Power Calc: Find Current & Power in Circuit - Resistors Absorbing?

In summary, the voltage drop across the resistor is 10 V and the power absorbed by the resistor is 10 W. The voltage drop across the voltage source is also 10 V, but the power absorbed by the voltage source is 20 W, indicating that it is absorbing power. The current source is supplying power to both elements, with a power output of 20 W. The polarity of the voltage drop across the resistor cannot be determined without additional information.
  • #1
kexanie
11
0

Homework Statement


Concider the circuit shown below. Find the current iR flowing through the resitor. Find the power for each element in the circuit. Which elements are absorbing power?

Homework Equations


Ohm's law: U = IR;
Kirchhoff's Current Law and Kirchhof's Voltage Law.

The Attempt at a Solution


My instructor told me that the resistor is supplying power and the voltage source is absorbing power. But I just didn't get it.
By the diagram, the current iR flows in a clockwise direction. With KCL, iR = 2A, and with Ohm's law, vR = iRR = 10V. But How can I know whether the resistor is absorbing power or supplying power? Because I can't determine the polarity between the resitor with KVL.
It's clear that the voltage source is absorbing power, and if it were a battery, it would be charging.
But I can't understand how can a resistor SUPPLY power to other elements in the circuit as my instructor told me.
 

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  • #2
Is that 2A a current source or ammeter?
 
  • #3
PhysicoRaj said:
Is that 2A a current source or ammeter?
It's an independent current source, and 10V is an independent voltage source.
 
  • #4
kexanie said:

Homework Statement


Concider the circuit shown below. Find the current iR flowing through the resitor. Find the power for each element in the circuit. Which elements are absorbing power?

Homework Equations


Ohm's law: U = IR;
Kirchhoff's Current Law and Kirchhof's Voltage Law.

The Attempt at a Solution


My instructor told me that the resistor is supplying power and the voltage is absorbing power. But I just didn't get it.
By the diagram, the current iR flows in a clockwise direction. With KCL, iR = 2A, and with Ohm's law, vR = iRR = 10V. But How can I know whether the resistor is absorbing power or supplying power? Because I can't determine the polarity between the resitor with KVL.
It's clear that the voltage source is absorbing power, and if it were a battery, it would be charging.
But I can't understand how can a resistor SUPPLY power to other elements in the circuit as my instructor told me.
Hi kexanie.

http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

I believe what your instructor intended you to understand is that through the resistor is power delivered to the voltage source. It is the constant current source that supplies this power. Resistors always dissipate heat, according to the I2•R formula, but there is nothing to say that the current can't then go on to deliver power to another element.
 
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  • #5
NascentOxygen said:
Hi kexanie.

http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

I believe what your instructor intended you to understand is that through the resistor is power delivered to the voltage source. It is the constant current source that supplies this power. Resistors always dissipate heat, according to the I2•R formula, but there is nothing to say that the current can't then go on to deliver power to another element.

so can we tell that the power of the current source, the resitor and the voltage source are respectively -20W, 0W and 20W since the resitor just give out what it absorb from the current source?
 
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  • #6
kexanie said:

Homework Statement


Concider the circuit shown below. Find the current iR flowing through the resitor. Find the power for each element in the circuit. Which elements are absorbing power?

Homework Equations


Ohm's law: U = IR;
Kirchhoff's Current Law and Kirchhof's Voltage Law.

The Attempt at a Solution


My instructor told me that the resistor is supplying power and the voltage source is absorbing power. But I just didn't get it.
By the diagram, the current iR flows in a clockwise direction. With KCL, iR = 2A, and with Ohm's law, vR = iRR = 10V. But How can I know whether the resistor is absorbing power or supplying power? Because I can't determine the polarity between the resitor with KVL.
It's clear that the voltage source is absorbing power, and if it were a battery, it would be charging.
But I can't understand how can a resistor SUPPLY power to other elements in the circuit as my instructor told me.

attachment.php?attachmentid=67436&d=1394358223.png
What is the voltage drop across the resistor? (in the direction of current flow)

What is the voltage drop across the voltage source? (in the direction of current flow)
 
  • #7
kexanie said:
so can we tell that the power of the current source, the resitor and the voltage source are respectively -20W, 0W and 20W since the resitor just give out what it absorb from the current source?
All electrical power a resistor 'absorbs' is turned into heat.
 
  • #8
SammyS said:
What is the voltage drop across the resistor? (in the direction of current flow)

What is the voltage drop across the voltage source? (in the direction of current flow)

1)By Ohm's law, the voltage drop across the resitor is 10 V? How can I know the polarity of that voltage?

2)Is it just 10 V? I am not sure. But by the definition of an independent voltage source, it should be 10 V.
 
  • #9
kexanie said:
1)By Ohm's law, the voltage drop across the resistor is 10 V? How can I know the polarity of that voltage?

2)Is it just 10 V? I am not sure. But by the definition of an independent voltage source, it should be 10 V.
Going around the circuit clockwise, the same direction in which the current is forced to flow, yes the voltage drops by 10 Volts as it flows through the resistor. (Now you can find the power )

Continuing in the same direction, how much does the voltage change as current passes through the voltage source?
 
  • #10
SammyS said:
Going around the circuit clockwise, the same direction in which the current is forced to flow, yes the voltage drops by 10 Volts as it flows through the resistor. (Now you can find the power )

Continuing in the same direction, how much does the voltage change as current passes through the voltage source?

Sorry for grammar mistakes.

1) I can easily tell that the magnitude of that power is 10 W, but should it be negative or positive? I can't tell the polarity of the voltage dropped across the resistor. That's why I am asking.

2) I don't know. Do you mean electric potential by voltage?
 
  • #11
kexanie said:
1) I can easily tell that the magnitude of that power is 10 W, but should it be negative or positive? I can't tell the polarity of the voltage dropped across the resistor. That's why I am asking
Current flows through the resistor from the higher potential end to the lower potential end, e.g., from positive towards negative, and this is described as a voltage drop.
 

1. What is the purpose of "Power Calc: Find Current & Power in Circuit - Resistors Absorbing"?

The purpose of this tool is to calculate the current and power in a circuit with resistors that are absorbing energy. This can be useful in understanding the behavior and efficiency of an electrical circuit.

2. How do I use this tool to calculate current and power in a circuit?

To use this tool, you will need to input the values of the resistors in the circuit, as well as the voltage and current values of the power source. The tool will then use Ohm's Law and the power formula to calculate the current and power in the circuit.

3. Can this tool be used for circuits with multiple resistors?

Yes, this tool can be used for circuits with multiple resistors. You will need to input the values for each individual resistor in the circuit, and the tool will calculate the total current and power for the entire circuit.

4. What units does this tool use for current and power?

This tool uses the SI units of Ampere (A) for current and Watt (W) for power. Make sure to input your values in these units to get accurate results.

5. Is this tool accurate for all types of circuits?

This tool is accurate for circuits with only resistors that are absorbing energy. It may not be accurate for circuits with other types of components, such as capacitors or inductors, as they have different equations for calculating current and power.

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