Question

The peak value of an alternating emf $e$ given by $e = e_0 \cos \omega t$ is $10\text{ V}$ and its frequency is $50\text{ Hz}$. At time $t = \frac{1}{600}\text{ s}$, the instantaneous e.m.f is $\left[\cos \frac{\pi}{6} = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\right]$

• A. $10\text{ V}$
• B. $\frac{1}{\sqrt{3}}\text{ V}$
• C. $5\text{ V}$
• D. $5\sqrt{3}\text{ V}$

Hint:

When evaluating alternating current phase arguments with a frequency of $50\text{ Hz}$, the term $2\pi f$ becomes exactly $100\pi$. Multiplying this instantly by your given time component ($1/600$) simplifies directly to a standard reference angle fraction ($\pi/6$), saving valuable test time.

Answer 1

The Correct Option is D

Step 1: Write the emf equation.
The alternating emf is $e = e_0\cos\omega t$ with peak $e_0 = 10\,\text{V}$ and frequency $f = 50\,\text{Hz}$.
Step 2: Find the angular frequency.
\[ \omega = 2\pi f = 2\pi(50) = 100\pi\,\text{rad/s}. \]
Step 3: Compute the phase angle.
At $t = \dfrac{1}{600}\,\text{s}$, \[ \omega t = 100\pi\times\frac{1}{600} = \frac{100\pi}{600} = \frac{\pi}{6}. \]
Step 4: Substitute into the emf.
\[ e = 10\cos\frac{\pi}{6}. \]
Step 5: Use the given value.
Since $\cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}$, \[ e = 10\times\frac{\sqrt{3}}{2}. \]
Step 6: Compute.
\[ e = 5\sqrt{3}\,\text{V}. \] That is option (D). \[ \boxed{e = 5\sqrt{3}\,\text{V}} \]