1. The given alternating current is $I = 5\sin(120\pi t)\,\text{A}$

2. Compare with the standard form of AC current: $I = I_0 \sin(\omega t)$ So, $I_0 = 5\,\text{A}, \qquad \omega = 120\pi\,\text{rad/s}$

3. The current starts from zero when $t=0$ because $I=5\sin(0)=0$

4. The current reaches its peak value when $\sin(\omega t)=1$ This happens first at $\omega t = \frac{\pi}{2}$

5. Substitute $\omega = 120\pi$: $120\pi t = \frac{\pi}{2}$

6. Solve for $t$: $t = \frac{\pi/2}{120\pi} = \frac{1}{240}\,\text{s}$

7. Therefore, the time taken to reach peak current from zero is $\boxed{\frac{1}{240}\,\text{s}}$

8. Checking options:

   • A: $\frac{1}{60}$ s ❌
   • B: $60$ s ❌
   • C: $\frac{1}{120}$ s ❌
   • D: $\frac{1}{240}$ s ✅

So the correct option is D.