When an AC voltage, of variable frequency is applied to series L-C-R circuit, the current in the cirucit is the same at 4kHz. The current in the ciruit is maximum at (x)kHz. Find the value of x

Answer

To solve the problem step by step, we need to analyze the behavior of a series L-C-R circuit when an AC voltage of variable frequency is applied.

### Step 1: Understanding the Circuit Behavior
In a series L-C-R circuit, the current is dependent on the impedance, which varies with frequency. The impedance \( Z \) of the circuit is given by:
\[ Z = R + j(\omega L - \frac{1}{\omega C}) \]
where \( \omega = 2\pi f \) is the angular frequency, \( R \) is the resistance, \( L \) is the inductance, and \( C \) is the capacitance.

### Step 2: Condition for Same Current
The problem states that the current is the same at 4 kHz. This implies that the impedance at this frequency is equal to the impedance at some other frequency. Therefore, we can write:
\[ Z(4 \text{ kHz}) = Z(x \text{ kHz}) \]
This means that the difference between the inductive reactance and capacitive reactance must be the same at both frequencies.

### Step 3: Impedance Equality
At two different frequencies \( f_1 \) and \( f_2 \), the condition for the same current can be expressed as:
\[ \omega_1 L - \frac{1}{\omega_1 C} = \omega_2 L - \frac{1}{\omega_2 C} \]
Substituting \( \omega_1 = 2\pi \times 4000 \) and \( \omega_2 = 2\pi \times x \times 1000 \), we can simplify this equation.

### Step 4: Resonance Condition
The current in the circuit is maximum at resonance frequency, which occurs when:
\[ \omega L = \frac{1}{\omega C} \]
This can be rearranged to find the resonance frequency \( f_0 \):
\[ f_0 = \frac{1}{2\pi\sqrt{LC}} \]

### Step 5: Solving for Maximum Current Condition
Given that the current is the same at 4 kHz and another frequency \( x \) kHz, we can set up the equation:
\[ \frac{1}{\omega_1 C} - \omega_1 L = \frac{1}{\omega_2 C} - \omega_2 L \]
This leads us to a quadratic equation in terms of frequencies.

### Step 6: Finding the Value of x
From the problem, we know that the current is maximum at resonance. Since the current is the same at 4 kHz and another frequency, we can assume that the two frequencies are symmetric about the resonance frequency. Thus, if we denote the resonance frequency as \( f_0 \), we can express it as:
\[ f_0 = \frac{4 + x}{2} \]
Given that the current is maximum at resonance, we can set:
\[ f_0 = 6 \text{ kHz} \]
Thus, we can solve for \( x \):
\[ 6 = \frac{4 + x}{2} \]
Multiplying both sides by 2 gives:
\[ 12 = 4 + x \]
Subtracting 4 from both sides results in:
\[ x = 8 \text{ kHz} \]

### Final Answer
The value of \( x \) is 8 kHz.