#### Answer

The least possible value of $t$ is $\frac{1}{1440}$ seconds.

#### Work Step by Step

$$i=I_{max}\sin(2\pi ft)$$
For $i=50,I_{max}=100,f=120$, we have
$$50=100\sin(2\pi\times120t)$$
$$50=100\sin(240\pi t)$$
$$\sin(240\pi t)=\frac{1}{2}$$
$t$ refers to time, so as a rule, $t\in[0,\infty)$
Also $t$ is minimum when $240\pi t$ is minimum, as $240\pi$ is a constant.
Therefore, we would be able to find the least of $t$ as we find the first value of $240\pi t$ that would have $\sin(240\pi t)=\frac{1}{2}$ over the interval $[0,2\pi)$.
Such a value of $240\pi t$ can be found using the inverse function for sine:
$$240\pi t=\sin^{-1}\frac{1}{2}$$
$$240\pi t=\frac{\pi}{6}$$
$$t=\frac{\pi}{6\times240\pi}$$
$$t=\frac{1}{1440}(seconds)$$
Therefore, the least value of $t$ satisfying the given data is $\frac{1}{1440}$ seconds.