Exercise 1.6 corrections

Chapter1.tex

@@ -60,8 +60,9 @@
     \ex{1.6}
     \begin{enumerate}
         \item
-        The total current required by New York City that will flow through the cable is $I = \frac{P}{V} = \frac{10^{10}\;\W}{115\;\V} = 869.6\;\M\A$. Therefore, the total power lost per foot of cable can be calculated by:
-        \[P = I^2R = (869.6 \times 10^6\;\A)^2 \times \left(5\times10^{-8}\;\frac{\Ohm}{\ensuremath{\text{ft}}}\right) = \mans{3.78\times10^{8} \frac{\W}{\ensuremath{\text{ft}}}} \] 
+        The total current required by New York City that will flow through the
+            cable is $I = \frac{P}{V} = \frac{10^{10}\;\W}{115\;\V} = 86.96\;\M\A$. Therefore, the total power lost per foot of cable can be calculated by:
+        \[P = I^2R = (86.96 \times 10^6\;\A)^2 \times \left(5\times10^{-8}\;\frac{\Ohm}{\ensuremath{\text{ft}}}\right) = \mans{3.78\times10^{8} \frac{\W}{\ensuremath{\text{ft}}}} \] 
         \item
         The length of cable over which all $10^{10}\;\W$ will be lost is:
         \[L = \frac{10^{10}\;\W}{3.78 \times 10^{8}\;\frac{\W}{\ensuremath{\text{ft}}}} = \mans{26.45\;ft}\]
@@ -70,7 +71,7 @@
         \[A = \pi DL = \pi \times 30.48\;{\ensuremath{\text{cm}}}\;\times 806.196\;{\ensuremath{\text{cm}}} = 7.72 \times 10^4\;{\ensuremath{\text{cm}}}^2\]
         Therefore,
         \[T = \sqrt[4]{\frac{P}{A\sigma}} = \sqrt[4]{\frac{10^{10}\;\W}{7.72 \times 10^4\;{\ensuremath{\text{cm}}}^2 \times 6 \times 10^{-12}\;\frac{\W}{\text{K}^4\text{cm}^2}}} = \mans{12,121\;K} \]
-        This is indeed a preposterous temperature, more than twice that at the surface of the Sun! The ``solution'' to this problem is to look at the melting point of copper, which is $\sim$1358 K at standard pressure. The copper cable will melt long before such a temperature is reached.
+        This is indeed a preposterous temperature, more than twice that at the surface of the Sun! The solution to this problem is that power should be transmitted along long distances at high voltage. This greatly reduces $I^2R$ losses. For example, a typical high voltage line voltage is $115 \k\V$. At this voltage, the power loss per foot of cable is only 378 W per foot. Intuitively, we know that reducing current allows for lower power dissipation. We can deliver the same amount of power with a lower current by using a higher voltage.
     \end{enumerate}
 
 	\ex{1.7}