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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.

Category: Answers, Electricity and Magnetism, Maple

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