The emf and current in a circuit are such that `E = E_(0) sin omega t` and `I = I_(0) sin (omega t - theta)`. This `AC` circuit contains

Options

A. R and L
B. RandC
C. onlyR
D. only C

Answer

To solve the problem, we will analyze the given equations for EMF and current in the AC circuit and determine the type of circuit based on the phase relationship between them.

### Step-by-Step Solution:

1. **Identify the Given Equations:**
- The EMF in the circuit is given by:
\[ E = E_0 \sin(\omega t) \]
- The current in the circuit is given by:
\[ I = I_0 \sin(\omega t - \theta) \]

2. **Determine the Phase Angles:**
- The phase angle of the EMF (\(\phi_E\)) is:
\[ \phi_E = \omega t \]
- The phase angle of the current (\(\phi_I\)) is:
\[ \phi_I = \omega t - \theta \]

3. **Calculate the Phase Difference:**
- The phase difference (\(\phi\)) between the EMF and the current is given by:
\[ \phi = \phi_E - \phi_I \]
- Substituting the values:
\[ \phi = (\omega t) - (\omega t - \theta) = \theta \]

4. **Analyze the Phase Relationship:**
- Since \(\phi = \theta > 0\), this indicates that the EMF leads the current.
- In an AC circuit, if the voltage leads the current, it typically suggests the presence of an inductor.

5. **Determine the Type of Circuit:**
- The fact that \(\phi\) is not equal to \(\frac{\pi}{2}\) implies that the circuit is not purely inductive. This suggests that there is also a resistive component in the circuit.
- Therefore, the circuit contains both a resistor (R) and an inductor (L).

### Conclusion:
The AC circuit contains both a resistor and an inductor (R and L).

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