If the graph is directed, that implies that no cycle can contain the same edge twice.
If the graph is simple (assuming you really meant a simple graph), that implies that there are no "loop" edges starting and ending at the same vertex, and that any pair of distinct vertices has at most one edge between them.
From those assumptions, we can easily show that any circuit (induced or otherwise) must contain at least three vertices and at least three edges. A circuit with one vertex would have to use a loop edge; a circuit with two vertices would have to reuse a single edge or have two edges between the same two vertices; etc.
Now, given any two distinct circuits (induced or otherwise), those two circuits cannot share all three vertices. If they did, they would either be the same circuit, or there would again be two different edges between the same pair of vertices. By essentially the same logic, any two distinct circuits cannot share more than one edge.
The result that distinct circuits share at most two vertices and at most one edge implies that for any graph G(V,E), there cannot be more than V-2 circuits, and there cannot be any more than (E-1)/2 circuits.
Now I can prove this upper bound is not only a bound but also a maximum by presenting a simple example that reaches it. Let G contain vertices A and B, with edge X from A to B. There are currently 0 cycles, 2 vertices and 1 edge. Now add a vertex C1, plus an edge Y1 from B to C1 and another edge Z1 from C1 to A. Now we have 1 cycle (which is obviously induced), 3 vertices and 3 edges. Similarly, add vertex C2 and edges Y2 and Z2 so that we have 2 induced cycles, 4 vertices and 5 edges. Continue the pattern, and by induction, when we add CN, YN and ZN, we'll have N induced cycles, 2+N vertices and 1+2N edges. That means N=V-2 and N=(E-1)/2, which was our theoretical upper bound.
Thus, the maximum number of induced circuits/cycles in a graph G(V,E) is the minimum of V-2 and (E-1)/2.