j\sin(\theta)$, we can rewrite the numerator as $2j\sin(10-\omega)$.
The expression simplifies to $\frac{2j\sin(10-\omega)}{j(10-\omega)} = \frac{2\sin(10-\omega)}{10-\omega}$. Since $\sin(-x) = -\sin(x)$, this is equivalent to $\frac{2\sin(\omega-10)}{\omega-10}$.

First, let's establish the true power consumed by the load, which is the ideal value we want to measure. This is calculated as $P_{true} = VI \cos\phi = 30 \times 25 \times 0.8 = 600$ W.

Next, we analyze the wattmeter's connection. The pressure coil is connected in parallel with the load, so it experiences the full load voltage. This means the wattmeter's reading is the sum of the true load power and the power dissipated by its own pressure coil.

The power loss in the pressure coil is $P_{pc} = \frac{V^2}{R_p} = \frac{30^2}{1000} = 0.9$ W.

Therefore, the power indicated by the wattmeter is $P_{measured} = P_{true} + P_{pc} = 600 + 0.9 = 600.9$ W.

The percentage error is the difference between the measured and true values, relative to the true value.
$\%\text{ Error} = \frac{P_{measured} - P_{true}}{P_{true}} \times 100 = \frac{600.9 - 600}{600} \times 100 = 0.15\%$.

The inductor current, $i_L$, is considered "just continuous" when the circuit operates at the boundary between continuous and discontinuous conduction modes. At this critical point, the minimum value of the inductor current waveform is exactly zero. This condition defines a specific relationship between the circuit components.

For a buck converter, the resistance $R$ that results in this boundary condition is given by the formula:
$R = \frac{2Lf}{1-D}$
Here, $L$ is the inductance, $f$ is the switching frequency, and $D$ is the duty ratio.

Substituting the given values into the equation:
$R = \frac{2 \times (5 \times 10^{-3} \text{ H}) \times (100 \times 10^3 \text{ Hz})}{1 - 0.6}$
$R = \frac{1000}{0.4} = 2500 \, \Omega$