Problem 159

When the slope of line $l$ does not exist, the equation of line $l$ is $x=1$,

By solving the system of equations $\begin{cases} x=1 \\ x^{2}+y^{2}=4 \end{cases}$, we get A(1, $-\sqrt{3}$) and B(1, $\sqrt{3}$),

In this case, $|AB|=2\sqrt{3}$, which holds true;

When the slope of line $l$ exists, let the equation of line $l$ be $y-1=k(x-1)$, which simplifies to $kx-y-k+1=0$,

The center of the circle $x^2+y^2=4$ is O(0, 0) with radius $r=2$,

The distance $d$ from the center of the circle to line $l$ is $d=\frac{|-k+1|}{\sqrt{k^2+1}}$,

Since $|AB|=2\sqrt{3}$,

By using the formula $d^2+(\frac{|AB|}{2})^2=r^2$, we get $(\frac{|-k+1|}{\sqrt{k^2+1}})^2+(\frac{2\sqrt{3}}{2})^2=4$,

Solving this gives $k=0$. Therefore, the equation of line $l$ is $y=1$.

Thus, the equation of line $l$ is either $x=1$ or $y=1$.

Hence, the answer is: $\boxed{x=1 \text{ or } y=1}$.

When the slope of line $l$ does not exist, the equation of line $l$ is $x=1$, which holds true; when the slope of line $l$ exists, let the equation of line $l$ be $kx-y-k+1=0$, by finding the center of the circle $x^2+y^2=4$ is O(0, 0) with radius $r=2$, and the distance $d$ from the center of the circle to line $l$ is $d=\frac{|-k+1|}{\sqrt{k^2+1}}$, by using the formula $d^2+(\frac{|AB|}{2})^2=r^2$, we can find the equation of line $l$.

This problem tests the method of finding the equation of a line, which is a medium-level question. When solving, it is important to carefully read the problem and correctly apply the formula for the distance from a point to a line and the properties of a circle.