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26th December 2019, 15:58 #1
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Clarification on power consumption
Common circuit i have seen but never thought let us say two power sources 10v and 5v in series with 5 ohm resistor in between. The current is 1amp, the net power by 10v source is 10w delivered but for the second source it is 5watt absorbed. The 5v being source how it can absorb ? I understand the resistor will dissipate what about source ?

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26th December 2019, 17:11 #2
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Re: Clarification on power consumption
kindly show the circuit you are talking about.
if the two sources are in series, there will be 15V across the resistor
at 5 ohm, the current is 3 amp
power dissipated by resistor is 45 W
power provided by 10V source is 30W
power provided by 5V source is 15W
if the two sources are in antiseries, there will be 5V across the resistor
at 5 ohm, the current is 1 amp
power dissipated by resistor is 5 W
power provided by 10V source is 10W
power absorbed by 5V source is 5W
the order of the components in the loop is not significant
the relative orientation of the two sources is

26th December 2019, 17:15 #3
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Re: Clarification on power consumption
The anti series is what i am referring to the doubt is how can a source absorb power?

26th December 2019, 17:17 #4
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Re: Clarification on power consumption
It's possible with ideal voltage sources in a circuit simulator or a circuit analysis exercise problem. Technical sources, e.g. a lab power supply are usually not capable of sinking current (or consuming power). There are special electronic loads that can sink but not source current and rarely bidirectional power supplies which handle both directions. These devices either burn the consumed power in a resistor or are feeding it back to the mains (regenerative power suppy).
A rechargeable battery can be basically used in this place although its voltage isn't exactly constant.

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26th December 2019, 17:33 #5
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Re: Clarification on power consumption
the lead acid battery in a car provides power to operate starter motor, lights, radio, etc when engine is off
when engine is on, said battery is charged to the level determined by the voltage regulator
when fully charged, the alternator provides less power
so it provides power (discharge) and absorbs power (charging)
other than homework problems to enforce understanding of Kirchoff's Laws,
i have never heard of anyone putting power supplies in antiparallel, as noted by FvM
what is behind your question?

27th December 2019, 06:42 #6
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27th December 2019, 08:50 #7
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Re: Clarification on power consumption
The example is of course not wrong. V2 is consuming power according to physical law. If you like simple technical implementations, think of an accumulator being charged.

27th December 2019, 09:08 #8
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Re: Clarification on power consumption
Hi,
To each part in your circuit:
Draw a voltage vector and a current vector.
* If both vectors of a device point in the same direction: it is a power sink, it consumes energy
* If both vectors of a device point in opposite direction: it is a power source, it delivers energy
KlausPlease donīt contact me via PM, because there is no time to respond to them. No friend requests. Thank you.

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29th December 2019, 04:29 #9
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Re: Clarification on power consumption
was referring to a online course as an example for power calculation he referred the circuit shown...
The two negative ends of the voltage source are connected to ground; they are at zero potential.
Potential at V1 is +10 and the potential at V2 is +5V respectively.
The resistor has 10V at one end and 5V at the other end. The current flows from the higher potential to the lower. The voltage and current directions are same.
The resistor dissipates 5V*5A=25W.
The same current must flow through the complete circuit in the steady state. The voltage source on the left is dissipating power (voltage and current are in the same way) total of 50W (10V and 5A).
The voltage source on the right is having current and voltage in the opposite direction. The voltage source on the right is therefore sinking (absorbing) power of 25W.
Essentially 50W of power is delivered by the voltage source on the left; of this 25W is dissipated in the resistor and and another 25W is stored in the voltage source on the right.
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