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 step by step, we will analyze the current given in the AC circuit, calculate the RMS value, the mean value, and determine the type of circuit based on the phase difference between current and voltage.

### Step 1: Rewrite the Current Expression
The current in the AC circuit is given by:
\[ I = 3 \sin(\omega t) + 4 \cos(\omega t) \]

To simplify this expression, we can convert it into a single sine function using the formula:
\[ R \sin(\omega t + \phi) = R \sin(\omega t) \cos(\phi) + R \cos(\omega t) \sin(\phi) \]

Where:
- \( R = \sqrt{a^2 + b^2} \)
- \( a = 3 \) and \( b = 4 \)

### Step 2: Calculate the Amplitude \( R \)
Calculate \( R \):
\[ R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 3: Determine the Phase Angle \( \phi \)
Now, we need to find the phase angle \( \phi \):
\[ \tan(\phi) = \frac{b}{a} = \frac{4}{3} \]

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

### Step 4: Rewrite the Current
Now we can rewrite the current as:
\[ I = 5 \sin(\omega t + \phi) \]
Where \( \phi = \tan^{-1}\left(\frac{4}{3}\right) \).

### Step 5: Calculate the RMS Value
The RMS (Root Mean Square) value of the current is given by:
\[ I_{\text{RMS}} = \frac{I_0}{\sqrt{2}} \]
Where \( I_0 = 5 \):
\[ I_{\text{RMS}} = \frac{5}{\sqrt{2}} \, \text{A} \]

### Step 6: Calculate the Mean Value
The mean value of the current over one complete cycle is given by:
\[ I_{\text{avg}} = \frac{1}{T} \int_0^T I \, dt \]
For a sine wave:
\[ I_{\text{avg}} = \frac{I_0}{\pi} \]

Substituting \( I_0 = 5 \):
\[ I_{\text{avg}} = \frac{5}{\pi} \, \text{A} \]

### Step 7: Determine the Type of Circuit
Since the current leads the voltage by an angle \( \phi \) (which is positive), this indicates that the circuit is capacitive. In a capacitive circuit, the current leads the voltage.

### Summary of Results
- **RMS Value**: \( I_{\text{RMS}} = \frac{5}{\sqrt{2}} \, \text{A} \)
- **Mean Value**: \( I_{\text{avg}} = \frac{5}{\pi} \, \text{A} \)
- **Type of Circuit**: Capacitive circuit (current leads voltage).