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Power consumed in a circuit problem

by reddvoid
Tags: current, newtwork, power, resistor, voltage
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reddvoid
#1
Jun14-11, 09:47 PM
P: 99
power consumed in network is 10W find x. . .
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MisterX
#2
Jun14-11, 10:29 PM
P: 581
You can do it using Ohm's law (V = IR), and Power = I2R
reddvoid
#3
Jun14-11, 11:04 PM
P: 99
Quote Quote by MisterX View Post
You can do it using Ohm's law (V = IR), and Power = I2R
but they've given total power consumed in circuit is 10W
that's what bothering me .

Carl Pugh
#4
Jun14-11, 11:19 PM
P: 384
Power consumed in a circuit problem

The vertical resistor has 2 volt across it. Thus the vertical resistor draws 1 amp so wattage is 2 watt.
This leaves 8 watt for horizontal resistor.
Watt = I^2*R
So what is correct answer?
reddvoid
#5
Jun15-11, 01:07 AM
P: 99
Quote Quote by Carl Pugh View Post
The vertical resistor has 2 volt across it. Thus the vertical resistor draws 1 amp so wattage is 2 watt.
This leaves 8 watt for horizontal resistor.
Watt = I^2*R
So what is correct answer?
what about the power consumed by voltage source ?
vk6kro
#6
Jun15-11, 04:19 AM
Sci Advisor
P: 4,032
Power = I2* R and this is a perfect voltage source with no resistance.

So, how much power would it dissipate?
reddvoid
#7
Jun15-11, 04:42 AM
P: 99
Quote Quote by vk6kro View Post
Power = I2* R and this is a perfect voltage source with no resistance.

So, how much power would it dissipate?
it doesn't dissipate any power but it consumes or delivers some power depending on whether the current is entering it's +ve terminal or leaving It's +ve terminal.
PajoTheDwarf
#8
Jun15-11, 10:18 AM
P: 9
The Resistance parallel to the 2V has 1 Amp and uses 2Watts.

The other resistor needs to use the remaining 8 Watts. 2Amps.


Answer is B
gneill
#9
Jun15-11, 09:23 PM
Mentor
P: 11,682
Quote Quote by PajoTheDwarf View Post
The Resistance parallel to the 2V has 1 Amp and uses 2Watts.

The other resistor needs to use the remaining 8 Watts. 2Amps.


Answer is B
If one of those two amps supplied by the current source is flowing through the vertical resistor, where does the other amp go?

It must flow INTO the voltage source via its + terminal, making it a power sink. So choice B will make the circuit consume more than 10W !
PajoTheDwarf
#10
Jun16-11, 08:00 AM
P: 9
But you also need to work out either the Voltage and Current sources or the loads but not both.

Sources and loads should cancel each other out.

The power absorbed by the 2V at 1Amp is 2 Watts.

The power from the current source is 6Volts and 2 Amps equals 12 Watts.

10 Watts total for circuit = 12 Watt - 2 Watt

EDIT: This is an example of how ideal components are misleading. If you add some internal resitance then the 2Volt would be a load and a source.
reddvoid
#11
Jun18-11, 12:12 AM
P: 99
here only current source is delivering power
the two resistors and voltage source are consuming power
so if we take X=2A total power delivered=total power consumed=12W
-------but at x=1.8A
we ll get total power delivered=total power consumed=10W
]]]]]] so answer is NONE [[[[[[[[
vk6kro
#12
Jun18-11, 12:43 AM
Sci Advisor
P: 4,032
Remember that energy can't be created or destroyed. It can be converted to another form like heat from electrical energy.

A power source cannot consume power without converting it to another form, so the energy, except that which is dissipated as heat in the resistors, is still in the circuit .

One way this could happen is if the voltage source was a rechargeable battery. The energy would be converted to chemical energy and could be recovered later as electrical energy.

So, the only power dissipated or consumed is that which goes to heat up the resistors.

we ll get total power delivered=total power consumed=10A

10 watts ?
reddvoid
#13
Jun18-11, 01:54 AM
P: 99
Quote Quote by vk6kro View Post
Remember that energy can't be created or destroyed. It can be converted to another form like heat from electrical energy.

A power source cannot consume power without converting it to another form, so the energy, except that which is dissipated as heat in the resistors, is still in the circuit .

One way this could happen is if the voltage source was a rechargeable battery. The energy would be converted to chemical energy and could be recovered later as electrical energy.

So, the only power dissipated or consumed is that which goes to heat up the resistors.

we ll get total power delivered=total power consumed=10A

10 watts ?
sorry its W, my mistake :):)
so,if we consider "the only power dissipated or consumed is that which goes to heat up the resistors (which is 10W here)." but total power produced by current source is 12W
then where is the remaining 2W going . . .
Voltage source is consuming it right?
maybe it is storing it as chemical energy but still it is consuming power right ???
vk6kro
#14
Jun18-11, 05:16 AM
Sci Advisor
P: 4,032
Maybe we have a problem with definitions.

Storing does not mean the same thing as consuming.

Consuming means that energy has left the circuit permanently. Storing means it is still in the circuit and available for future use.

It is just like a charged capacitor. If it is loss-free, it can return as much power to the circuit as was stored in it.

Discuss it with your teacher, but I doubt that you will get much support.
reddvoid
#15
Jun18-11, 08:02 AM
P: 99
Quote Quote by vk6kro View Post
Maybe we have a problem with definitions.

Storing does not mean the same thing as consuming.

Consuming means that energy has left the circuit permanently. Storing means it is still in the circuit and available for future use.

It is just like a charged capacitor. If it is loss-free, it can return as much power to the circuit as was stored in it.

Discuss it with your teacher, but I doubt that you will get much support.
Ok


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