In a given series LCR circuit `R=4Omega, X_(L)=5Omega` and `X_(C)=8Omega`, the current

Options

A. Leads the voltage by `tan^(-1)(3//4)`.
B. Leads the voltage by `tan^(-1)(5//8)`.
C. Lags the voltage by `tan^(-1)(3//4).`
D. Lags the voltage by `tan^(-1)(5//8)`.

Answer

To solve the problem regarding the phase relationship between current and voltage in a series LCR circuit, we will follow these steps:

### Step 1: Identify the given values
We have the following values from the problem:
- Resistance, \( R = 4 \, \Omega \)
- Inductive reactance, \( X_L = 5 \, \Omega \)
- Capacitive reactance, \( X_C = 8 \, \Omega \)

### Step 2: Calculate the net reactance
The net reactance \( X \) in a series LCR circuit is given by:
\[ X = X_C - X_L \]
Substituting the values:
\[ X = 8 \, \Omega - 5 \, \Omega = 3 \, \Omega \]

### Step 3: Calculate the phase angle \( \theta \)
The phase angle \( \theta \) can be calculated using the formula:
\[ \tan \theta = \frac{X}{R} \]
Substituting the values we found:
\[ \tan \theta = \frac{3 \, \Omega}{4 \, \Omega} \]

### Step 4: Find \( \theta \)
To find the angle \( \theta \), we take the arctangent:
\[ \theta = \tan^{-1}\left(\frac{3}{4}\right) \]

### Step 5: Determine the phase relationship
Since \( X \) is positive (3 Ω), this indicates that the circuit is capacitive, meaning that the current leads the voltage.

### Step 6: Conclusion
Thus, the current leads the voltage by an angle of \( \theta = \tan^{-1}\left(\frac{3}{4}\right) \).

### Final Answer
The current leads the voltage by an angle of \( \tan^{-1}\left(\frac{3}{4}\right) \).

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