Solving Simple Circuit: Find Voltage Drop & Power Dissipation

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In summary, the current source in the circuit is providing 0.5A and the resistors are providing a total resistance of 28.13 Ohms. The voltage drop across the current source is 28.13 - 0.5 = 25.13 volts. The voltage drop across resistor 1 is V=I^2*R and the voltage drop across resistors 2, 3, 4, and 5 is V=I^2*(28.13+25.13)
  • #1
MagLok
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Hello everyone. I have some brief questions about a simple circuits problem in my homework.

First off, here is the image of the circuit (sorry..drawn in MSPaint.) http://img211.imageshack.us/img211/3271/circuitgh9.jpg

The Current Source = 0.5A
R1 = 5 Ohms
R2 = 25 Ohms
R3 = 10 Ohms
R4 = 50 Ohms
R5 = 5 Ohms

It asks to find the total resistance of resistors 2, 3, 4, and 5. I've done this by counting R2 as part of a series circuit, and figuring R3,R4, and R5 as a parallel.

So, 25 + (1/((1/10)+(1/50)+(1/5))) = 28.13 Ohms (assuming this is correct..)


Now it asks to find the Voltage drop across the current source..I'm not sure how to figure this since I don't have distance..

Next it asks to find voltage drop across resistor 1, How much power is dissipated by this resistor (V=I^2*R)...But once again...voltage drop?

And finally it asks for the voltage drop across R2,3,4,5...and once again, voltage drop?

I've read and re-read all of the notes taken in class, and have no idea how to figure voltage drop for this type of circuit, I checked out an Ugly book and the NEC book, but everything there is to do with voltage drop requires distance.


Thanks in advance,

-MagLok
 
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  • #2
Your initial resistance calculation is not correct. To simplify the resistive network, use several steps of combining series and parallel resistances. For example, what do the resistors R3 and R4 combine to make? What do R2 and R5 combine to make? And how do you combine those two resistive combinations in the next step?
 
  • #3
Hello, you are making some mistakes in combining the resistors. To see why this is we need to examine some basic terminology. First we need to know what an element is, basically a battery, resistor, capacitor, etc are all elements, so basically any circuit component is an element. Now a node is a place where two elements meet. This is where your problem lies as the definition of being in parallel means that two elements are connected to the same two nodes. The resistors you have described above aren't in parallel yet, but we can get them to that point.

The first thing we need to recognize is that R2 and R5 are in series and so are R3 and R4 so we can directly add them to get equivalent resistors of 30 ohms and 60 ohms respectively. Now these two equivalent resistors are in parallel, because they share the same two nodes, visually this means that between where you had R3 and R4 you have an equivalent resistor (obviously you take the two resistors you combined to form these new resistors out of the drawing) of 60 ohms and in the spot of R5 is an equivalent resistor having 30 ohms. The nodes they share are the entire length of the bottom (the current source is attached to it too you see) and the other node is the point where the wires containing R1 and the two parallel resistors meet.

Now combine the two equivalent resistors you just created using the formula for parallel resistance and you get a new equivalent resistor of 20 ohms which is in series with the last remaining resistor, R1. Now you can obviously add the two to get the final equivalent resistance of 25 ohms.

I don't have time to help you with the rest of your questions, or post what I've done graphically so I apologize if the above visual description wasn't clear, but now that you have the equivalent resistance you can now find the voltage drop across the entire circuit using Ohm's law and you should be able to use what you know physically about the circuit to answer the rest of your questions, good luck!
 
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Related to Solving Simple Circuit: Find Voltage Drop & Power Dissipation

What is voltage drop in a circuit?

Voltage drop is the decrease in voltage that occurs as electricity flows through a circuit due to resistance. It is measured in volts (V) and can be calculated using Ohm's Law (V=IR), where V is voltage, I is current, and R is resistance.

How do you calculate voltage drop in a circuit?

Voltage drop can be calculated using Ohm's Law (V=IR), where V is voltage, I is current, and R is resistance. You can also use Kirchhoff's Voltage Law, which states that the sum of voltage drops in a closed loop circuit is equal to the applied voltage.

What is power dissipation in a circuit?

Power dissipation is the amount of energy that is converted into heat in a circuit due to the resistance of the components. It is measured in watts (W) and can be calculated using the formula P=IV, where P is power, I is current, and V is voltage.

How do you calculate power dissipation in a circuit?

Power dissipation can be calculated using the formula P=IV, where P is power, I is current, and V is voltage. You can also use Ohm's Law (P=I²R) or Joule's Law (P=I²Rt), where R is resistance and t is time.

Why is it important to calculate voltage drop and power dissipation in a circuit?

Calculating voltage drop and power dissipation is important because it allows us to understand how much energy is being lost due to resistance in a circuit. This information can help us design and optimize circuits for better efficiency and prevent overheating of components. It is also crucial for safety purposes as high voltage drops and power dissipation can lead to damage or failure of the circuit.

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