In the circuit shown in the figure `X_(L)=(X_(C))/(2))=R` the peck value current `i_(0)` is

Options

A. a.An inductor of 0.103H
B. b.An inductor of 0.25H
C. A resistance of `6.Omega`
D. A resistance of `100Omega`

Answer

To solve the problem step by step, we need to analyze the given circuit conditions and apply the relevant formulas for alternating current circuits.

### Step 1: Understand the given conditions
We are given that:
- \( X_L = \frac{X_C}{2} = R \)
From this, we can derive:
- \( X_L = R \)
- \( X_C = 2R \)

### Step 2: Calculate the impedance \( Z \)
The total impedance \( Z \) in an AC circuit with resistance \( R \), inductive reactance \( X_L \), and capacitive reactance \( X_C \) can be calculated using the formula:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Substituting the values we have:
\[ Z = \sqrt{R^2 + (R - 2R)^2} \]
\[ Z = \sqrt{R^2 + (-R)^2} \]
\[ Z = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2} \]

### Step 3: Calculate the peak current \( I_0 \)
The peak current \( I_0 \) can be calculated using Ohm's law for AC circuits:
\[ I_0 = \frac{V_0}{Z} \]
Substituting the expression for \( Z \):
\[ I_0 = \frac{V_0}{R\sqrt{2}} \]

### Step 4: Simplify the expression
To express \( I_0 \) in a more manageable form, we can rewrite it as:
\[ I_0 = \frac{V_0 \sqrt{2}}{2R} \]

### Conclusion
Thus, the peak value of the current \( I_0 \) is:
\[ I_0 = \frac{V_0 \sqrt{2}}{2R} \]