A voltage at a frequency of 500kHz is maintained across a coil in series with a capacitor. When the capacitor is set to 455pF, the current has its maximum value, but the capacitance is reduced to 422pF, the current is 0,707 of its maximum value. Find the:

Question

1) inductance of the coil
2) inductive reactive
3) capacitive reactance at 424pF
4) resistance of the coil

Answer 1

To solve these questions involving a resonant circuit consisting of an inductor and a capacitor, we will follow these steps:

Given:

• Frequency, \( f = 500 , \text{kHz} = 500 \times 10^3 , \text{Hz} \)
• Maximum capacitance, \( C_1 = 455 , \text{pF} = 455 \times 10^{-12} , \text{F} \)
• Capacitance when current is 0.707 of maximum, \( C_2 = 422 , \text{pF} = 422 \times 10^{-12} , \text{F} \)

1. Find the Inductance of the Coil

At resonance, the inductive reactance \( X_L \) equals the capacitive reactance \( X_C \).

The capacitive reactance \( X_C \) is given by: \[ X_C = \frac{1}{2\pi f C} \]

Calculating \( X_C \) at \( C_1 \): \[ X_{C1} = \frac{1}{2\pi \times 500 \times 10^3 \times 455 \times 10^{-12}} \] \[ = \frac{1}{2\pi \times 500000 \times 455 \times 10^{-12}} \] \[ \approx \frac{1}{1.4307 \times 10^{-6}} \approx 698.56 , \Omega \]

At resonance, \( X_L = X_{C1} \), thus \[ X_L = 698.56 , \Omega \]

Now, the inductive reactance \( X_L \) is given by: \[ X_L = 2\pi f L \] Setting \( X_L = 698.56 \): \[ 698.56 = 2\pi \times 500000 \times L \] \[ L = \frac{698.56}{2\pi \times 500000} \] \[ \approx \frac{698.56}{3141592.65} \approx 0.0002224 , H \approx 222.4 , \mu H \]

2. Find the Inductive Reactance

Using the inductance found: \[ X_L = 2\pi f L = 2\pi \times 500 \times 10^3 \times 222.4 \times 10^{-6} \] \[ X_L = 698.56 , \Omega \]

Thus, \[ X_L \approx 698.56 , \Omega \]

3. Find the Capacitive Reactance at 424pF

Calculating \( X_C \) at \( C_3 = 424 , \text{pF} \): \[ X_{C3} = \frac{1}{2\pi f C_3} = \frac{1}{2\pi \times 500 \times 10^3 \times 424 \times 10^{-12}} \] \[ = \frac{1}{2\pi \times 500000 \times 424 \times 10^{-12}} \approx \frac{1}{1.3327 \times 10^{-6}} \approx 751.4 , \Omega \]

4. Find the Resistance of the Coil

It's given that when the capacitance is reduced to \( C_2 \), the current is \( 0.707 \) of its maximum value.

Using the formula for current in a series RLC circuit: \[ I = \frac{V}{\sqrt{R^2 + (X_L - X_C)^2}} \] At maximum current \( C_1 \): \[ I_{max} = \frac{V}{R} \] At \( C_2 \) when current is \( 0.707 I_{max} \): \[ 0.707 I_{max} = \frac{V}{\sqrt{R^2 + (X_L - X_{C2})^2}} \] We can set the two equations in terms of \( I_{max} \): \[ 0.707 \frac{V}{R} = \frac{V}{\sqrt{R^2 + (X_L - X_{C2})^2}} \] Cancelling \( V \) and squaring both sides: \[ 0.707^2 R^2 = R^2 + (X_L - X_{C2})^2 \] \[ 0.499849 R^2 = R^2 + (X_L - X_{C2})^2 \] \[ 0.499849 R^2 - R^2 = (X_L - X_{C2})^2 \] \[ -0.500151 R^2 = (X_L - X_{C2})^2 \] Now substituting the values of \( X_L \) and \( X_{C2} \): \[ X_{C2} = 751.4 ,\Omega \] \[ X_L - X_{C2} = 698.56 - 751.4 = -52.84 ,\Omega \] Thus: \[ -0.500151 R^2 = (-52.84)^2 \] Calculating: \[ -0.500151 R^2 = 2798.1456 \rightarrow R^2 = \frac{2798.1456}{0.500151} \approx 5583.28 \] \[ R \approx \sqrt{5583.28} \approx 74.7 , \Omega \]

Summary of Results

1. Inductance of the coil: \( L \approx 222.4 , \mu H \)
2. Inductive reactance at resonance frequency: \( X_L \approx 698.56 , \Omega \)
3. Capacitive reactance at 424pF: \( X_{C3} \approx 751.4 , \Omega \)
4. Resistance of the coil: \( R \approx 74.7 , \Omega \)