Let's solve the problem step-by-step. --- ## **Given Data** - Resistance, \( R = 10 \Omega \) - Inductance, \( L = 159 \mu H = 159 \times 10^{-6} \, H \) - Supply voltage, \( V = 50 \, mV = 0.05 \, V \) - Frequency, \( f = 1 \, MHz = 1 \times 10^6 \, Hz \) The circuit is a **series RLC circuit**. --- ## **(a) Capacitance for Resonance** **Resonance in a series RLC circuit occurs when:** \[ \omega L = \frac{1}{\omega C} \] Where \( \omega = 2\pi f \) **Solving for \( C \):** \[ \omega^2 = \frac{1}{LC} \implies C = \frac{1}{\omega^2 L} \] **Calculate \( \omega \):** \[ \omega = 2\pi f = 2\pi \times 10^6 = 6.2832 \times 10^6 \, \text{rad/s} \] **Now substitute values:** \[ C = \frac{1}{(6.2832 \times 10^6)^2 \times 159 \times 10^{-6}} \] Calculate \( (6.2832 \times 10^6)^2 \): \[ (6.2832 \times 10^6)^2 = 39.478 \times 10^{12} \] \[ C = \frac{1}{39.478 \times 10^{12} \times 159 \times 10^{-6}} \] \[ = \frac{1}{39.478 \times 159 \times 10^{12} \times 10^{-6}} \] \[ = \frac{1}{6279 \times 10^{6}} \] \[ = \frac{1}{6.279 \times 10^9} \] \[ C \approx 1.593 \times 10^{-10} \, F = 159.3 \, pF \] **Final Answer (a):** \[ \boxed{C = 159 \, pF} \] --- ## **(b) Current at Resonance** At resonance, **impedance \( Z = R \)** (since reactances cancel): \[ I_{res} = \frac{V}{R} \] \[ I_{res} = \frac{0.05}{10} = 0.005 \, A = 5 \, mA \] **Final Answer (b):** \[ \boxed{I_{res} = 5 \, mA} \] --- ## **(c) Capacitance for Current Reduced to 10% of Resonance Value** Let \( I = 0.1 I_{res} \). The impedance at any value of \( C \) is: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] Where \[ X_L = \omega L,\quad X_C = \frac{1}{\omega C} \] At resonance, \( X_L = X_C \) and \( Z = R \). For the current to be \( 0.1 \) times the resonance current: \[ I = \frac{V}{Z} \implies Z = \frac{V}{I} = \frac{V}{0.1 I_{res}} = \frac{V}{0.1 \frac{V}{R}} = 10R \] So, \[ \sqrt{R^2 + (X_L - X_C)^2} = 10R \] \[ R^2 + (X_L - X_C)^2 = 100 R^2 \] \[ (X_L - X_C)^2 = 99 R^2 \] \[ |X_L - X_C| = \sqrt{99} R \approx 9.95 R \] \[ |X_L - X_C| \approx 99.5 \, \Omega \] Since \( X_L = 1000 \Omega \): \[ |1000 - X_C| = 99.5 \implies X_C = 1000 \pm 99.5 = 1100 \ \text{or} \ 900 \ \Omega \] So, for \( X_C = 1100 \Omega \) and \( X_C = 900 \Omega \): \[ X_C = \frac{1}{\omega C} \implies C = \frac{1}{\omega X_C} \] For \( X_C = 1100 \Omega \): \[ C_1 = \frac{1}{6.2832 \times 10^6 \times 1100} = \frac{1}{6.9115 \times 10^9} \approx 1.446 \times 10^{-10} \, F = 144.6 \, pF \] For \( X_C = 900 \Omega \): \[ C_2 = \frac{1}{6.2832 \times 10^6 \times 900} = \frac{1}{5.6549 \times 10^9} \approx 1.77 \times 10^{-10} \, F = 177 \, pF \] **Final Answer (c):** \[ \boxed{C = 144.6 \, pF \ \text{or} \ 177 \, pF} \] --- ## **Summary of Answers** **a) Capacitance at resonance:** \(\boxed{159 \, pF}\) **b) Current at resonance:** \(\boxed{5 \, mA}\) **c) Capacitance for current = 10% of resonance:** \(\boxed{144.6 \, pF \ \text{or} \ 177 \, pF}\)