Voltage Drop over Constant Current Circuit

In summary, the problem is to find the voltage drop across a 1000 mA current source and two 2000Ω resistors in series. The attempt at solving the problem involves using Ohm's law to calculate the total resistance and the voltages across each resistor, which are then added together to get the total voltage across the resistors. It is noted that the current source provides the voltage for the resistors and therefore the voltage across the current source is equal to the total voltage across the resistors.
  • #1
chronotied
2
0

Homework Statement


So I have a problem that I am working on that requires me to find the voltage drop across a 1000 mA current source, and two 2000Ω resistors that are in parallel. I've added a picture of a circuit diagram for this below.

Homework Equations



V = I * R

The Attempt at a Solution



I am not particularly familiar with how constant current power supplies work, but from my understanding, as opposed to a regular battery, current sources push a constant current throughout the entire circuit, and have variable voltage to accommodate different resistances it passes through. So, for the question above, I would think that you could just use ohm's law to solve for the voltage correct? For example, since the resistors are in series, you add them together to find total resistance:

Rtotal = R1 + R2...

Rtotal = 2000Ω + 2000Ω

Rtotal = 4000Ω

We know the current is at 1000mA, so:

V = I * R1
V = I * R2

V = 1A * 2000Ω = 2000V

Am i doing this correctly, or am I not thinking it through right?...

Also, how would I find the Voltage going through the current source itself?

Thank you for any and all answers. New to these boards but have glanced at other problems on here throughout the past couple quarters. This is seriously an invaluable resource.
 

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  • #2
Are you getting 2000 v across battery?
 
  • #3
chronotied said:

Homework Statement


So I have a problem that I am working on that requires me to find the voltage drop across a 1000 mA current source, and two 2000Ω resistors that are in parallel. I've added a picture of a circuit diagram for this below.

HINT: From the diagram shown, the two resistors are in SERIES, not PARALLEL.
 
  • #4
chronotied said:
I am not particularly familiar with how constant current power supplies work, but from my understanding, as opposed to a regular battery, current sources push a constant current throughout the entire circuit, and have variable voltage to accommodate different resistances it passes through.
That's about right.
V = 1A * 2000Ω = 2000V
That's the voltage across each resistor. So voltage across both must be 4000V.
Also, how would I find the Voltage going through the current source itself?
The current source provides the voltage that you measure across the two resistors. So the voltage across the current source = the voltage across the combined resistances.

When describing voltage across a circuit element, it is usually a good idea to point out which end is more positive than the other, too. Given that current is flowing into the resistor string from the left and emerging from the right, which end must have the higher potential?
 
  • #5
HINT: From the diagram shown, the two resistors are in SERIES, not PARALLEL.

Totally meant in series. I have basically the same problem but them in parallel as well, so I got mixed up when typing it out. Thanks though!

The current source provides the voltage that you measure across the two resistors. So the voltage across the current source = the voltage across the combined resistances.

That makes sense since the voltage should drop from its initial (from the power source) to zero after it passes through the resistors so it should just be the the sum of the voltages through each resistor as you said. Thank you so much for your help! I knew it had to be a fairly simply problem, I was just getting tripped up by the fact that it was a current source as opposed to a regular battery that produces constant voltage.

Thanks everyone!
 

FAQ: Voltage Drop over Constant Current Circuit

1. What is voltage drop over a constant current circuit?

Voltage drop over a constant current circuit refers to the decrease in electrical potential energy as current flows through the circuit. This drop in voltage is typically caused by the resistance of the conductive materials in the circuit, which results in a loss of energy.

2. How is voltage drop calculated in a constant current circuit?

Voltage drop is calculated by multiplying the current in the circuit by the resistance of the conductive materials. This is known as Ohm's Law, which states that voltage drop (V) is equal to current (I) multiplied by resistance (R). This calculation can also be represented as V = IR.

3. What factors affect voltage drop in a constant current circuit?

The main factor that affects voltage drop in a constant current circuit is the resistance of the conductive materials. Other factors that may contribute to voltage drop include the length and thickness of the wires used, the temperature of the materials, and the type of material used.

4. Why is voltage drop important to consider in a constant current circuit?

Voltage drop is important to consider because it affects the overall performance and efficiency of the circuit. Excessive voltage drop can result in a decrease in the amount of usable energy, which can lead to malfunctions or underperformance of the circuit. It is also important to consider voltage drop in order to properly size and design the circuit for optimal performance.

5. How can voltage drop be reduced in a constant current circuit?

Voltage drop can be reduced by using materials with lower resistance, such as copper instead of aluminum. Additionally, using thicker wires or shorter lengths of wire can also help reduce voltage drop. Properly sizing and designing the circuit can also help minimize voltage drop.

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