Measuring the current of a Wheatstone Bridge, given four known resistances and supply voltage
It depends on if Vg represents an open circuit or a resistive load of some kind.   


If it's an open circuit, then you have the trivial case: there is no current and \$I_{g}=0\$. This case might be useful if you have voltage probes hooked up to nodes D and B.
If there is a resistive load, then you must have prior knowledge of the load's resistance, \$R_{g}\$, to calculate the current. Knowing \$V_{g}\$ and \$R_{g}\$, you have enough information to use Ohm's Law to solve for \$I_{g}\$.


A common classroom experiment with a Wheatstone bridge is to put a light bulb  between nodes D and B. When there is an imbalance in the bridge, current will flow between D and B the light bulb will light up. R2 is adjusted until the light goes out, indicating that the bridge is balanced and D and B are at the same potential. How much current is going through the bulb depends entirely on the properties of the bulb itself.