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deepakpallerla committed Jun 14, 2021
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132 changes: 66 additions & 66 deletions Chapter1.tex
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Expand Up @@ -25,34 +25,34 @@
to[R=$R_2$,v=$V_{2}$] (2,0)
to (0,0)
\end{circuit}
By KVL and Ohm's law \[ V = V_{1} + V_{2} = R_{1}\cdot I + R_{2} \cdot I = (R_{1}+R_{2}) \cdot I = R \cdot I \]
where \[\mans{R = R_{1} + R_{2}}\] is the resistance of $R_{1}$ and $R_{2}$ in series. Now, consider a simple parallel resistor circuit.
\begin{circuit}{fig:1.3.2}{A basic parallel circuit.}
(0,0) to[V=$V$,invert] (0,3)
to[short,i=$I$] (2,3)
to[R=$R_1$,i>^=$I_{1}$] (2,0);
\draw (2,3) to[short] (4,3)
to[R=$R_2$,i>^=$I_{2}$] (4,0)
to (0,0)
\end{circuit}
By KCL and Ohm's law \[ I = I_{1} + I_{2} = \frac{V}{R_{1}} + \frac{V}{R_{2}} = \left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)\cdot V \]
solving for V as a function of I we get
\[V = \dfrac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2}}}\cdot I = \frac{R_{1}R_{2}}{R_{1}+R_{2}}\cdot I = R\cdot I \]
where \[\mans{R = \dfrac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2}}} = \frac{R_{1}R_{2}}{R_{1}+R_{2}}}\] is the resistance of $R_{1}$ and $R_{2}$ in parallel.
By KVL and Ohm's law \[ V = V_{1} + V_{2} = R_{1}\cdot I + R_{2} \cdot I = (R_{1}+R_{2}) \cdot I = R \cdot I \]
where \[\mans{R = R_{1} + R_{2}}\] is the resistance of $R_{1}$ and $R_{2}$ in series. Now, consider a simple parallel resistor circuit.

\begin{circuit}{fig:1.3.2}{A basic parallel circuit.}
(0,0) to[V=$V$,invert] (0,3)
to[short,i=$I$] (2,3)
to[R=$R_1$,i>^=$I_{1}$] (2,0);
\draw (2,3) to[short] (4,3)
to[R=$R_2$,i>^=$I_{2}$] (4,0)
to (0,0)
\end{circuit}
By KCL and Ohm's law \[ I = I_{1} + I_{2} = \frac{V}{R_{1}} + \frac{V}{R_{2}} = \left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)\cdot V \]
solving for V as a function of I we get
\[V = \dfrac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2}}}\cdot I = \frac{R_{1}R_{2}}{R_{1}+R_{2}}\cdot I = R\cdot I \]
where \[\mans{R = \dfrac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2}}} = \frac{R_{1}R_{2}}{R_{1}+R_{2}}}\] is the resistance of $R_{1}$ and $R_{2}$ in parallel.
\ex{1.4}

We known that the resistance $R_{12}$\footnote{Here we have only assigned a name to the resistance in parallel between $R_{1}$ and $R_{2}$.} of two resistors $R_{1}$ and $R_{2}$ in parallel is given by \[R_{12} = \dfrac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2}}}\]

Now, the resistance $R_{123}$ of three resistors $R_{1}$, $R_{2}$ and $R_{3}$ in parallel is equal to the resistance of two resistors $R_{12}$ (the resistance between $R_{1}$ and $R_{2}$ in parallel) and $R_{3}$ in parallel, then \[R_{123} = \frac{1}{\frac{1}{R_{12}}+\frac{1}{R_{3}}} = \frac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}}\]

We will prove by induction that the resistance $R_{1\cdots n}$ of $n$ resistances $R_{1}, R_{2}, \ldots, R_{n}$ in parallel is given by \[R_{1\cdots n} = \dfrac{1}{\sum_{i=1}^{n}\frac{1}{R_{i}}}\]
First, it's trivial to show that with $n = 1$ the equality holds. Now, we will assume that the equality is satisfied for $n = k$, that is
\[R_{1\cdots k} = \dfrac{1}{\sum_{i=1}^{k}\frac{1}{R_{i}}}\]
Then, we must show that equality holds for $n = k+1$. Thus, the resistance $R_{1\cdots (k+1)}$ of $(k+1)$ resistances $R_{1}, R_{2}, \ldots, R_{k+1}$ in parallel is equal to the resistance of two resistors $R_{1\cdots k}$ and $R_{k+1}$ in parallel, then \[R_{1\cdots (k+1)} = \frac{1}{\frac{1}{R_{1\cdots k}}+\frac{1}{R_{k+1}}} = \frac{1}{\sum_{i=1}^{k}\frac{1}{R_{i}}+\frac{1}{R_{k+1}}} = \frac{1}{\sum_{i=1}^{k+1}\frac{1}{R_{i}}}\]
where we have proved that equality holds for $n = k+1$. Finally, the resistance of $n$ resistors in parallel is given by \[\mans{R_{1\cdots n} = \dfrac{1}{\sum_{i=1}^{n}\frac{1}{R_{i}}} = \frac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2}}+\ldots+\frac{1}{R_{n}}}}\]

First, it's trivial to show that with $n = 1$ the equality holds. Now, we will assume that the equality is satisfied for $n = k$, that is
\[R_{1\cdots k} = \dfrac{1}{\sum_{i=1}^{k}\frac{1}{R_{i}}}\]

Then, we must show that equality holds for $n = k+1$. Thus, the resistance $R_{1\cdots (k+1)}$ of $(k+1)$ resistances $R_{1}, R_{2}, \ldots, R_{k+1}$ in parallel is equal to the resistance of two resistors $R_{1\cdots k}$ and $R_{k+1}$ in parallel, then \[R_{1\cdots (k+1)} = \frac{1}{\frac{1}{R_{1\cdots k}}+\frac{1}{R_{k+1}}} = \frac{1}{\sum_{i=1}^{k}\frac{1}{R_{i}}+\frac{1}{R_{k+1}}} = \frac{1}{\sum_{i=1}^{k+1}\frac{1}{R_{i}}}\]
where we have proved that equality holds for $n = k+1$. Finally, the resistance of $n$ resistors in parallel is given by \[\mans{R_{1\cdots n} = \dfrac{1}{\sum_{i=1}^{n}\frac{1}{R_{i}}} = \frac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2}}+\ldots+\frac{1}{R_{n}}}}\]
\ex{1.5}
Given that $P = \dfrac{V^2}{R}$, we know that the maximum voltage we can achieve is 15V and the smallest resistance we can have across the resistor in question is $1\k\Ohm$. Therefore, the maximum amount of power dissipated can be given by \[P = \frac{V^2}{R} = \frac{(15\V)^2}{1\k\Ohm} = \mans{0.225\W}\]
This is less than the 1/4W power rating.
Expand All @@ -74,51 +74,51 @@
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}
A $20,000 \Ohm/\V$ meter read, on its $1 \V$ scale, puts an $20,000 \Ohm/\V \cdot 1 \V = 20,000 \Ohm = 20\k\Ohm$ resistor in series with an ideal ammeter (ampere meter). Also, a voltage source with an internal resistance is equivalent to an ideal voltage source with its internal resistance in series.
\begin{enumerate}
\item In the first question, we have the following circuit:
\begin{circuit}{fig:1.7.1}{A voltage source with internal resistance and a $20,000 \Ohm/\V$ meter read in its $1 \V$ scale.}
(0,0) to[V=$1 \V$,invert] (0,4)
to[R=$10 \k\Ohm$,i>^=$I$] (5,4);
\draw[blue] (5,4) to[R=$20 \k\Ohm$,*-,color=blue] (5,2)
to node[draw,circle,fill=white] {A} (5,0);
\draw[blue] node[draw,circle,fill=blue,inner sep=1pt] at(5,0) {};
\draw (5,0) to (0,0)
\end{circuit}
Then, we have that the current in the ideal ammeter and the voltage in the meter resistance are given by\footnote{When a meter only measures currents, it puts a resistance in series to measures the current through that resistance and internally converts that current into voltage to \textit{measure voltages}.}
\[I = \frac{1 \V}{10\k\Ohm + 20\k\Ohm} = \mans{0.0333 \m\A} \quad \text{ and } \quad V = 0.0333 \m\A \cdot 20\k\Ohm = \mans{0.666 \V}\]
\item In the second question, we have the following circuit:
\begin{circuit}{fig:1.7.2}{A $10\k\Ohm-10\k\Ohm$ voltage divider and a $20,000 \Ohm/\V$ meter read in its $1 \V$ scale.}
(0,0) to[V=$1 \V$,invert] (0,4)
to[short] (3,4)
to[R=$10 \k\Ohm$] (3,2)
to[R=$10 \k\Ohm$] (3,0);
\draw[blue] (3,2) to node[draw,circle,fill=white] {A} (5,2)
to[R=$20 \k\Ohm$,color=blue] (5,0)
to[short,-*,color=blue] (3,0);
\draw[blue] node[draw,circle,fill=blue,inner sep=1pt] at(3,2) {};
\draw (3,0) to (0,0)
\end{circuit}
Now, we can to obtain the Thévenin equivalent circuit of circuit in Figure \ref{fig:1.7.2} with
\[R_{\Th} = \frac{10\k\Ohm \cdot 10\k\Ohm}{10\k\Ohm + 10\k\Ohm} = 5\k\Ohm\]
and
\[V_{\Th} = 1\V \cdot \frac{10\k\Ohm}{10\k\Ohm + 10\k\Ohm} = 0.5\V\]
Then, we have the following equivalent circuit:
\begin{circuit}{fig:1.7.3}{Thévenin equivalent circuit of circuit in Figure \ref{fig:1.7.2}.}
(0,0) to[V=$V_{\Th}$,invert] (0,4)
to[R=$R_{\Th}$,i>^=$I$] (5,4);
\draw node[draw,circle,fill=blue,inner sep=1pt] at(5,4) {};
\draw[blue] (5,4)
to node[draw,circle,fill=white] {A} (5,2)
to[R=$20 \k\Ohm$,-*,color=blue] (5,0);
\draw (5,0) to (0,0)
\end{circuit}
Finally, we have that the current in the ideal ammeter and the voltage in the meter resistance are given by
\[I = \frac{0.5 \V}{5\k\Ohm + 20\k\Ohm} = \mans{0.02 \m\A} \quad \text{ and } \quad V = 0.02 \m\A \cdot 20\k\Ohm = \mans{0.4 \V}\]
\end{enumerate}
\ex{1.8}
\ex{1.7}
A $20,000 \Ohm/\V$ meter read, on its $1 \V$ scale, puts an $20,000 \Ohm/\V \cdot 1 \V = 20,000 \Ohm = 20\k\Ohm$ resistor in series with an ideal ammeter (ampere meter). Also, a voltage source with an internal resistance is equivalent to an ideal voltage source with its internal resistance in series.
\begin{enumerate}
\item In the first question, we have the following circuit:
\begin{circuit}{fig:1.7.1}{A voltage source with internal resistance and a $20,000 \Ohm/\V$ meter read in its $1 \V$ scale.}
(0,0) to[V=$1 \V$,invert] (0,4)
to[R=$10 \k\Ohm$,i>^=$I$] (5,4);
\draw[blue] (5,4) to[R=$20 \k\Ohm$,*-,color=blue] (5,2)
to node[draw,circle,fill=white] {A} (5,0);
\draw[blue] node[draw,circle,fill=blue,inner sep=1pt] at(5,0) {};
\draw (5,0) to (0,0)
\end{circuit}
Then, we have that the current in the ideal ammeter and the voltage in the meter resistance are given by\footnote{When a meter only measures currents, it puts a resistance in series to measures the current through that resistance and internally converts that current into voltage to \textit{measure voltages}.}
\[I = \frac{1 \V}{10\k\Ohm + 20\k\Ohm} = \mans{0.0333 \m\A} \quad \text{ and } \quad V = 0.0333 \m\A \cdot 20\k\Ohm = \mans{0.666 \V}\]
\item In the second question, we have the following circuit:
\begin{circuit}{fig:1.7.2}{A $10\k\Ohm-10\k\Ohm$ voltage divider and a $20,000 \Ohm/\V$ meter read in its $1 \V$ scale.}
(0,0) to[V=$1 \V$,invert] (0,4)
to[short] (3,4)
to[R=$10 \k\Ohm$] (3,2)
to[R=$10 \k\Ohm$] (3,0);
\draw[blue] (3,2) to node[draw,circle,fill=white] {A} (5,2)
to[R=$20 \k\Ohm$,color=blue] (5,0)
to[short,-*,color=blue] (3,0);
\draw[blue] node[draw,circle,fill=blue,inner sep=1pt] at(3,2) {};
\draw (3,0) to (0,0)
\end{circuit}
Now, we can to obtain the Thévenin equivalent circuit of circuit in Figure \ref{fig:1.7.2} with
\[R_{\Th} = \frac{10\k\Ohm \cdot 10\k\Ohm}{10\k\Ohm + 10\k\Ohm} = 5\k\Ohm\]
and
\[V_{\Th} = 1\V \cdot \frac{10\k\Ohm}{10\k\Ohm + 10\k\Ohm} = 0.5\V\]
Then, we have the following equivalent circuit:
\begin{circuit}{fig:1.7.3}{Thévenin equivalent circuit of circuit in Figure \ref{fig:1.7.2}.}
(0,0) to[V=$V_{\Th}$,invert] (0,4)
to[R=$R_{\Th}$,i>^=$I$] (5,4);
\draw node[draw,circle,fill=blue,inner sep=1pt] at(5,4) {};
\draw[blue] (5,4)
to node[draw,circle,fill=white] {A} (5,2)
to[R=$20 \k\Ohm$,-*,color=blue] (5,0);
\draw (5,0) to (0,0)
\end{circuit}
Finally, we have that the current in the ideal ammeter and the voltage in the meter resistance are given by
\[I = \frac{0.5 \V}{5\k\Ohm + 20\k\Ohm} = \mans{0.02 \m\A} \quad \text{ and } \quad V = 0.02 \m\A \cdot 20\k\Ohm = \mans{0.4 \V}\]
\end{enumerate}

\ex{1.8}
\begin{enumerate}
\item In the first part, we have the following circuit:
\begin{circuit}{fig:1.8.1}{50\u A ammeter with 5k\Ohm\ internal
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