Ok, i managed with Robin Chapman's help compute the following:
Let $X_1,X_2,\ldots,X_N$ be the random variables defined as follows:
$$
X_i=
\begin{cases} 1 \quad\text{if there is a run of heads starting at the position } i,\\
0 \quad\text{otherwise}.
\end{cases}
$$
Then $X=X_1+X_2+...+X_N$. By linearity of expectation we have $E(X)=E(X_1)+\ldots+E(X_N)$.
Now, a run of heads can start at the position $i$ only by the condition "a tail at the position $i-1$". That means $P(X_i=1)=(1-p)p=qp$. But at $i=1$ there is no $i=0$, and hence $P(X_1=1)=p$.
Last step: $E(X_i)=1\cdot P(X_i=1)+0\cdot P(X_i=0)=P(X_i=1)$ and finally:
$$
E(X)=p+(N-1)pq
$$
In case of $p=1/2$ we have $E(X)=(N+1)/4$, and if $N=>\infty$ then $E(X)=>N/4$.
Congrats leonbloy! Your approach was very fruitful. Now, to get exact variance seems to be rather tricky. But in any case, many thanks anyway.