A latex template based on the "Exam" class for creating paper-based exams, with lots of nice features. See the README for details.
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 \question[10] These are some long form" questions. A PV Panel is found to have a maximum power point of 34.1V and 9.83A when tested at STC (1kW/m\textsuperscript{2}), and has a stated efficiency of 19.6\%. Estimate the active area of this panel (i.e. the area of semiconductor that light falls on) in \si{\meter\squared}.   \begin{solutionorbox}[7.75in] At Standard test conditions, the incident radiant energy is 1kW/m\textsuperscript{2}. Assume that the area of the panel is $A$. The radiant power falling on this panel, i.e. incident power, is therefore: \begin{equation*} P_\text{incident}= 1\text{kW}/m^2 * A \end{equation*} The output power is just the incident power times the efficiency: \begin{equation*} P_\text{output}= 1\text{kW}/m^2 * A * \eta \end{equation*} Where efficiency $\eta=(19.6/100)$.   The maximum output power can be determined from the maximum power point details: Pout \begin{equation*} P_\text{out}= V_{oc}*I_{sc}. \end{equation*} Thus, equating the two values of P\textsubscript{out}, we can calculate $A$:   \begin{equation*} A= (V_{oc}*I_{sc})/(\eta * 1kW/m^2) = 1.71 m^2 = 18\text{\ square feet}. \end{equation*} \end{solutionorbox}   \clearpage   \question[10] This is a question that requres a graph / plot as the response. The graph can be generated in \TeX. A PV Panel has a maximum power point of 34.1V and 9.83A when tested at STC (standard testing conditions) and a fill factor of 83.8\%. The open circuit voltage is found to be 40\si{\volt}. Compute the short circuit current for this panel and then sketch the VI curve, and label the maximum power point. \fillwithgrid{8in} \clearpage   \question This is an example of a complex, multi-part question with multiple types of sub-parts. A user wants to connect an inductive load (Z) with a rating of 10kW and a power factor of 0.5 to the utility supply, as shown in the figure below. The supply voltage is $v_g(t) = 170\sin({\omega t + 0^\circ})$ , with a frequency of 60Hz. \begin{figure}[h]  \centering  \begin{circuitikz}[scale=0.7]  \draw  (0,0) to[sinusoidal voltage source, , v_<=$v_g(t)$] (0,4)  to[short,i=${i_g(t)}$] (5,4)  %(5,0) to[I, color=blue, *-*, l=$i_c(t)$] (5,4)  (5,4) -- (7,4)  to[european resistor, l=$Z$] (7,0) -- (0,0);  \end{circuitikz}  %\caption*{Problem 1}  \label{fig:prob1} \end{figure} \begin{parts}  \part[5] Calculate values of P, Q and S (with the appropriate units): \begin{solutionorbox}[4in]  Given that P is 10kW and $cos\theta=0.5$ therefore current lags voltage by $60^\circ$.    Magnitude of apparent power is therefore P/$\cos60^\circ$=20kVA.    Therefore $Q=S\sin\theta=17.32\text{kVAR}$.    Since the load is inductive, Q has a positive value. \end{solutionorbox} \vspace{5mm} P = \fillin[10kW][2in] \vspace{2mm}   Q = \fillin[17.32kVAr][2in] \vspace{2mm}   S = \fillin[20kVA][2in]   \part[5]Draw the power triangle for this load. You can change the spacing of the grid too: \vspace{5mm} \setlength{\gridsize}{\dimexpr.025\linewidth-41\gridlinewidth}  \fillwithgrid{3in} \clearpage   \part[5] Calculate the net impedance now. \begin{solutionorbox}[3.5in]  The net impedance is the parallel combination of the capacitor's impedance and the existing load.  We found that $Z=0.181 +0.313j$. The impedance of the newly added capacitor is:  $X_C = \cfrac{1}{j\omega C} = -0.4171j \Omega$  Therefore net impedance is:  $X_C || Z = \cfrac{ZX_C}{Z+X_C} = 0.7222 - 0.0027j\Omega \approx 0.7222\Omega$ \end{solutionorbox}   \part[5] What is the power factor seen by the grid after the capacitor is installed? \begin{solutionorbox}[2in]  The power factor is now nearly unity.    Phase angle is about 0.2 degrees which for all practical purposes is almost zero. \end{solutionorbox} \end{parts}