Question

Using an AC voltmeter the potential difference in the electrical line in a house is read to be \(234\,V\). If line frequency is known to be \(50\) cycles/s, the equation for the line voltage is

• A. \(V = 165\sin(100\pi t)\)
• B. \(V = 331\sin(100\pi t)\)
• C. \(V = 220\sin(100\pi t)\)
• D. \(V = 440\sin(100\pi t)\)

Hint:

AC voltmeter shows RMS value. Peak value is \(V_0=\sqrt{2}V_{rms}\). Voltage equation is \(V=V_0\sin(2\pi ft)\).

Answer 1

The Correct Option is B

Step 1: RMS and peak relation.
AC voltmeter reads RMS value:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} \] Step 2: Find peak voltage.
\[ V_0 = V_{rms}\sqrt{2} = 234\sqrt{2} \approx 331\,V \] Step 3: Write angular frequency.
\[ \omega = 2\pi f = 2\pi(50)=100\pi \] Step 4: Equation of voltage.
\[ V = V_0\sin(\omega t)=331\sin(100\pi t) \] Final Answer: \[ \boxed{V = 331\sin(100\pi t)} \]