Update kcl-kvl.tex

Author: DrakeLin

eecs16a/topics/kcl-kvl.tex

@@ -189,7 +189,7 @@ \section{Kirchhoff's Laws}
     \end{center}
 \end{frame}
 
-\begin{frame}{Practice: KVL [olution]}
+\begin{frame}{Practice: KVL [Solution]}
     \begin{center}
         \begin{circuitikz}[scale=0.75, transform shape]
             \draw (-0.5, 3) to[V=$V_1$] (-0.5, 0)
@@ -243,4 +243,4 @@ \section{Kirchhoff's Laws}
         i_3 = V/R_2
     \end{align*}
     Notice that $i_3$ is \textit{equivalent to the current through resistor $R_2$} because no current goes through $R_3$. And since the voltage at node $N_1$ is $V$ and the voltage at $N_2$ is 0, the current $i_3 = V/R_2$.
-\end{frame}
+\end{frame}