\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}	