The Feynman Lectures on Physics Volume II Chapter 22 AC Circuits (Exercise)

http://file.buturi.blog.shinobi.jp/Feynmantoiengl4_1.pdf
Chapter 22 AC Circuits (Exercise)


1-1
Let
$R=1\Omega $.

1)
Let $A,B,C,D$ be the vertices.
Let $V$ be the potential difference between $A$ and $C$.
Let $I_1$ be the electric current through $ABC$.
Let $I_2$ be the electric current through $ADC$.

\[
V=2RI_1=2RI_2,I=I_1+I_2,
\]
Hence
\[
V=RI.
\]

2)

Let $V$ be the potential difference between $A$ and $B$.

Let $I_1$ be the electric current through $AB$.
Let $I_2$ be the electric current through $ADCB$.

\[
V=RI_1=3RI_2, I=I_1+I_2,
\]
Hnece
\[
R=\frac{V}{I_1+I_2}=\frac{V}{\frac{V}{R}+\frac{V}{3R}}=\frac{3R}{4}.
\]


1-2

a)

Let $I_1$ be the electric current through the condenser, and $I_2$ be the electric current through the inductor.
By Kirchhoff's rules
\[
I_1+I_2=I,
\]
\[
-\mathcal{E}+Ii\omega L+I_1\frac{1}{i\omega C}=0,
\]
\[
-I_1\frac{1}{i\omega C}+I_2i\omega L=0.
\]


(By the first and the second equation,
\[
-\mathcal{E}+(I_1+I_2)i\omega L+I_1\frac{1}{i\omega C}=0.
\]
This equation minus the third equation
\[
-\mathcal{E}+I_1\left( i\omega L+\frac{2}{i\omega C}\right) =0.
\]
\[
I_1=\frac{-\mathcal{E}}{i\omega L+\frac{2}{i\omega C}}.
\]
Hence
\begin{eqnarray*}
I&=&I_1+I_2\\
&=&I_1(1-\frac{1}{\omega ^2}CL)\\
&=&\frac{-\mathcal{E}}{i\omega L+\frac{2}{i\omega C}}\frac{\omega ^2CL-1}{\omega ^2CL}\\
&=& \frac{-i\mathcal{E}}{\omega L}\frac{1-\omega ^2CL}{2-\omega ^2 CL}.
\end{eqnarray*}
Let $\mathcal{E}=V_0e^{i\omega t}$.
Take the real part of $I$
\[
\Re I=\frac{V_0\sin \omega t}{\omega L}\frac{1-\omega ^2CL}{2-\omega ^2 CL}
\]
)

b)
\[
I_1+I_2=I,
\]
\[
-\mathcal{E}+Ii\omega L+I_2i\omega M+I_1\frac{1}{i\omega C}=0,
\]
\[
-I_1\frac{1}{i\omega C}+I_2i\omega L+Ii\omega M=0.
\]


(Substituting the first and the second equation into the third equation
\[
-\mathcal{E}+(I_1+I_2)i\omega L+I_2i\omega M+I_1\frac{1}{i\omega C}=0,
\]
\[
-I_1\frac{1}{i\omega C}+I_2i\omega L+(I_1+I_2)i\omega M=0.
\]
The difference ofabove equations is
\[
-\mathcal{E}+I_1\left( i\omega L+\frac{2}{i\omega C}-i\omega M\right) .
\]
By the tird equation
\[
I_1\left( -\frac{1}{i\omega C}+i\omega M\right) =-I_2i\omega (L+M).
\]
Hence
\begin{eqnarray*}
I&=&I_1+I_2\\
&=&I_1\left( 1+ \frac{\frac{1}{i\omega C}-i\omega M}{i\omega (L+M)})\right) \\
&=&\frac{-\mathcal{E}}{i\omega L+\frac{2}{i\omega C}-i\omega M}
\frac{ i\omega (L+M)+ \frac{1}{i\omega C}-i\omega M}{i\omega (L+M)} \\
&=&
\frac{-i\mathcal{E}}{\omega (L+M)}\frac{1-\omega ^2CL}{2-\omega ^2 C(L-M)}
\end{eqnarray*}
)

\[
\Re I=\frac{V_0\sin \omega t}{\omega (L+M)}\frac{1-\omega ^2CL}{2-\omega ^2 C(L-M)}.
\]


1-3

1-4

Let $I_1$ be the electric current through the left hand side.
Let $I_2$ be the electric current through the right hand side.
By Kirchhoff's rules
\[
I=I_1+I_2,
\]
\[
-\mathcal{E}+I_1R+\frac{1}{i\omega C}I_1=0,
\]
\[
-\mathcal{E}+I_2\frac{1}{i\omega C}+I_2R=0.
\]
By the second and the third equation
\[
I_1=I_2=\frac{-\mathcal{E}}{R+\frac{1}{i\omega C}}.
\]

The potential difference between $b$ and $a$ is

\begin{eqnarray*}
V_b-V_a&=&\frac{I_2}{i\omega C}-R I_1\\
&=&\left( \frac{1}{i\omega C}-R\right) \frac{\mathcal{E}}{R+\frac{1}{i\omega C}}\\
&=&-\frac{R+\frac{i}{\omega C}}{R-\frac{i}{\omega C}}V_0e^{i\omega t}\\
&=&-e^{i(\theta +\omega t)}V_0,
\end{eqnarray*}
where
\[
\tan \frac{\theta }{2}=\frac{1}{R\omega C}.
\]

1-5


\begin{eqnarray*}
I
&=&\left( \frac{1}{R}+i\left( \omega C-\frac{1}{\omega L}\right) \right) \mathcal{E}\\
\end{eqnarray*}

In the series connection case
\[
-\mathcal{E}=\left( R+i(\omega L-\frac{1}{\omega C})\right) I.
\]

1-6

Let
\[
L_1=R_a+\mathcal{L},
\]
\[
L_2=R,
\]
\[
L_3=R,
\]
\[
L_4=\frac{\frac{1}{i\omega C}R_b}{R_b+\frac{1}{i\omega C}}=\frac{R_b}{i\omega C R_b+1}
\]
.

If there is no electric current through $R_D$, then by the fact that
the left and the right hand side of $R_D$ is equipotential and Kirchhoff's rules



\[
-\mathcal{E}+I_1L_1+I_1L_3,
\]
\[
-\mathcal{E}+I_2L_2+I_2L_4
\]
\[
I_1L_1=I_2L_2.
\]

We eliminate $I_1$ and $I_2$

\[
L_1=\frac{I_2}{I_1}L_2=\frac{L_1+L_3}{L_2+L_4}L_2.
\]
Hence
\[
(L_2+L_4)L_1=(L_1+L_2)L_2.
\]
Hence
\[
L_1=\frac{L_2L_3}{L_4}=\frac{R^2}{\frac{R_b}{i\omega CR_b+1}}
\]
Hence by the definition of $\mathcal{L}$
\[
\mathcal{L}=L_1-Ra=\frac{R^2}{R_b}(i\omega CR_b+1)-R_a.
\]


1-7

Let
\[
L_1=r_1,
\]
\[
L_2=R_1+\frac{1}{i\omega C_1},
\]
\[
L_3=r_3,
\]
\[
L_4=\frac{\frac{R_2}{i\omega C}}{R_2+\frac{1}{i\omega C}}.
\]

Similar to the Excercise 1-6
\[
L_1L_4=L_2L_3.
\]

Hence
\[
r_1 \frac{R_2}{R_2i\omega C_2+1}=\left( R_1+\frac{1}{i\omega C_1}\right) r_2.
\]
Hence
\begin{eqnarray*}
\frac{r_1}{r_2}&=&\frac{ R_1+\frac{1}{i\omega C_1}}{\frac{R_2}{R_2i\omega C_2+1}}\\
&=&\left( R_1+\frac{1}{i\omega C_1}\right) (R_2i\omega C_2+1)\frac{1}{R_2}\\
&=&\frac{R_1}{R_2}+\frac{C_2}{C_1}+i\left( R_1\omega C_2-\frac{1}{\omega C_1R_2}\right) \hspace{1cm}(1)
\end{eqnarray*}
By comparing the real part of equation (1).

\[
\frac{r_1}{r_2}=\frac{R_1}{R_2}+\frac{C_2}{C_1},
\]
By comparing the imaginary part of equation (1).
\[
\left( R_1\omega C_2-\frac{1}{\omega C_1R_2}\right) =0.
\]
Hence
\[
\omega =\frac{1}{\sqrt{C_1C_2R_1R_2}}.
\]

1-8

a)

\[
-\mathcal{E}+\left( \frac{R\frac{1}{i\omega C}}{R+\frac{1}{i\omega C}}+\frac{Ri\omega L}{R+i\omega L}\right) I.
\]
\[
I=\frac{\mathcal{E}}{\frac{R\frac{1}{i\omega C}}{R+\frac{1}{i\omega C}}+\frac{Ri\omega L}{R+i\omega L}}
\]
By substituting $RC=L/R$.
\[
\frac{R\frac{1}{i\omega C}}{R+\frac{1}{i\omega C}}+\frac{Ri\omega L}{R+i\omega L}=\frac{L}{RC}.
\]

b)


Let $I_1, I_2$ be the electric current through
above registance and capacitor respectively.

\[
I_1+I_2=I,
\]
\[
I_1R=I_2\frac{1}{i\omega C}.
\]
by the second equation
\[
I_2=i\omega CRI_1.
\]
Substituting this to the first equation
\[
I_1(1+i\omega CR)=I.
\]
Hence
\[
I_1=\frac{I}{1+i\omega CR}=\frac{1}{1+i\omega CR}\frac{CR}{L}\mathcal{E}.
\]


Let $V_{RC}$ be the voltage of registance and capacitor .

\[
V_{RC}=I_1R=\frac{1}{1+i\omega CR}\frac{CR}{L}\mathcal{E}
\]


Hence
\[
\tan \delta =i\omega CR,
\]
where $\delta $ is phase difference.

1-9


a)
\[
I=\frac{\mathcal{E}}{R+i\omega L+\frac{m}{i\omega C}}.
\]

\[
\hat{I}=\frac{V_0}{R+i\omega L+\frac{m}{i\omega C}}.
\]

Hence the average rate of energy loss in $R$ is
\[
=\frac{R|\hat{I}|^2}{2}=\frac{RV_0^2}{R^2+\left( \omega L-\frac{m}{\omega C}\right) }.
\]

b)

i)

By $\omega L-\frac{m}{\omega C}=0$,

\[
m=\omega ^2LC=2.
\]

ii)

The voltage of
$P_0, P_1$ is
\[
\frac{1}{i\omega C}I=\frac{1}{i\omega C}\frac{V_0 e^{i\omega t}}{R}=50e^{i\omega t}V.
\]
The voltage of $R$ is
\[
RI=\mathcal{E}=100e^{i\omega t}V.
\]