Is the work done by a constant voltage source positive or negative when current is opposite to emf?

 

If the current is against the EMF then you are applying energy to the source of the EMF, such as occurs when you charge a battery.

 

Is the work done by a constant voltage source positive or negative when current is opposite to emf?

This sounds like a homework question.

If by "current is opposite to emf" you mean that conventional current is flowing into the positive terminal of the source, then ask yourself how this can happen. If you hook up such a source to a resistor, which direction does current flow and is the work done by the source positive or negative. Let's say the voltage source is a battery -- is it gaining or losing energy? If current is flowing in the opposite direction, then ask yourself if the battery is gaining or losing energy and what that means for the work done BY the source (as opposed to the work done ON the source).

 

This sounds like a homework question.

If by "current is opposite to emf" you mean that conventional current is flowing into the positive terminal of the source, then ask yourself how this can happen. If you hook up such a source to a resistor, which direction does current flow and is the work done by the source positive or negative. Let's say the voltage source is a battery -- is it gaining or losing energy? If current is flowing in the opposite direction, then ask yourself if the battery is gaining or losing energy and what that means for the work done BY the source (as opposed to the work done ON the source).

when current enters positive terminal of voltage source, voltage source charges up , so it gains energy , its power(energy derivative) is positive
now what separates "work done BY the source" from"work done ON the source"?
and which is positive in this case?

 

Positive power = dissipated. Negative - generated

e.g. A power plant is actually -16 MW in order for the math to work.

 

when current enters positive terminal of voltage source, voltage source charges up , so it gains energy , its power(energy derivative) is positive
now what separates "work done BY the source" from"work done ON the source"?
and which is positive in this case?

Like many quantities, what is positive and what is negative completely depends on how the polarity is defined.

Work done ON something is positive if the energy of that thing is increased at the expense of whatever is doing the work. Work done BY something is positive if it is giving up energy to something else. Whenever work is done, it is always work done BY one thing ON another thing and thus the net work done by both combined is equal to zero.

When you define the voltage and current for a source (or a load) you have to assign a reference direction for both. For a passive load, there is only one reasonable choice and that is to define the positive current as entering the load at the positive voltage terminal. When you then multiply the voltage across the load and the current through the load, the result will be positive if the device is gaining energy and negative if it is not -- this is known as the passive sign convention. But for a source you can make a good argument for assigning positive current as either entering the positive terminal (so as to be consistent with the passive sign convention) or as leaving the positive terminal (so that the normal situation, namely the source providing energy to the circuit, results in a positive number). Either is reasonable and both are widely used, which is why it is important to clearly defined your reference directions in your work.