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Answer: Voltage and Current of a Circuit

Saturday, 2 of February , 2008 at 12:45 pm

The circuit in this question is an example of a Wheatstone Bridge (for the most part). In order to make this circuit a real Wheatstone Bridge the center resistor has to be replaced with a galvanometer. This circuit was introduced to me in a lab for my first E&M course in university which we used to accurately measure the resistance of an unknown resistor.

Question: Calculate the current and voltage difference between points D and B in the following circuit.

Answer: The easiest way to solve this problem is to use Kirchhoff’s Laws. We will start off by using Kirchhoff’s first law which is defined as:

The current entering a junction must equal the current out of the junction.

\sum I_{in} = \sum I_{out}

In order to use this law, we first need to define the direction the current is traveling. It really doesn’t matter how you define the current direction. If you picked the wrong direction, the current will just be a negative value which means that it flows the opposite way. This is how I defined the current direction for the circuit.

If we apply Kirchhoff’s first law to junction B, we get the following:

I_{DB} + I_{AB} = I_{BC}

If we this law again on junction D, we get:

I_{AD} = I_{DB} + I_{DC}

Next, we will need to use Kirchhoff’s Second law which is defined as:

The potential differences across all elements around any closed circuit loop must be zero.

\displaystyle\sum_{Closed~Path} \Delta V = 0

If we apply Kirchhoff’s 2nd law and Ohm’s Law to the loop ADB, we get:

V_{AD} + V_{DB} - V_{AB} = 0
RI_{AD} + RI_{DB} - 3RI_{AB} = 0
I_{AD} + I_{DB} - 3I_{AB} = 0 because R = 1~Ohm

If we do the same thing to the loop DBC, we get:

V_{DB} + V_{BC} - V_{DC} = 0
RI_{DB} + RI_{BC} - 2RI_{DC} = 0
I_{DB} + I_{BC} - 2I_{DC} = 0 because R = 1~Ohm

Now, we will use these laws on the loop ADC, we get:

5 - V_{AD} - V_{DC} = 0
5 - RI_{AD} - 2RI_{DC} = 0
I_{AD} + 2I_{DC} = 5 because R = 1~Ohm

Therefore, we have the following system of linear equations.

I_{DB} + I_{AB} - I_{BC}= 0
I_{AD} - I_{DC} - I_{DB} = 0
I_{AD} + I_{DB} - 3I_{AB} = 0
I_{DB} + I_{BC} - 2I_{DC} = 0
I_{AD} + 2I_{DC} = 5

which we can put in matrix form.

\left( \begin{array}{ccccc}0 & 1 & 1 & 0 & -1 \\1 & 0 & -1 & -1 & 0 \\1 & -3 & 1 & 0 & 0 \\0 & 0 & 1 & -2 & 1 \\1 & 0 & 0 & 2 & 0 \end{array} \right)\left( \begin{array}{c}I_{AD} \\I_{AB}\\I_{DB} \\I_{DC} \\I_{BC} \end{array} \right) = \left( \begin{array}{c}0 \\0\\0 \\0\\5\end{array} \right)

This system of equations can easily be solved with Maple.

> with(Student[LinearAlgebra]):
> A := Matrix([[0, 1, 1, 0, -1, 0],
		[1, 0, -1, -1, 0, 0],
		[1, -3, 1, 0, 0, 0],
		[0, 0, 1, -2, 1, 0],
		[1, 0, 0, 2, 0, 5]])



> ReducedRowEchelonForm(A)

Therefore, this means that the current going through the point D to B is \frac{25}{29}~Amps. If we use Ohm’s law, we can also determine the voltage difference between the points D and B to be \frac{25}{29}~Volts.

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Category: Answers, Electricity and Magnetism, Maple

Question: Voltage and Current of a Circuit

Saturday, 26 of January , 2008 at 12:20 am

Question: Calculate the current and voltage difference between points D and B in the following circuit.

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Category: Electricity and Magnetism, Questions

Answer: Particle in electric and magnetic field

Saturday, 1 of December , 2007 at 11:53 am

Question: Write down the equation of motion for a charged particle in uniform, parallel electric and magnetic fields, both in the z-direction, and solve it, given that the particle starts from the origin with the velocity (v_0,0,0). A screen is placed at x=a where a \ll \frac{mv}{qB}. Show that the locus of points of arrival of particles with given m and q, but different speeds v_0, is approximately a parabola.

Answer: The equation of motion is given by the Lorentz Force equation:

m\ddot{r} = q \vec{E} + q \dot{r} \times \vec{B}

However, since \vec{E} = E \hat{k} and \vec{B} = B \hat{k} for this problem, the equation of motion becomes

m\ddot{r} = q E \hat{k} + q \dot{r} \times B \hat{k}

which reduces to

m \ddot{x} = q\dot{y}B, m\ddot{y} = - q\dot{x}B, m \ddot{z} = qE

First, lets solve \ddot{z} = \frac{qE}{m}. If we integrate \ddot{z}, we get

\dot{z}(t) = \int \frac{qE}{m}dt = \frac{qE}{m}t + C
since the initial velocity at t=0 is (v_0,0,0), this means \dot{z}(0) = 0.

Therefore, \dot{z} = \frac{qE}{m}t.
If we integrate again, we get:

z(t) = \int \frac{qE}{m}tdt = \frac{qE}{2m}t^{2} +C
since the initial position at t=0 is (0,0,0), this means z(0) = 0.

Therefore, z = \frac{qE}{2m}t^{2}.

Next, we will reduce the other two equations

\ddot{x} = \frac{qB}{m}\dot{y}
\dot{x} = \int \frac{qB}{m}\dot{y}dt = \frac{qB}{m}y + C
since \dot{x}(0) = v_0, therefore \dot{x} = \frac{qB}{m}y + v_0.

We can similar reduce \ddot{y}

\ddot{y} = -\frac{qB}{m}\dot{x}
\dot{y} = -\int\frac{qB}{m}\dot{x}dt = -\frac{qB}{m}x +c
since \dot{y}(0) = 0, therefore \dot{y} = -\frac{qB}{m}x.

If we put these equations into the original equations of motion, we can decouple the x and y terms.

\ddot{x} = \frac{qB}{m}(-\frac{qB}{m}x) = - (\frac{qB}{m})^{2}x

which has a general solution of

x = Asin(\frac{qB}{m}t) + Bcos(\frac{qB}{m}t)

However, since x(0) = 0, therefore x = Asin(\frac{qB}{m}t).
In order to calculate A, we need to take the derivative.

\dot{x} = \frac{AqB}{m}cos(\frac{qB}{m}t)
since \dot{x}(0) =v_0, the following reduction can be made \frac{AqB}{m} = v_0 \Rightarrow A = \frac{m v_0}{qB}.

Therefore, x = \frac{m v_0}{qB}sin(\frac{qB}{m}t) and \dot{x} =v_0 cos(\frac{qB}{m}t).

Also, since \dot{x} = \frac{qB}{m}y +v_0, we can solve for y

v_0 cos(\frac{qB}{m}t) = \frac{qB}{m}y + v_0
y = \frac{v_0 m}{qB}(cos(\frac{qB}{m}t) - 1)

This means the equation of motion is:

x = \frac{m v_0}{qB}sin(\frac{qB}{m}t)
y = \frac{v_0 m}{qB}(cos(\frac{qB}{m}t) - 1)
z = \frac{qE}{2m}t^{2}

Next we will analysis what happens when x=a

x = a = \frac{m v_0}{qB}sin(\frac{qB}{m}t)
t = \frac{m}{qB}sin^{-1}(\frac{aqB}{m v_0})

Next take the taylor expansion

t = \frac{m}{qB} $\displaystyle\sum_{n=1}^{\infty}\frac{(2n)!}{4^{n}(n!)^{2}(2n+1)}(\frac{aqB}{mv_0})^{2n+1}$
t = \frac{m}{qB} ((\frac{aqB}{m v_0} + \frac{1}{66}(\frac{aqB}{m v_0})^{3}+\cdots))

Since a \ll \frac{mv}{qB}, this means \frac{aqB}{m v_0} \ll 1. This means that a good approximation of t would be the first term:

t = \frac{m}{qB} (\frac{aqB}{m v_0}) = \frac{a}{v_0}

Next. lets find y(\frac{a}{v_0})

y = \frac{v_0 m}{qB}(cos(\frac{qBA}{m v_0}) - 1)

Again, we will take the taylor expansion

y = \frac{v_0 m}{qB}($\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(2n+1)!}(\frac{qBa}{m v_0})^{2n+1}$ - 1)
y = \frac{v_0 m}{qB}((1 - \frac{1}{6}(\frac{qBa}{m v_0})^{2}+\cdots) - 1)

Since a \ll \frac{mv}{qB}, this means \frac{aqB}{m v_0} \ll 1. This means that a good approximation of y would be the first two term:

y = \frac{v_0 m}{qB}(1 - \frac{1}{6}(\frac{qBa}{m v_0})^{2} - 1)
y = -\frac{1}{6}\frac{qBa^{2}}{m}(\frac{1}{v_0})

Next, lets find z(\frac{a}{v_0})

z = \frac{q E}{2m}(\frac{a}{v_0})^{2} = \frac{q E a^{2}}{2m}(\frac{1}{v_0})^{2}

If we merge the results from y(\frac{a}{v_0}) and z(\frac{a}{v_0}), we get the following relation:

z = \frac{q E a^{2}}{2m}(\frac{1}{v_0})^{2} = \frac{q E a^{2}}{2m}(-6\frac{m}{qBa^{2}}y)^{2}
z = 18 \frac{mE}{qa^{2}B^{2}}y^{2}

This relation shows that the locus of points on the screen with given m and q, but different speeds v_0, is approximately a parabola.

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Category: Answers, Electricity and Magnetism

Question: Particle in electric and magnetic field

Saturday, 24 of November , 2007 at 10:37 pm

Question: Write down the equation of motion for a charged particle in uniform, parallel electric and magnetic fields, both in the z-direction, and solve it, given that the particle starts from the origin with the velocity (v_0,0,0). A screen is placed at x=a where a \ll \frac{mv}{qB}. Show that the locus of points of arrival of particles with given m and q, but different speeds v_0, is approximately a parabola.

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Category: Electricity and Magnetism, Questions

Answer: Electric field of a nonconducting sphere with a spherical cavity

Wednesday, 24 of October , 2007 at 10:03 pm

Answer: The electric field inside the cavity is going to be the superposition of the field due to the uncut sphere plus the field due to a sphere the size of the cavity with a uniform charge density of -\rho. The key to solve this problem is to calculate the electric field of each sphere in a different coordinate systems.

First, lets deal with the electric field of the large sphere of charge density \rho. To simplify Gauss’s Law, I am going to use a spherical coordinate system with the origin at the center of the sphere. The coordinate system is going to defined as:

\vec{r} = x \hat{i} + y \hat{j} + z \hat{k} = r \hat{r}

We can now use Gauss’s Law to calculate the electric field of the sphere.

\oint {\vec{E} \cdot dA = } \frac{1}{{\varepsilon _0 }}Q_{inside}

\oint {E dA = } \frac{1}{{\varepsilon _0 }} \frac{4}{3}\pi r^{3}\rho

E 4 \pi r^{2} = \frac{1}{{\varepsilon _0 }} \frac{4}{3}\pi r^{3}\rho

E = \frac{\rho}{{3 \varepsilon _0 }} r

Since the electric field is radial:

\vec{E} = \frac{\rho}{{3 \varepsilon _0 }} r \hat{r}

Now that we have calculated the electric field for the big sphere, we can calculate the field of the small sphere of charge density -\rho. We will, again, use a spherical coordinate system with the origin at the center of the sphere. The coordinate system is going to defined as:

\vec{r^{\prime}} = x^{\prime} \hat{i}^{\prime} + y^{\prime} \hat{j}^{\prime} + z^{\prime} \hat{k}^{\prime} = r^{\prime} \hat{r}^{\prime}

When we apply Gauss’s Law to the small sphere, we will get \vec{E^{\prime}} = \frac{-\rho}{{3 \varepsilon _0 }} r^{\prime} \hat{r^{\prime}}

Next we need to take the superposition of both the electric fields. In order to do so, we will need to relate the follow coordinate systems.


Using simple vector addition, we find that \vec{r^{\prime}} + z \hat{j} = \vec{r}. Also because \vec{r} = r \hat{r}, we can reduce this formula to: r^{\prime} \hat{r}^{\prime} = r \hat{r} - z \hat{j}

Summing the electric field of the two spheres we will get:

\vec{E} = \frac{\rho}{{3 \varepsilon _0 }} r \hat{r} - \frac{\rho}{{3 \varepsilon _0 }} r^{\prime} \hat{r^{\prime}}

\vec{E} = \frac{\rho}{{3 \varepsilon _0 }} r \hat{r} - \frac{\rho}{{3 \varepsilon _0 }} (r \hat{r} - z \hat{j})

\vec{E} = \frac{\rho}{{3 \varepsilon _0 }} z \hat{j}

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Category: Answers, Electricity and Magnetism