A capacitor of capacity 1 μF, an inductance of 0.2 H, and a resistance of 800 Ω are connected in series. Is the circuit oscillatory? If yes, find the frequency of oscillation.

Steps to Solve

1. Identify given values

The given parameters are:

• Capacitance, $C = 1 ,\mu F = 1 \times 10^{-6} ,F$
• Inductance, $L = 0.2 ,H$
• Resistance, $R = 800 ,\Omega$

2. Check the condition for oscillation

To determine if the circuit will oscillate, we use the condition:

$$ \frac{R^2}{4L^2} < \frac{1}{LC} $$

We need to calculate $ \frac{R^2}{4L^2} $ and $ \frac{1}{LC} $.

3. Calculate $ \frac{R^2}{4L^2} $

First, substitute the values into the equation:

[ R^2 = (800)^2 = 640000 ]

Now calculate:

[ 4L^2 = 4 \times (0.2)^2 = 4 \times 0.04 = 0.16 ]

Then,

[ \frac{R^2}{4L^2} = \frac{640000}{0.16} = 4000000 ]

4. Calculate $ \frac{1}{LC} $

Now we calculate $LC$:

[ LC = 0.2 \times 1 \times 10^{-6} = 2 \times 10^{-7} ]

Now find:

[ \frac{1}{LC} = \frac{1}{2 \times 10^{-7}} = 5 \times 10^{6} ]

5. Compare the results

Now compare the two calculated results:

• $ \frac{R^2}{4L^2} = 4000000$
• $ \frac{1}{LC} = 5000000$

Since $4000000 < 5000000$, the condition for oscillation is satisfied.

6. Calculate the frequency of oscillation

The angular frequency of oscillation is given by:

$$ \omega = \frac{1}{\sqrt{LC}} $$

Substituting the values gives:

[ \omega = \frac{1}{\sqrt{0.2 \times 1 \times 10^{-6}}} = \frac{1}{\sqrt{2 \times 10^{-7}}} ]

7. Calculate the frequency in Hz

The frequency $f$ is related to angular frequency $\omega$ by:

$$ f = \frac{\omega}{2\pi} $$

Now we can substitute for $\omega$:

[ \omega = \frac{1}{\sqrt{2 \times 10^{-7}}} \implies f = \frac{1/\sqrt{2 \times 10^{-7}}}{2\pi} ]

Calculating this gives us the frequency of oscillation.