GATE EE 2013

This circuit operates with negative feedback. Since the op-amp is ideal and its non-inverting (+) input is grounded, a virtual ground is established at the inverting (-) input, making its voltage $V_{-} = 0 \text{ V}$.

The feedback path contains a PNP transistor, identifiable by the emitter arrow pointing inward. This transistor is turned on by the current from the 5V source. The transistor's emitter is connected to the inverting input ($V_E = 0 \text{ V}$), while its base is connected to the output ($V_B = V_{out}$).

For a silicon PNP transistor to be forward-biased, the emitter-to-base voltage drop must be approximately $0.7 \text{ V}$. Therefore, we have the relationship $V_{EB} = V_E - V_B = 0.7 \text{ V}$.

Substituting the known node voltages into this equation yields $0 \text{ V} - V_{out} = 0.7 \text{ V}$. Solving for the output voltage gives $V_{out} = -0.7 \text{ V}$.

To determine the transfer function, we can analyze the circuit as a voltage divider in the Laplace domain. The impedance of the resistor is $R$, and the impedance of each capacitor is $\frac{1}{sC}$.

The output voltage $V_2(s)$ is taken across the series combination of the resistor and the lower capacitor. The transfer function is the ratio of this output impedance to the total impedance of all three series components:
$\frac{V_2(s)}{V_1(s)} = \frac{R + \frac{1}{sC}}{\frac{1}{sC} + R + \frac{1}{sC}} = \frac{R + \frac{1}{sC}}{R + \frac{2}{sC}}$
To simplify this complex fraction, we multiply the numerator and denominator by $sC$, yielding $\frac{sCR + 1}{sCR + 2}$.
Let's calculate the time constant $RC$: $RC = (10 \times 10^3 \, \Omega)(100 \times 10^{-6} \, \text{F}) = 1 \text{ s}$.
Substituting $RC = 1$ into our expression gives the final transfer function:
$\frac{s(1) + 1}{s(1) + 2} = \frac{s+1}{s+2}$

This system's transfer function, which relates the output to the input, is $G(s) = \frac{1}{s}$. The input is a unit step function, $u(t)$, and its Laplace transform is $U(s) = \frac{1}{s}$.

To find the output in the Laplace domain, we multiply the transfer function by the input transform:
$Y(s) = G(s)U(s) = \frac{1}{s} \cdot \frac{1}{s} = \frac{1}{s^2}$

Finally, we find the time-domain response, $y(t)$, by taking the inverse Laplace transform of $Y(s)$. The inverse transform of $\frac{1}{s^2}$ is the ramp function, $tu(t)$.

To find the system's output, we can use the Laplace transform method. First, we find the system's transfer function, $H(s)$, which is the Laplace transform of the impulse response $h(t)=tu(t)$. This gives us $H(s) = 1/s^2$.

Next, we find the Laplace transform of the input signal, $x(t) = u(t-1)$, which is $X(s) = \frac{e^{-s}}{s}$.

The output in the Laplace domain, $Y(s)$, is the product of the transfer function and the input transform:
$Y(s) = H(s)X(s) = \left(\frac{1}{s^2}\right) \left(\frac{e^{-s}}{s}\right) = \frac{e^{-s}}{s^3}$.

To find the output $y(t)$, we take the inverse Laplace transform. We know the transform pair for a ramp squared is $L^{-1}\{\frac{1}{s^3}\} = \frac{t^2}{2}u(t)$. The $e^{-s}$ term in $Y(s)$ signifies a time delay of 1, so we apply the time-shifting property.

This results in the final output: $y(t) = \frac{(t-1)^2}{2}u(t-1)$.

For a continuous-time LTI system to be both causal and stable, all of its poles must lie strictly in the left-half of the s-plane. This means the real part of every pole must be negative ($Re(s) < 0$). This fundamental rule confirms that statements A and D are true, as the poles of a system are the roots of its characteristic equation. Statement B is also correct, as the location of zeros does not affect a system's stability.

Statement C, however, is incorrect. It suggests that poles must lie within a circle of radius 1 in the s-plane, defined by $|s|=1$. This is not the stability region for continuous-time systems. This condition, $|z|<1$, is actually the stability requirement for discrete-time systems in the z-plane.

When two systems are connected in cascade, the output of the first system serves as the input to the second. In the frequency domain, this relationship simplifies to the multiplication of their respective transfer functions. If $H_1(s)$ and $H_2(s)$ are the transfer functions, the overall system's transfer function is $H(s) = H_1(s) \cdot H_2(s)$.

A fundamental property of the Laplace transform is the convolution theorem, which states that multiplication in the frequency domain is equivalent to convolution in the time domain.

Therefore, to find the overall impulse response, $h(t)$, we take the inverse Laplace transform of this product, resulting in the convolution of the individual impulse responses: $h(t) = h_1(t) * h_2(t)$.

To deliver maximum power from a source with complex internal impedance to a purely resistive load, we apply a specific case of the maximum power transfer theorem. This theorem states that the load resistance, $R_L$, must be equal to the magnitude of the source's internal impedance, $Z_s$.

The given source impedance is $Z_s = (4+j3)\Omega$.

We find the magnitude of this complex impedance to determine the required value for the load resistor:
$R_L = |Z_s| = |4+j3|\Omega$
$R_L = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\Omega$.

To understand the load's behavior, we calculate the complex power $S = P + jQ$. The key is to be consistent with sign conventions. The given current, $I = 10\angle -150^{\circ}$ A, flows out of the load. We can use this to find the power supplied by the load using the formula $S_{supplied} = V \cdot I^*$.

The conjugate of the current is $I^* = 10\angle 150^{\circ}$ A. The calculation is:
$S_{supplied} = (100\angle 60^{\circ}) \cdot (10\angle 150^{\circ}) = 1000\angle 210^{\circ}$ VA.

In rectangular form, this is $S_{supplied} \approx -866 - j500$ VA. A negative value for supplied power means that power is actually being absorbed. Since both the supplied real power ($P$) and reactive power ($Q$) are negative, the load absorbs both.

The no-load loss corresponds to the constant core loss, so $P_{core} = 64$ W.

The short-circuit test reveals the copper loss, which varies with the square of the current. At 90% of the rated current, the copper loss is 81 W. To find the copper loss at full load ($P_{cu,fl}$), we can scale this value:
$P_{cu,fl} = \frac{81 \text{ W}}{(0.90)^2} = \frac{81}{0.81} = 100$ W.

Maximum efficiency occurs when the constant core loss equals the variable copper loss. Let $x$ be the fraction of the rated current where this happens. The condition is:
$P_{core} = x^2 P_{cu,fl}$

Solving for $x$ gives the optimal operating point:
$x = \sqrt{\frac{P_{core}}{P_{cu,fl}}} = \sqrt{\frac{64}{100}} = 0.8$.

This means the transformer reaches maximum efficiency at 80% of its rated current.

A valid magnetic flux density field must satisfy Gauss's law for magnetism, which states that its divergence is zero. This is one of Maxwell's equations, written as $\nabla \cdot \vec{B} = 0$. This condition reflects the non-existence of magnetic monopoles.

For the given field $\vec{B} = 4x\hat{a}_{x} + 2ky\hat{a}_{y} + 8\hat{a}_{z}$, we must compute its divergence and set it to zero.

The divergence in Cartesian coordinates is given by $\nabla \cdot \vec{B} = \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z}$.

Applying this to our field gives:
$\frac{\partial}{\partial x}(4x) + \frac{\partial}{\partial y}(2ky) + \frac{\partial}{\partial z}(8) = 0$

Evaluating the partial derivatives results in the equation $4 + 2k + 0 = 0$. Solving for the constant $k$, we find $2k = -4$, which means $k = -2$.

To find the probability that the random variable $X$ is greater than 1, we must find the area under the probability density function, $f(x) = e^{-x}$, from $x=1$ to infinity. This area is calculated using a definite integral.

The probability is given by:
$P(X > 1) = \int_{1}^{\infty } e^{-x} \, dx$

First, we find the antiderivative of $e^{-x}$, which is $-e^{-x}$. Then, we evaluate this result at the limits of integration:
$[-e^{-x}]_{1}^{\infty } = \lim_{b \to \infty }(-e^{-b}) - (-e^{-1})$

Since $e^{-b}$ approaches 0 as $b$ goes to infinity, the expression simplifies to $0 - (-e^{-1})$, which equals $e^{-1}$. The value of $e^{-1}$ is approximately 0.368.

This problem asks for the curl of the gradient of a scalar field, represented as $\nabla \times (\nabla V)$. A fundamental identity in vector calculus is that the curl of the gradient of any scalar field is always the zero vector. This is a universal property, often written as $\nabla \times (\nabla V) = \vec{0}$.

The underlying reason is that this operation involves second partial derivatives, and for smooth functions, the order of differentiation doesn't matter (e.g., $\frac{\partial^2 V}{\partial x \partial y} = \frac{\partial^2 V}{\partial y \partial x}$), causing all terms in the curl calculation to cancel out. Because of this powerful identity, we don't need to perform any calculations on the specific function $V$; the result is guaranteed to be zero.

The circuit shown employs voltage-series feedback. In this configuration, the input impedance is increased and the output impedance is decreased by the desensitivity factor, $(1+A_o k)$.

The input impedance with feedback is given by $R_{if} = R_i(1 + A_o k)$.
The output impedance with feedback is given by $R_{of} = \frac{R_o}{1 + A_o k}$.

From these equations, we can see that if the feedback factor $k$ increases, the term $(1 + A_o k)$ also increases. Consequently, the input impedance $R_{if}$ will increase, while the output impedance $R_{of}$ will decrease.

A PMMC voltmeter is designed to measure the average (DC) value of a voltage waveform. Let's analyze the circuit's behavior over one cycle of the input signal.

During the positive half-cycle, the ideal diode is forward-biased, acting as a short circuit to ground. This means the voltage across the 100 kΩ resistor is 0 V.

During the negative half-cycle, the diode is reverse-biased and acts as an open circuit. The circuit becomes a voltage divider. The voltmeter thus sees a half-wave rectified signal. The peak voltage across the load is $V_{p,out} = 14.14 \text{ V} \times \frac{100 \text{ k}\Omega}{1 \text{ k}\Omega + 100 \text{ k}\Omega}$.

The average value of a half-wave rectified sine wave is its peak value divided by $\pi$. Therefore, the voltmeter reading will be:
$V_{avg} = \frac{V_{p,out}}{\pi} = \frac{1}{\pi} \left( 14.14 \times \frac{100}{101} \right) \approx 4.46 \text{ V}$

The straight line on the Bode plot, along with the fact that the phase is always negative, points to a transfer function with only poles at the origin. We can model this as $G(s) = \frac{K}{s^n}$.

First, let's use the given point at $\omega = 1$ rad/s, where the gain is 32 dB. The gain equation is $20 \log|G(j\omega)|$. Plugging in the values gives us $20 \log|\frac{K}{(j1)^n}| = 32$, which simplifies to $20 \log K = 32$. Solving for $K$ gives $K = 10^{(32/20)} \approx 39.8$.

Next, we use the second point: a gain of -8 dB at $\omega = 10$ rad/s. This gives the equation $20 \log|\frac{39.8}{(j10)^n}| = -8$. Using the properties of logarithms, we can expand this to $20 \log(39.8) - 20 \log(10^n) = -8$.

From our first step, we know that $20 \log(39.8) = 32$. Substituting this into our equation gives $32 - 20n \log(10) = -8$. This simplifies to $32 - 20n = -8$, which means $20n = 40$, and therefore $n=2$. Combining our results for $K$ and $n$, the transfer function is $G(s) = \frac{39.8}{s^2}$.

Let's model the two switches as logical inputs, say $A$ and $B$, and the bulb as the output. We can use $0$ to represent one state (e.g., "down" or "off") and $1$ for the other ("up" or "on"). The key feature is that flipping either switch changes the bulb's state.

If both switches are in the same position (both $A=0, B=0$ or both $A=1, B=1$), the bulb will be in a certain state (let's say OFF, or $0$). If you then flip just one switch, making their positions different ($A=0, B=1$ or $A=1, B=0$), the bulb must turn ON (output $1$).

This behavior, where the output is $1$ if and only if the inputs are different, is the defining characteristic of an Exclusive OR (XOR) gate. The operation $A \oplus B$ perfectly describes the logic of the staircase switches.

The given signal $v(t)$ is a superposition of three sinusoids. The angular frequencies of these individual components are $\omega_1 = 100$ rad/s, $\omega_2 = 300$ rad/s, and $\omega_3 = 500$ rad/s. For the combined signal to be periodic, all component frequencies must be integer multiples of a fundamental frequency, $\omega_0$. This fundamental frequency is the greatest common divisor (GCD) of the component frequencies.

Calculating the GCD:
$\omega_0 = \text{GCD}(100, 300, 500)$

We can see that $100 = 1 \times 100$, $300 = 3 \times 100$, and $500 = 5 \times 100$. The largest number that divides all three is 100. Thus, the fundamental frequency is $\omega_0 = 100$ rad/s.

The Nyquist sampling theorem provides the rule for perfectly reconstructing a signal from its samples. It states that the sampling frequency, $f_s$, must be at least twice the maximum frequency component of the signal, $f_m$.

This relationship is expressed as:
$f_s \geq 2f_m$

For the given signal, the maximum frequency is $f_m = 5$ kHz. Plugging this into the theorem gives us the minimum required sampling frequency:
$f_s \geq 2 \times 5 \text{ kHz}$
$f_s \geq 10 \text{ kHz}$

Therefore, any valid sampling frequency must be 10 kHz or greater. The frequency of 5 kHz fails to meet this essential condition.

Let's analyze the relationship between the delta and star networks using the standard conversion formulas. The resistance of any element in the equivalent star connection is found by dividing the product of the two adjacent delta resistors by the sum of all three delta resistors.

Let's pick one of the star resistors, say $R_A$. Its value is given by:
$R_A = \frac{R_b R_c}{R_a + R_b + R_c}$

Now, suppose we scale each resistor in the delta connection by a factor of $k$. Let the new delta resistances be $R'_a = kR_a$, $R'_b = kR_b$, and $R'_c = kR_c$. The new equivalent star resistor, $R'_A$, will be:
$R'_A = \frac{R'_b R'_c}{R'_a + R'_b + R'_c} = \frac{(kR_b)(kR_c)}{kR_a + kR_b + kR_c}$

Factoring out $k$ from the numerator and denominator gives:
$R'_A = \frac{k^2 (R_b R_c)}{k (R_a + R_b + R_c)} = k \left( \frac{R_b R_c}{R_a + R_b + R_c} \right)$

Since the expression in the parenthesis is just the original resistance $R_A$, we find that $R'_A = k R_A$. Thus, the elements of the star connection are scaled by the same factor $k$.

The swing equation, $M \frac{d^2\delta}{dt^2} = P_m - P_e$, describes the rotational dynamics of a synchronous generator's rotor. To analyze this motion, we establish a reference frame that rotates at the constant synchronous speed, $\omega_s$, which is determined by the electrical grid's frequency. The angle $\delta$ in this equation is the rotor angle, which represents the angular displacement of an axis physically fixed to the rotor with respect to this synchronously rotating reference frame. An imbalance between the mechanical input power ($P_m$) and electrical output power ($P_e$) causes the rotor to accelerate or decelerate, leading to a change, or "swing," in the angle $\delta$.

The total flux produced by a winding (e.g., the stator) can be divided into two parts. The useful part is the mutual flux ($\phi_m$), which crosses the air gap and links both the stator and rotor windings, enabling torque production. The remaining part is the leakage flux ($\phi_l$), which does not follow the main magnetic path across the air gap. This leakage flux links only its own winding (either stator or rotor) and does not contribute to the mutual energy transfer. Therefore, leakage flux is the flux that links the stator winding or the rotor winding, but not both.

As this is an AC circuit with reactive components, we must analyze the voltages using phasors, but the voltmeters will read the RMS magnitudes of these voltages.

Let the RMS current flowing through the series circuit be $I$. The voltage readings across the individual components are calculated by multiplying the current by the magnitude of the impedance.
Reading on voltmeter $V_1$ (across capacitor): $V_1 = I \times |-j1\Omega| = I \times 1 = I$.
Reading on voltmeter $V_2$ (across inductor): $V_2 = I \times |j2\Omega| = I \times 2 = 2I$.

The voltmeter $V$ reads the total voltage across the series combination. First, we find the total equivalent impedance: $Z_{total} = -j1\Omega + j2\Omega = j1\Omega$.
The total voltage reading is $V = I \times |Z_{total}| = I \times |j1\Omega| = I$.

Now, let's compare the three scalar readings: $V_1 = I$, $V_2 = 2I$, and $V = I$.
From these results, we can see that $V_2 - V_1 = 2I - I = I$.
Since $V=I$, we can conclude that the relationship is $V = V_2 - V_1$.

To find the square roots of a complex number, we first express it in polar form. The number $-i$ is on the negative imaginary axis, so its magnitude is $1$ and its principal angle is $-\frac{\pi}{2}$. Thus, we can write $-i = \cos(-\frac{\pi}{2}) + i\sin(-\frac{\pi}{2})$.

We use De Moivre's Theorem for roots, which states that the $n$-th roots are given by $\cos(\frac{\theta + 2k\pi}{n}) + i\sin(\frac{\theta + 2k\pi}{n})$ for $k = 0, 1, ..., n-1$.

For the square roots ($n=2$), we set $k=0$ and $k=1$.

For $k=0$, the first root is $\cos(\frac{-\pi/2}{2}) + i\sin(\frac{-\pi/2}{2}) = \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4})$.

For $k=1$, the second root is $\cos(\frac{-\pi/2 + 2\pi}{2}) + i\sin(\frac{-\pi/2 + 2\pi}{2}) = \cos(\frac{3\pi}{4}) + i\sin(\frac{3\pi}{4})$.

To calculate the line integral, we first need to define the path and the differential displacement vector $d\mathbf{l}$. The path is a segment on the x-axis, which means the coordinates $y$ and $z$ are always zero. Consequently, there is no change in $y$ or $z$, so $dy=0$ and $dz=0$.

The general expression for the dot product $\mathbf{F} \cdot d\mathbf{l}$ is:
$\mathbf{F} \cdot d\mathbf{l} = (y^2x\,a_x - yz\,a_y - x^2\,a_z) \cdot (dx\,a_x + dy\,a_y + dz\,a_z) = y^2x\,dx - yz\,dy - x^2\,dz$

Now, we apply the conditions of our specific path ($y=0$, $dy=0$, $dz=0$) to this expression:
$\mathbf{F} \cdot d\mathbf{l} = (0)^2x\,dx - (0)z\,(0) - x^2(0) = 0$

Since the value of the integrand $\mathbf{F} \cdot d\mathbf{l}$ is zero for every point along the specified path, the definite integral from $x=1$ to $x=2$ is also zero.

This matrix equation represents a system of linear equations. Let's write it out by performing the matrix multiplication:
$2x_1 - 2x_2 = 0$
$x_1 - x_2 = 0$

The first equation is just the second equation multiplied by 2. This means the two equations are not independent; they both describe the same relationship.

Simplifying either equation gives us the single condition $x_1 = x_2$. Any vector where the two components are equal will be a solution. For example,

,

, and

are all solutions. Since there are infinitely many such vectors, the equation has multiple solutions.

First, we must determine the maximum permissible excitation voltage, $V_i$. This is limited by the 20 mA maximum current through the strain gauge's branch. Under no-load conditions, the resistance of this branch is $R_s + R_3 = 300\Omega + 300\Omega = 600\Omega$. Thus, the maximum voltage is $V_i = (20 \text{ mA}) \times (600\Omega) = 12 \text{ V}$.

Next, the strain causes the gauge resistance to increase by 1%, so its new value is $R_s' = 300\Omega \times 1.01 = 303\Omega$.

The output voltage $V_0$ is the difference between the voltages at the midpoints of the two parallel branches. Using the voltage divider rule for each branch:
$V_0 = V_i \left| \frac{R_2}{R_1+R_2} - \frac{R_3}{R_s'+R_3} \right| = 12 \left| \frac{300}{300+300} - \frac{300}{303+300} \right|$
$V_0 = 12 \left| 0.5 - \frac{300}{603} \right| \approx 12 \times 0.002487 = 0.02985 \text{ V}$, which is $29.85 \text{ mV}$.

The Zener diode regulates the voltage across the load to $V_Z = 5$ V. This leaves a voltage drop of $10V - 5V = 5V$ across the 100 $\Omega$ resistor. The total current flowing from the source is therefore constant at $I_s = \frac{5V}{100\Omega} = 50$ mA.

To find the minimum value of $R_L$, we must maximize the current flowing through it, $I_{Load}$. This occurs when the Zener current $I_Z$ is at its minimum for regulation, which is the knee current of $10$ mA. The maximum load current is $I_{Load,max} = I_s - I_{Z,min} = 50 \text{ mA} - 10 \text{ mA} = 40$ mA. The corresponding minimum load resistance is $R_{L,min} = \frac{V_Z}{I_{Load,max}} = \frac{5V}{40mA} = 125 \Omega$.

The minimum power rating for the Zener is determined by the maximum power it will ever dissipate. This worst-case scenario happens when the load current is zero (i.e., $R_L$ is disconnected). In this case, the entire source current of 50 mA flows through the Zener. The maximum power dissipation is $P_{Z,max} = V_Z \times I_{s} = 5V \times 50mA = 250$ mW.

First, we identify the time constant of the open-loop system. The transfer function $\frac{10}{1+10s}$ is in the standard first-order form $\frac{K}{1+\tau s}$, so the open-loop time constant is $\tau_{OL} = 10$ s.

Next, we find the transfer function of the closed-loop system using the feedback formula, $G_{CL}(s) = \frac{G(s)}{1+G(s)}$, where $G(s) = K_a \frac{10}{1+10s}$.
$G_{CL}(s) = \frac{\frac{10K_a}{1+10s}}{1 + \frac{10K_a}{1+10s}} = \frac{10K_a}{1+10s+10K_a}$
To find the new time constant, we rearrange this into the standard form $\frac{K_{CL}}{1+\tau_{CL}s}$:
$G_{CL}(s) = \frac{\frac{10K_a}{1+10K_a}}{1 + \frac{10}{1+10K_a}s}$
From this, we see the closed-loop time constant is $\tau_{CL} = \frac{10}{1+10K_a}$. The problem requires this to be 100 times smaller than the open-loop time constant, so $\tau_{CL} = \frac{\tau_{OL}}{100} = \frac{10}{100} = 0.1$.

Setting the two expressions for $\tau_{CL}$ equal gives $\frac{10}{1+10K_a} = 0.1$. Solving for $K_a$, we find $100 = 1+10K_a$, which gives $K_a=9.9$. The closest available option is 10.

To find the Thevenin equivalent voltage, $V_{Th}$, we calculate the open-circuit voltage across the load resistor's terminals. This open-circuit condition means the current $I_2$ in the right-hand loop must be zero.

When $I_2 = 0$, the dependent voltage source $j40I_2$ in the left loop becomes 0 V, effectively a short circuit. The left loop then simplifies to the source $V_s$ in series with a resistor and an inductor. We can find the control voltage $V_{L_1}$ using the voltage divider rule:
$V_{L_1} = V_s \left( \frac{j4}{3+j4} \right) = (100\angle 53.13^{\circ}) \left( \frac{4\angle 90^{\circ}}{5\angle 53.13^{\circ}} \right) = 80\angle 90^{\circ} \text{ V}$

Now, we find the voltage across the open terminals. Since $I_2=0$, there is no voltage drop across the $j6\,\Omega$ and $5\,\Omega$ impedances. The Thevenin voltage is therefore equal to the voltage of the dependent source in that loop.
$V_{Th} = 10V_{L_1} = 10 \times (80\angle 90^{\circ} \text{ V}) = 800\angle 90^{\circ} \text{ V}$

To find the maximum safe operating conditions, we must first identify the "weakest link" in the circuit. Let's calculate the maximum charge each capacitor can hold before breakdown using $Q_{max} = C \times V_{breakdown}$. This gives us $Q_{1,max} = 100 \mu C$, $Q_{2,max} = 25 \mu C$, and $Q_{3,max} = 4 \mu C$.

Capacitors $C_2$ and $C_3$ are connected in series, meaning they must store the same amount of charge. The maximum charge this series branch can hold is limited by the capacitor with the smallest charge capacity, which is $C_3$. Therefore, the maximum charge for the series branch is $Q_{series,max} = 4 \mu C$.

The equivalent capacitance of this series branch is $C_{23} = \frac{C_2 C_3}{C_2 + C_3} = \frac{5 \times 2}{5+2} = \frac{10}{7} \mu F$. The voltage across this branch at maximum charge is $V_{max} = \frac{Q_{series,max}}{C_{23}} = \frac{4 \mu C}{10/7 \mu F} = 2.8 V$. Since this branch is in parallel with $C_1$, this is the maximum safe voltage for the entire circuit (as $2.8V$ is well below $C_1$'s $10V$ limit).

To find the total charge, we sum the charges on the parallel branches at this voltage. The charge on $C_1$ is $Q_1 = C_1 V_{max} = (10 \mu F)(2.8 V) = 28 \mu C$. The total charge is therefore $Q_{total} = Q_1 + Q_{series,max} = 28 \mu C + 4 \mu C = 32 \mu C$.

To understand the voltage across WX, let's analyze the circuit's behavior during the positive and negative phases of the input AC voltage, assuming the diodes are ideal.

For the positive half-cycle, when $V_Y > V_Z$, the potential difference causes all four ideal diodes to become forward-biased and behave as short circuits. The bottom diodes (D3 and D4) effectively connect terminals W and X directly to terminal Z. This forces the potentials $V_W$ and $V_X$ to both be equal to $V_Z$, resulting in a potential difference $V_{WX} = V_W - V_X = 0$.

For the negative half-cycle, when $V_Z > V_Y$, the top two diodes (D1 and D2) become forward-biased, while the bottom two are reverse-biased. As D1 and D2 act as short circuits, they connect both W and X to the same node above them (the junction with the 1k$\Omega$ resistor). This ensures that $V_W = V_X$, making the potential difference $V_{WX}$ zero once again.

Since the voltage across WX is zero throughout the entire cycle, the measured voltage is always 0 Volts.

First, let's find the motor's back EMF ($E_b$) when it operates at rated speed and voltage. Using the DC motor's voltage equation with the rated values:
$V_{a,rated} = E_b + I_{a,rated} R_a$
$150V = E_b + (20A)(1\Omega)$, which means $E_b = 130V$.

The problem requires the motor to run at this same rated speed, so the back EMF remains $130V$. For a separately excited motor, torque is proportional to armature current. To get 50% of the rated torque, we need 50% of the rated armature current: $I_{a,new} = 0.5 \times 20A = 10A$.

Now, calculate the new armature voltage ($V_a$) needed for this condition:
$V_a = E_b + I_{a,new} R_a = 130V + (10A)(1\Omega) = 140V$.

This voltage is supplied by the chopper from a 200V source. The duty ratio ($D$) is the ratio of the required output voltage to the input source voltage:
$D = \frac{V_a}{V_s} = \frac{140V}{200V} = 0.7$.

To find the current drawn by the capacitor, we can analyze the system from the perspective of node 3 by creating a Thevenin equivalent circuit. The Thevenin voltage ($V_{th}$) is the initial voltage at the node, $V_{th} = 1.3 \angle -10^{\circ}$ per unit. The Thevenin impedance ($Z_{th}$) is the driving point impedance at node 3, given by the bus impedance matrix element $Z_{33} = j0.5$ per unit.