A device connected to a current source and then resistor

In summary: So don't make that assumption.In summary, The given device can be modeled as a current source in parallel with a resistance. When connected to a 5[mA] current source, it delivers 296[mW] to the source and when connected to a 10[V] voltage source, it absorbs 16.5[mW]. To find the power absorbed when connected to a 10[kΩ] resistor, the device can also be modeled as a voltage source in series with a resistance. Care must be taken with the signs of potential and current in the circuit equations to ensure accurate results.
  • #1
rock42
5
0
Forgive the title, I was unsure of how to briefly describe the problem.
Attached are diagrams.

Homework Statement



The device shown in Figure P2.5 can be modeled as a current source in parallel with a resistance. When this device is connected to a 5[mA] current source, so that 5[mA] enters the device at terminal A and leaves at terminal B, then the device delivers 296[mW] to the 5[mA] current source.
Then, when the same device is disconnected from the current source and is connected to a 10[V] voltage source, so that there is a 10[V] voltage at terminal B with respect to terminal A, then the device absorbs 16.5[mW].
Find the power absorbed by the device when the device is connected to a 10[kΩ] resistor.

Homework Equations


v=ir, p=iv, KVL, KCL

The Attempt at a Solution



First with the Current Source [Circuit one]
power.del.by.device = power.abs.by.5ma = .296[W]
p=iv so...
v(5ma current source) = 59.2 [V]

KCL on top node...
-.005 + (59.2/Rx) - ix = 0

Now with the Voltage Source [Circuit two]
power.abs.by.device = power.del.by.voltagesoruce = .0165 [W]
p=iv so...
i(voltage source) = .00165[A]

KCL on top node...
-.00165 + (10/Rx) - ix = 0

Solve the linear equations...
Rx = 14686.57 [Ω]
ix = -.000969

Third condition (resistor)
I forgot to make a diagram for this one. The new resistor has a current flowing upward. Its poles are positive on top and negative below.

power.abs.by.device = power.del.by.10kresistor
p=i^2 (r)
we need i at the resistor...

KCL on top node...
-ix - i(10k resistor) +i(middle resistor) = 0

KVL on left loop
-i(10k resistor) +Rx*i(middle resistor) = ...
-i(10k resistor)*10000+14686.57*i(middle resistor) = ...
-i(10k resistor)*10000+14686.57*(ix + i(10k resistor))= 0 (Solvable)

i(10k resistor) = .003037[A]

p= i^2 (r) = .003037^2 (10000)
p.abs.by.dev = 92.2 [mW]

My main concern is that their is something wrong with my understanding of current flow, when to flip signs etc. I know that in all my circuits I have the current iQ (which i refer to as the current across the introduced component) which goes the opposite direction of what it "should" be going (positive to negative). There are several questions like this on my HW which is another reason I want to be sure I get this one right. If something was left unclear please do ask.
 

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  • #2
As you've discovered, you need to be a bit careful about the signs of things.

Consider a simple circuit consisting of a current supply driving a resistor:

attachment.php?attachmentid=43491&stc=1&d=1328390957.gif


The resistor is dissipating power, so the current source must be producing it. Note that the potential that appears across the current source is positive. So the current source is delivering power at a rate p = IV. By extrapolation then, the current supply would dissipate power if that potential is negative.

In your first circuit, where the current source is said to be absorbing power this implies that the potential across the current source should be negative. It's taking charge carriers from a higher potential at its bottom and delivering them to a lower potential at its top, and absorbing the energy involved in doing so.

For the second case, where the voltage source is placed across the device, pay attention to the stated polarity. Which polarity is connected to terminal B?

You might also want to consider that if the device can be modeled as a current source in parallel with a resistance, then via Norton and Thevenin, it can also be modeled by a voltage source in series with a resistance. This can be handy if you're more comfortable dealing with series circuits.

Another hint: The problem does not state that the "resistance" in the device model is in fact a typical resistor.
 

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Related to A device connected to a current source and then resistor

1. What is a current source?

A current source is an electronic component that is designed to provide a constant flow of electric current in a circuit. It can be either an ideal current source, which provides a fixed current regardless of the load, or a practical current source, which has a limited range of output currents.

2. How does a current source work?

A current source works by maintaining a constant voltage across a load, resulting in a constant current flow. It achieves this by adjusting its output voltage according to changes in the load's resistance, thus keeping the current at a constant level.

3. What is a resistor?

A resistor is an electronic component that is used to control the flow of electric current in a circuit. It is designed to have a specific resistance value, which determines how much current can pass through it. Resistors are commonly used to limit current, reduce voltage, and adjust signal levels in electronic circuits.

4. How does a resistor affect a current source?

A resistor placed in series with a current source will limit the amount of current flowing through the circuit. The resistance of the resistor will determine how much current is allowed to pass through, and the overall voltage drop across the circuit will be affected. This can be used to control the current and voltage levels in a circuit.

5. What is the purpose of connecting a device to a current source and then resistor?

The purpose of connecting a device to a current source and then a resistor is to regulate the flow of current through the device. The combination of a current source and resistor can be used to control the voltage and current levels in a circuit, ensuring that the device receives the correct amount of power and operates within safe limits.

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