Voltage drop over an independent current source?

In summary, the conversation is about understanding a circuit and its thevenin equivalent. The person is confused about a specific node and wire, but is reminded that current sources will force current to flow regardless of potential difference. This is similar to how voltage sources work.
  • #1
nickadams
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Homework Statement



I am supposed to give a thevenin equivalent circuit to the one posted in the picture.

Homework Equations



V=IR
KVL: "voltage drops around a closed loop sums to zero
KCL: "current into a node equals current out of a node"

The Attempt at a Solution



The part I'm confused about is the node and wire I've circled in red. I know since the wire coming down from the node is connected to ground that the bottom of the wire has V=0... I also know that the voltage drop across the current source is = Voc... but Voc has the higher voltage on top which would mean the current source is forcing current to flow from low voltage to high voltage...

Can someone please help me understand?
 

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  • #2
nickadams said:
The part I'm confused about is the node and wire I've circled in red. I know since the wire coming down from the node is connected to ground that the bottom of the wire has V=0... I also know that the voltage drop across the current source is = Voc... but Voc has the higher voltage on top which would mean the current source is forcing current to flow from low voltage to high voltage...
Current sources will make their specified current flow regardless of the potential difference. That's how they can supply power to a circuit. If you think about it, voltage sources do the same thing, moving charges from a lower potential to a higher potential.
 
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  • #3
gneill said:
Current sources will make their specified current flow regardless of the potential difference. That's how they can supply power to a circuit. If you think about it, voltage sources do the same thing, moving charges from a lower potential to a higher potential.

Okay! That makes sense; thank you.
 

1. What is voltage drop over an independent current source?

Voltage drop over an independent current source refers to the decrease in voltage that occurs when a current flows through the source. This drop in voltage is a result of the internal resistance of the source and can affect the overall performance of the circuit.

2. How is voltage drop over an independent current source calculated?

The voltage drop over an independent current source can be calculated using Ohm's Law, which states that V=IR, where V is the voltage drop, I is the current, and R is the internal resistance of the source. Alternatively, it can also be calculated by measuring the voltage across the source and the current flowing through it.

3. What factors influence voltage drop over an independent current source?

The main factors that influence voltage drop over an independent current source are the internal resistance of the source, the amount of current flowing through it, and the load resistance connected to the source. Higher internal resistance and current will result in a larger voltage drop.

4. How does voltage drop over an independent current source affect circuit performance?

Voltage drop over an independent current source can affect circuit performance in several ways. It can lead to a decrease in the overall voltage available to the circuit, which can cause a decrease in the current flowing through the circuit. This can affect the functioning of devices connected to the circuit and can also cause a decrease in the overall efficiency of the circuit.

5. How can voltage drop over an independent current source be minimized?

To minimize voltage drop over an independent current source, one can use a source with a lower internal resistance, reduce the amount of current flowing through the source, or use a load with a higher resistance. Additionally, using multiple sources in parallel can also help to reduce the overall voltage drop in a circuit.

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