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 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 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 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] \partDraw 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 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 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}