
#1
Jun1411, 09:47 PM

P: 99

power consumed in network is 10W find x. . .




#2
Jun1411, 10:29 PM

P: 536

You can do it using Ohm's law (V = IR), and Power = I^{2}R




#3
Jun1411, 11:04 PM

P: 99

that's what bothering me . 



#4
Jun1411, 11:19 PM

P: 383

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? 



#5
Jun1511, 01:07 AM

P: 99





#6
Jun1511, 04:19 AM

Sci Advisor
P: 4,003

Power = I^{2}* R and this is a perfect voltage source with no resistance.
So, how much power would it dissipate? 



#7
Jun1511, 04:42 AM

P: 99





#8
Jun1511, 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 



#9
Jun1511, 09:23 PM

Mentor
P: 11,428

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 ! 



#10
Jun1611, 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. 



#11
Jun1811, 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 [[[[[[[[ 



#12
Jun1811, 12:43 AM

Sci Advisor
P: 4,003

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 ? 



#13
Jun1811, 01:54 AM

P: 99

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 ??? 



#14
Jun1811, 05:16 AM

Sci Advisor
P: 4,003

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 lossfree, 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. 



#15
Jun1811, 08:02 AM

P: 99




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