How does voltage divided by EMF show efficiency? I get that electromotoric force (emf) is energy per coulomb supplied and that voltage is energy per coulomb used but I don't get what it means when we say that the voltage divided by the emf gives the efficiency.
 A: Your question could use a context to be more clear. But I think you mean that the battery supplies a certain $\mathrm{emf}$ to the circuit, and the circuit elements require a certain voltage $V$ for the current to run.
Now, if the voltage $V$ over all circuit elements (summed up) is less than the $\mathrm{emf}$ supplied, then some is lost. Meaning, some parts of the circuit that was not intended to have resistance, does have resistance. That could for example be the wires in between circuit elements. There will be a voltage drop across such unwanted resistances, and if I am right in assuming that these voltage drops are not included in the voltage $V$ you are talking about, then the ratio of $\mathrm{emf}$ to $V$ simply tells you how much was lost over such unwanted resistances:
$$\eta=\frac{V}{\mathrm{emf}}$$


*

*If $\eta=1$, then $V$ is equal to the $\mathrm{emf}$, and all $\mathrm{emf}$ supplied was "spent" as intended. This is an efficiency of $100\,\%$, when all added energy (per Coluomb) is spent in a useful (intended) manner.

*If $\eta<1$, then $V$ is less than the $\mathrm{emf}$, and some has been "lost" in unwanted resistances. Your efficiency is then less than $100\,\%$, since less than $100\,\%$ of the energy (per Coloumb) added to the circuit was spent in a useful (intended) manner.


In certain circuits where a specific "job" is to be done - for example in a flashlight where a specific circuit resistance is to convert added electrical energy into light, or in a toaster where a specific resistance is to convert added electrical energy into heat - then you might have an efficiency defined as energy used for this specific "job" divided with the energy (per Coloumb) supplied from the battery. Then it is suddenly not only unwanted small resistances in wires etc. that give loss, but also all other circuit elements apart from the heating resistance that are necessary for the circuit to work. Then the efficiency tells you how big a fraction of the energy (per Coloumb) that is used specifically for the "job" or the purpose of the device / the circuit.
Be sure what the voltage $V$ covers when you plug it into the formula - else you can't know what the efficiency tells you.
A: The electrical power delivered by a battery is $\mathcal{E} I$ where $\mathcal{E}$ is the emf of the cell and $I$ is the current passing through the cell.
This "input" power originates from a chemical reaction within the battery.  
The electrical power dissipated in an external circuit connected to the terminals of the battery is $VI$ where $V$ is the potential difference across the terminals of the battery and that you might call the "output" power.
So efficiency $\eta =  \dfrac{\text {electrical power dissipated in external circuit }}{\text{electrical power delivered by battery from a chemical reaction}}$
$\Rightarrow \eta =\dfrac {VI}{\mathcal{E}I} = \dfrac{\text {pd across terminals of battery }}{\text{emf of battery}}$
The efficiency is less than one because work has to be done moving charge through the battery due to the "internal resistance" of the battery.
A: Your question makes no sense.  Volts is merely a common unit EMF is measured in.
To say that volts means a different kind of energy per charge than EMF is just plain wrong, making your question non-sensical.
