Let's solve each part step by step as requested. --- ## (a) What is the **frequency** of the oscillations? ### **Step 1: Write the Formula** The angular frequency \( \omega \) of an LC circuit is: \[ \omega = \frac{1}{\sqrt{LC}} \] The frequency \( f \) is: \[ f = \frac{\omega}{2\pi} = \frac{1}{2\pi\sqrt{LC}} \] ### **Step 2: Substitute Values and Solve** Given \( L = 3.\,\text{mH} = 3. \times 10^{-3}\,\text{H} \) and \( C = 250\,\text{nF} = 250 \times 10^{-9}\,\text{F} \): \[ f = \frac{1}{2\pi \sqrt{3. \times 10^{-3} \times 250 \times 10^{-9}}} \] Calculate the denominator: \[ \sqrt{3. \times 10^{-3} \times 250 \times 10^{-9}} = \sqrt{750 \times 10^{-12}} = \sqrt{7.5 \times 10^{-10}} = 2.7386 \times 10^{-5} \] So, \[ f = \frac{1}{2\pi \times 2.7386 \times 10^{-5}} = \frac{1}{1.7203 \times 10^{-4}} \approx 5,812\,\text{Hz} \] **Explanation:** We used the LC circuit frequency formula and substituted the given values to find the oscillation frequency. --- ### **Final Answer (a):** \[ \boxed{f \approx 5.8\ \text{kHz}} \] --- If you want to continue with part (b), let me know!

Let's solve each part of the question step by step. --- ## (a) What is the **frequency** of the oscillations? ### **Step 1: Write the Formula** The angular frequency \( \omega \) of an LC circuit is given by: \[ \omega = \frac{1}{\sqrt{LC}} \] The frequency \( f \) can be expressed as: \[ f = \frac{\omega}{2\pi} = \frac{1}{2\pi\sqrt{LC}} \] ### **Step 2: Substitute Values and Solve** Given: - \( L = 3\,\text{mH} = 3 \times 10^{-3}\,\text{H} \) - \( C = 250\,\text{nF} = 250 \times 10^{-9}\,\text{F} \) Substituting into the formula: \[ f = \frac{1}{2\pi \sqrt{3 \times 10^{-3} \times 250 \times 10^{-9}}} \] Calculating the denominator: \[ \sqrt{3 \times 10^{-3} \times 250 \times 10^{-9}} = \sqrt{750 \times 10^{-12}} = \sqrt{7.5 \times 10^{-10}} \approx 2.7386 \times 10^{-5} \] Thus, \[ f = \frac{1}{2\pi \times 2.7386 \times 10^{-5}} \approx \frac{1}{1.7203 \times 10^{-4}} \approx 5,812\,\text{Hz} \] **Explanation:** We used the frequency formula for an LC circuit and substituted the values for the inductance and capacitance to find the oscillation frequency. --- ### **Final Answer (a):** \[ \boxed{f \approx 5.8\ \text{kHz}} \] --- ## (b) What is the **maximum current** in the inductor? ### **Step 1: Write the Formula for Maximum Current** The maximum current \( I_{\text{max}} \) in the inductor is related to the maximum charge \( Q_{\text{max}} \) on the capacitor by: \[ I_{\text{max}} = \frac{Q_{\text{max}}}{T/4} \] Where \( T \) is the period of oscillation: \[ T = \frac{1}{f} \] ### **Step 2: Calculate Maximum Current** First, find the period \( T \): \[ T = \frac{1}{f} \approx \frac{1}{5812} \approx 0.000172 \text{ s} \approx 172 \, \mu s \] Thus, \( T/4 \approx 43\,\mu s \). Now substitute the maximum charge \( Q_{\text{max}} = 500 \, \text{nC} = 500 \times 10^{-9} \, \text{C} \): \[ I_{\text{max}} = \frac{500 \times 10^{-9}}{43 \times 10^{-6}} \approx 11.63 \, \text{mA} \] **Explanation:** We calculated the period of oscillation and used the maximum charge to find the maximum current using the relationship between charge and time. --- ### **Final Answer (b):** \[ \boxed{I_{\text{max}} \approx 11.63 \, \text{mA}} \] --- ## (c) How much **time** does it take to attain that value? ### **Step 1: Write the Formula** The time to reach maximum current occurs at \( t = \frac{T}{4} \): \[ t = \frac{1}{4f} \] ### **Step 2: Calculate Time** Using the frequency from part (a): \[ t = \frac{1}{4 \times 5812} \approx \frac{1}{23248} \approx 43 \, \mu s \] **Explanation:** The maximum current is achieved at a quarter of the period of the oscillation, which we calculated from the frequency. --- ### **Final Answer (c):** \[ \boxed{t \approx 43 \, \mu s} \] --- ## (d) What is the **minimum resistor** that can be added in series so that it does not oscillate but decays exponentially? ### **Step 1: Write the Formula for Damping** For critical damping in an RLC circuit, the resistance \( R \) is given by: \[ R_c = 2\sqrt{\frac{L}{C}} \] ### **Step 2: Calculate Minimum Resistance** Substituting values: \[ R_c = 2\sqrt{\frac{3 \times 10^{-3}}{250 \times 10^{-9}}} \] Calculating: \[ R_c = 2\sqrt{12000} \approx 2 \times 109.54 \approx 219.08\, \Omega \] **Explanation:** We derived the critical resistance required for the circuit to be critically damped, preventing oscillation. --- ### **Final Answer (d):** \[ \boxed{R_c \approx 219.08 \, \Omega} \]