Voltage and current is an AC circuit are given by `V=sin(100pit-(pi)/6)` and `I=4sin(100pit+(pi)/6)`

Options

A. Voltage leads the current by `30^(@)`
B. Current leads the voltage by `30^(@)`
C. Current leads the voltage by `60^(@)`
D. Voltage leads the current by `60^(@)`

Explanation

To solve the problem, we need to determine the phase difference between the voltage and current in the given AC circuit. Let's break down the steps:

### Step 1: Identify the phase angles
The voltage \( V \) and current \( I \) are given by:
- \( V = \sin(100\pi t - \frac{\pi}{6}) \)
- \( I = 4\sin(100\pi t + \frac{\pi}{6}) \)

From these equations, we can identify the phase angles:
- For voltage \( V \), the phase angle \( \phi_1 = -\frac{\pi}{6} \)
- For current \( I \), the phase angle \( \phi_2 = +\frac{\pi}{6} \)

### Step 2: Calculate the phase difference
The phase difference \( \Delta \phi \) is calculated using the formula:
\[ \Delta \phi = \phi_2 - \phi_1 \]
Substituting the values:
\[ \Delta \phi = \frac{\pi}{6} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{6} + \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

### Step 3: Determine the leading or lagging relationship
Since \( \Delta \phi = \frac{\pi}{3} \), we can interpret this as follows:
- If \( \Delta \phi > 0 \), then the current \( I \) leads the voltage \( V \).
- If \( \Delta \phi < 0 \), then the current \( I \) lags behind the voltage \( V \).

In this case, since \( \Delta \phi = \frac{\pi}{3} \) (which is positive), we conclude that:
- The current \( I \) leads the voltage \( V \) by \( \frac{\pi}{3} \) radians.

### Step 4: Convert the phase difference to degrees
To express the phase difference in degrees:
\[ \frac{\pi}{3} \text{ radians} = 60 \text{ degrees} \]

### Final Conclusion
The current leads the voltage by \( 60 \) degrees.

### Answer
The current \( I \) leads the voltage \( V \) by \( 60 \) degrees.

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