Current in an ac circuit is given by `I= 3 sin omega t + 4 cos omega t, ` then

Options

A. rms value of current is 5A.
B. mean value of this current in any one half period will be `(6//pi)`.
C. If voltage applied is `V=V_(m) sin omega t`, then the circuit may be containing resistance and capacitance only.
D. If voltage applied is `V=V_(m) sin omega t`, then the circuit may be contain resostamce and inductance only.

Explanation

To solve the problem, we need to analyze the given current equation in an AC circuit:

\[ I = 3 \sin(\omega t) + 4 \cos(\omega t) \]

We will find the peak current, RMS value, average current, and phase difference step by step.

### Step 1: Combine the Sine and Cosine Terms
We can express the current in a single trigonometric function using the following identity:

\[ I = R \sin(\omega t + \phi) \]

Where \( R \) is the resultant amplitude and \( \phi \) is the phase angle.

To find \( R \) and \( \phi \), we can use the following relationships:

- \( R = \sqrt{A^2 + B^2} \)
- \( \tan(\phi) = \frac{B}{A} \)

Here, \( A = 3 \) and \( B = 4 \).

### Step 2: Calculate the Resultant Amplitude \( R \)
Using the formula for \( R \):

\[ R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 3: Calculate the Phase Angle \( \phi \)
Using the tangent formula:

\[ \tan(\phi) = \frac{4}{3} \]

To find \( \phi \):

\[ \phi = \tan^{-1}\left(\frac{4}{3}\right) \]

### Step 4: Calculate the RMS Value \( I_{RMS} \)
The RMS value for an AC current is given by:

\[ I_{RMS} = \frac{I_0}{\sqrt{2}} \]

Where \( I_0 \) is the peak current (which we found to be \( R = 5 \)):

\[ I_{RMS} = \frac{5}{\sqrt{2}} \, \text{A} \]

### Step 5: Calculate the Average Current \( I_{avg} \)
The average current over one complete cycle is given by:

\[ I_{avg} = \frac{I_0}{\pi} \]

Substituting \( I_0 = 5 \):

\[ I_{avg} = \frac{5}{\pi} \, \text{A} \]

### Step 6: Determine the Phase Difference \( \phi \)
From our earlier calculation, we have:

\[ \phi = \tan^{-1}\left(\frac{4}{3}\right) \approx 53^\circ \]

### Summary of Results
- Peak Current \( I_0 = 5 \, \text{A} \)
- RMS Current \( I_{RMS} = \frac{5}{\sqrt{2}} \, \text{A} \)
- Average Current \( I_{avg} = \frac{5}{\pi} \, \text{A} \)
- Phase Difference \( \phi \approx 53^\circ \)