\begin{solution}
First we find the total current in the circuit. The two resistances,
$R_1 = 650\U{\Ohm}$ and $R_2 = 2200\U{\Ohm}$, in series provide an
-effective resistance of $R_e = R_1 + R_2$. By Kirchoff's loop rule
+effective resistance of $R_e = R_1 + R_2$. By Kirchhoff's loop rule
\begin{align*}
V - I R_e &= 0 \\
I &= \frac{V}{R_e} = \frac{V}{R_1 + R_2}
\begin{solution}
Let $V = 110\U{V}$ be the source voltage, $P_1 = 7.0\U{W}$ be the
power of one bulb, and $R_1$ be the resistance of one bulb. By
-Kirchoff's loop rule
+Kirchhoff's loop rule
\begin{align*}
V - 8IR_1 &= 0 \\
I &= \frac{V}{8R_1}
\end{problem*}
\begin{solution}
-From Kirchoff's loop rule
+From Kirchhoff's loop rule
\begin{align*}
\mathcal{E}_1 - IR - \mathcal{E}_2 - Ir_2 - Ir_1 &= 0 \\
I(R+r_1+R_2) &= \mathcal{E}_1-\mathcal{E}_2 \\
Label the batteries from left to right: $V_1 = 6.0\U{V}$ and $V_2 =
3.0\U{V}$.
-Applying Kirchoff's junction rule to junction $a$ we have
+Applying Kirchhoff's junction rule to junction $a$ we have
$$I_1 + I_2 - I_3 = 0$$
-Applying Kirchoff's loop rule to the left-hand loop we have
+Applying Kirchhoff's loop rule to the left-hand loop we have
$$V_1 - I_1 (R_1 + R_2) + R_3 I_2 = 0$$
where we \emph{add} the voltage change over $R_3$ because we cross it
\emph{against} the direction of the current $I_2$.
-Applying Kirchoff's loop rule to the right-hand loop we have
+Applying Kirchhoff's loop rule to the right-hand loop we have
$$V_2 - R_4 I_3 - R_3 I_2 - R_5 I_5 = V_2 - I_3 (R_4 + R_5) - R_3 I_2 = 0$$
We now have three equations for three unknowns (the $I_i$).
wire(a, B.end, nsq);
\end{asy}
\end{center}
-Using Kirchoff's loop rule
+Using Kirchhoff's loop rule
\begin{align*}
V - IR_1 - IR_2 &= 0 \\
I &= \frac{V}{R_1+R_2}
$$P = IV$$
so the power supplied increases if the current $I$ increases (because $V$ remains constant for batteries).
-From Kirchoff's loop rule, we know the voltage drop across the
+From Kirchhoff's loop rule, we know the voltage drop across the
resistors is the same as the voltage gain across the battery.
$$V_b = V_R$$
We also know that the voltage across the resistors relates to the current via Ohm's law
Make a circuit using a known resistance $R$ to connect the two
terminals of the battery, and measure the current $I$.
-From Kirchoff's loop rule
+From Kirchhoff's loop rule
\begin{align*}
V - Ir - IR &= 0 \\
Ir &= V - IR \\
\begin{solution}
\Part{a}
-We're going to use Kirchoff's loop rule: the sum of voltage changes
+We're going to use Kirchhoff's loop rule: the sum of voltage changes
around a loop is $0$.
\begin{equation}
\sum_\text{loop} \Delta V = 0
Anyhow lets travel the loop along the path $a\rightarrow b\rightarrow
c\rightarrow d\rightarrow a$, adding up voltage changes as we go. By
-Kirchoff's loop rule, the total voltage change must be $0$. We'll
+Kirchhoff's loop rule, the total voltage change must be $0$. We'll
need to pick a direction for the current $I$ to be flowing as well, so
we know whether the potential increases or decreases going across
resistors. Let's pick the same direction (clockwise) as the path
\end{verbatim}
\end{center}
\Part{a}
-Labeling the currents as above, we use Kirchoff's junction rule
+Labeling the currents as above, we use Kirchhoff's junction rule
summing the currents entering node $a$.
\begin{align}
0 &= -I_1 - I_2 + I_3 &
\EMF_2 &= -(V_{ba}+I_2R_2) = \ans{7.00\U{V}}
\end{align}
-To find $\EMF_1$ we'll need $I_1$. Applying Kirchoff's junction rule
+To find $\EMF_1$ we'll need $I_1$. Applying Kirchhoff's junction rule
to $a$
\begin{align}
0 &= I_3 + I_1 - I_2 \\
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