Here is a step-by-step solution to the problems visible in the provided image:

Problem 3:

Task: Calculate the period $T$, natural frequency $f_0$, and angular frequency $\omega_0$ of an LC circuit where $L = 38 \, \text{mH}$ and $C = 2.5 \, \mu\text{F}$.

Solution:

1. Formula for period: $T = 2\pi \sqrt{LC}$ Convert units: $L = 38 \times 10^{-3} \, \text{H}, \quad C = 2.5 \times 10^{-6} \, \text{F}$ Substitute: $T = 2\pi \sqrt{(38 \times 10^{-3})(2.5 \times 10^{-6})}$ $T = 2\pi \sqrt{9.5 \times 10^{-8}}$ $T \approx 2\pi \times 9.75 \times 10^{-4} \approx 6.13 \times 10^{-3} \, \text{s} = 6.13 \, \text{ms}$

2. Natural frequency $f_0$: $f_0 = \frac{1}{T} = \frac{1}{6.13 \times 10^{-3}} \approx 163 \, \text{Hz}$

3. Angular frequency $\omega_0$: $\omega_0 = 2\pi f_0 = \frac{2\pi}{T} \approx 2\pi \times 163 \approx 1024 \, \text{rad/s}$

Problem 4:

Task: Calculate the maximum current in the circuit where $U_{\text{max}} = 0.22 \, \text{kV}$, $L = 0.22 \, \text{H}$.

Solution:

The maximum current is given by: $I_{\text{max}} = \frac{U_{\text{max}}}{\sqrt{L/C}}$

Convert values: $U_{\text{max}} = 0.22 \, \text{kV} = 220 \, \text{V}, \quad L = 0.22 \, \text{H}, \quad C = 2.5 \times 10^{-6} \, \text{F}$

Substitute: $I_{\text{max}} = \frac{220}{\sqrt{0.22 / (2.5 \times 10^{-6})}}$ $I_{\text{max}} = \frac{220}{\sqrt{88 \times 10^{3}}}$ $I_{\text{max}} = \frac{220}{296} \approx 0.743 \, \text{A}$

Problem 5:

Task: Calculate the effective values of current and voltage for a pendulum making 90 oscillations in 2 minutes.

Solution:

• Frequency of oscillations: $f = \frac{\text{Number of oscillations}}{\text{Time}} = \frac{90}{120} = 0.75 \, \text{Hz}$

• Since the RMS (effective) current and voltage depend on the characteristics of the system, further specifics would be required.

Problem 6:

Task: Determine the inductance $L$ needed to achieve an oscillation frequency $f = 3 \, \text{kHz}$ with a capacitor $C = 12 \, \mu\text{F}$.

Solution:

1. Frequency formula: $f = \frac{1}{2\pi \sqrt{LC}}$

2. Rearrange for $L$: $L = \frac{1}{(2\pi f)^2 C}$

3. Convert values: $f = 3 \, \text{kHz} = 3000 \, \text{Hz}, \quad C = 12 \, \mu\text{F} = 12 \times 10^{-6} \, \text{F}$

4. Substitute: $L = \frac{1}{(2\pi \times 3000)^2 \times 12 \times 10^{-6}}$ $L = \frac{1}{(6.28 \times 3000)^2 \times 12 \times 10^{-6}}$ $L = \frac{1}{142.1 \times 10^6 \times 12 \times 10^{-6}}$ $L \approx \frac{1}{1.705 \, \text{H}^{-1}} \approx 0.587 \, \text{H}$

Summary:

• Problem 3: $T \approx 6.13 \, \text{ms}, f_0 \approx 163 \, \text{Hz}, \omega_0 \approx 1024 \, \text{rad/s}$
• Problem 4: $I_{\text{max}} \approx 0.743 \, \text{A}$
• Problem 6: $L \approx 0.587 \, \text{H}$

Would you like more detailed steps for any part or further assistance?

Follow-up Questions:

1. How does the inductance value affect the frequency of the LC circuit?
2. What happens if the capacitor in Problem 3 is increased?
3. How can we interpret the RMS values of current and voltage in Problem 5?
4. Why does $I_{\text{max}}$ in Problem 4 depend on $\sqrt{L/C}$?
5. How would decreasing $f$ in Problem 6 affect $L$?

Tip: When solving LC circuit problems, always ensure the units are consistent before applying formulas.