Resistor and inductor circuit with a steady voltage source

Question:
A circuit that has a resistor, $R$, and an inductor, $L$, which are connected in series with a DC voltage source $V$. The initial value of the current is $I(t=0)=0$ A. Find the current as a function of time.

Answer:
From Kirchoff's law, the generated voltage is consumed by each element.
\[
V-RI-L\frac{dI}{dt}=0
\]
\begin{eqnarray}
RI+L\frac{dI}{dt} &=& V  \\
I' + \frac{1}{\tau}I &=& \frac{V}{L}
\end{eqnarray}
where $I'=\frac{dI}{dt}$ and $\tau=\frac{L}{R}$. Now, let $D\equiv \frac{d}{dt}$.
\begin{equation}
\left(D+\frac{1}{\tau}\right)I=\frac{V}{L}  \\
\end{equation}
The fundamental solution is
\begin{equation}
I_f = C_1e^{-\frac{t}{\tau}}
\end{equation}
The particular solution is
\begin{equation}
I_p = \frac{1}{D+\frac{1}{\tau}}\cdot \frac{V}{L}=\frac{1}{0+\frac{1}{\tau}}\cdot \frac{V}{L}=\frac{V}{R}
\end{equation}
Therefore, we have the general solution.
\begin{equation}
I = C_1e^{-\frac{t}{\tau}} + \frac{V}{R}
\end{equation}
When $t=0$, $I=0$. So $0=C_1+V/R$ and $C_1=-V/R$. Then, the current is expressed as
\[
I(t) = -\frac{V}{R}e^{-\frac{t}{\tau}}+\frac{V}{R}
\]
or
\[
I(t) = \frac{V}{R}(1-e^{-\frac{t}{\tau}})
\]

Powered by Hirophysics.com

Magnetic fields from current flows and resistor circuit

Question:
The wire splits into two ways, which are bent circularly with radius of $a$. The upper half wire has resistance $2R$ and the lower part has $R$. Find the magnetic fields at the center of the circle in terms of the total current $I$.

Answer:
The magnetic field created by current $I_1$ is denoted as $B_1$. From Biot-Savart law,
\begin{eqnarray}
\vec{B_1} &=& \oint \frac{\mu_0}{4\pi}\frac{I d\vec{l}\times \vec{r}}{r^2}  \\
    &=& \frac{\mu_0}{4\pi}\frac{I_1}{a^2}\int_{0\rightarrow \pi a}dl (-\vec{z})  \\
    &=& -\frac{\mu_0}{4\pi}\frac{I_1}{a^2}\pi a\vec{z} \\
    &=& -\frac{\mu_0 I_1}{4a}\vec{z}
\end{eqnarray}
Likewise, we can obtain the magnetic field created by the lower part.
\begin{equation}
\vec{B_2} = \frac{\mu_0 I_2}{4a}\vec{z}
\end{equation}
This is a parallel connection, and we can use
\[
\frac{I_1}{I_2}=\frac{R}{2R} \quad \rightarrow \quad I_1=\frac{1}{2}I_2
\]
The current is conserved.
\[
I = I_1 + I_2 \quad \rightarrow \quad I_2 = I - I_1
\]
Therefore,
\[
I_1=\frac{1}{2}(I - I_1) \quad \rightarrow \quad I_1 = \frac{I}{3}
\]
Hence, the other current will be
\[
I_2 = \frac{2I}{3}
\]
The magnetic field is expressed as
\[
\vec{B}=\vec{B_1}+\vec{B_2}=\frac{\mu_0}{4a}(I_2-I_2)\vec{z}=\frac{\mu_0}{4a}\left(\frac{2I}{3}-\frac{I}{3}\right)\vec{z}=\frac{\mu_0 I}{12a}\vec{z}
\]
The magnetic field is directed toward us.

Powered by Hirophysics.com

RC circuit: With a DC power supply

Question:
Consider an RC circuit with a DC voltage supply as shown in the figure. There are four identical resistors ($R$=1.00 mega $\mathrm{\Omega}$) connected to one capacitor ($C$= 1.00 micro F). The voltage of the power supply is 10$\times$10$^6$ V. If the capacitor is fully charged and then the power supply is removed, find the current at $t=$0.5 s.

Answer:
All the four resistors are connected in parallel. The equivalent resistance of multiple parallel resistors is given by
\[
\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R_{\mathrm{1}}}+\frac{1}{R_{\mathrm{2}}}+\frac{1}{R_{\mathrm{3}}}+\cdots
\]
In this case, we obtain
\[
\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{4}{R}
\]
Thus,
\[
R_{\mathrm{eq}}=\frac{R}{4}=\frac{1.00\times 10^6}{4}=2.50\times 10^5 \mathrm{\Omega}
\]
From Kirchoff's law, each circuit element consumes the voltage, and the total of it and supplied voltage must be zero. The capacitor consumes voltage, $\frac{Q}{C}$ where $Q$ is the total charge. The resistor consumes $RI$ because of Ohm's law. Since we consider the discharging process, voltage from the power supply is zero. Then, we have
\[
-RI-\frac{Q}{C}=0
\]
We know $I=\frac{dQ}{dt}$, so
\[
\frac{dQ}{dt}+\frac{Q}{RC}=0
\]
Solve this differential equation:
\[
Q=Q_0 e^{-\frac{t}{RC}}
\]
The charges and current flow are equivalent in terms of time, so can express it as
\[
I=I_0 e^{-\frac{t}{RC}}
\]
$RC$ is known as the time constant, $\tau=2.50\times 10^5 \times 1.00 \times 10^{-6}=0.25$ s. $I_0$ is the current right before the voltage source is removed; namely, the maximum current. It can be calculated by Ohm's law.
\[
I_0=\frac{V}{R}=\frac{10\times 10^6}{2.50\times 10^5}=40.0 \ \mathrm{A}
\]
Then, we can find the current at $t=0.5$ s.
\[
I=40.0 e^{-\frac{0.5}{0.25}}=5.41 \ \mathrm{A}
\]

Powered by Hirophysics.com

AC circuit: Frequency of the generator

Question:
There is a AC circuit with a resistor $R=50 \ \Omega$. When the time is zero, the voltage is equal to zero. After 1/720 s, the voltage becomes a half of the maximum
voltage. Find the frequency of the voltage generator.

Answer:
The voltage varies with time, so
\[
V = V_m \sin \omega t
\]
where $V_m$ is the peak voltage. From the condition, we can express
\[
0.5V_m = V_m \sin \left(\omega\cdot \frac{1}{720}\right)
\]
$V_m$ is cancelled out. Then take the arcsine.
\[
 \frac{\omega}{720} = \sin^{-1}0.5
\]
Therefore,
\[
\omega = 120 \pi \ \mathrm{rad/s}
\]
In order to get the frequency, divide it by $2\pi$.
\[
f = \frac{\omega}{2\pi} = 60 \ \mathrm{Hz}
\]
Note that we don't have to use the resistance.

Powered by Hirophysics.com

A parallel plate capacitor, dielectric constants

Question:
There is a parallel plate capacitor of area $L^2$ and separation distance of $d$. A half of the space, $d/2$, is filled with a material of dielectric constant, $\kappa_1$. The other half is filled with the other material of constant, $\kappa_2$. What is the capacitance of this capacitor if the free space capacitance is $C_0$?

Answer:
The capacitance of a parallel plate capacitor is
\[ C = \frac{\epsilon_0\kappa A}{d} \]
where $A=L^2$. When a capacitor does not have any dielectrics (free space between plates), it will be
\[ C_0 = \frac{\epsilon_0 A}{d} \]
We can consider the particular capacitor is like series connection of two different capacitors. Namely,
\[ \frac{1}{C_{\mathrm{Total}}} = \frac{1}{C_1}+\frac{1}{C_2} \\
\rightarrow C_{\mathrm{Total}} = \frac{C_1C_2}{C_1+C_2} \]
From the condition, we have each capacitance of $C_1$ and $C_2$.
\[ C_{1,2} =  \frac{\epsilon_0\kappa_{1,2} A}{\frac{1}{2}d} = \frac{2\epsilon_0\kappa_{1,2} A}{d} \]
Then, we can rewrite it in terms of $C_0$.
\[ C_{1,2} = 2\kappa_{1,2}C_0 \]

Plug in the above.
\[ C_{\mathrm{Total}} = \frac{2\kappa_1C_0\times 2\kappa_2 C_0}{2\kappa_1C_0+2\kappa_2 C_0}    \\
      = \frac{4\kappa_1\kappa_2 C^2_0}{2C_0(\kappa_1+\kappa_2)}   \\
      =  \frac{2\kappa_1\kappa_2 C_0}{\kappa_1+\kappa_2}  \]

Powered by Hirophysics.com