# Solution to Series LCR at Resonance Given: - Resonance current, \( I_ = .01 \) A (assuming "100A" is a typo, and meant to be **.01 A** or **10 mA**; if not, please clarify) - voltage, \( V = 2 \) mV \( = 2 \times 10^{-3} \) V - Frequency, \( f = 200 \) kHz \( = 2 \times 10^5 \) Hz - Inductance, \( L = 50 \) H We are to find: - (a) Resistance \( R \) - (b) Capacitance \( C \) --- ## Step 1: At Resonance For a series LCR circuit at resonance: \[ Z = R \quad \text{(since } X_L = X_C) \] \[ I_ = \frac{V}{R} \] --- ## Step 2: Calculate Resistance \( R \) \[ R = \frac{V}{I_} \] Plug in the values: \[ R = \frac{2 \times 10^{-3}}{.01} = .2~\Omega \] --- ## Step 3: Calculate Capacitance \( C \) At resonance: \[ \omega_ = 2\pi f = 2\pi \times 2 \times 10^5 = 4\pi \times 10^5~\text{rad/s} \] The resonance condition: \[ \omega_^2 = \frac{1}{LC} \implies C = \frac{1}{L \omega_^2} \] Plug in the values: \[ C = \frac{1}{50 \times (4\pi \times 10^5)^2} \] Calculate \( (4\pi \times 10^5)^2 \): \[ (4\pi \times 10^5)^2 = (4\pi)^2 \times (10^5)^2 = 16\pi^2 \times 10^{10} \] So, \[ C = \frac{1}{50 \times 16\pi^2 \times 10^{10}} \] \[ = \frac{1}{800\pi^2 \times 10^{10}} \] Now, \( \pi^2 \approx 9.87 \): \[ 800 \pi^2 \approx 800 \times 9.87 = 7896 \] \[ C \approx \frac{1}{7896 \times 10^{10}} \] \[ = \frac{1}{7.896 \times 10^{13}} \] \[ \approx 1.27 \times 10^{-14}~\text{F} \] \[ = 12.7~\text{fF (femtofarads)} \] --- ## Final Answers ### (a) Circuit Resistance \[ \boxed{R = .2~\Omega} \] ### (b) Circuit Capacitance \[ \boxed{C \approx 1.27 \times 10^{-14}~\text{F} = 12.7~\text{fF}} \] --- ## Summary Table | Quantity | Value | |--------------|-----------------------| | Resistance | \(.2~\Omega\) | | Capacitance | \(1.27 \times 10^{-14}~\text{F}\) | --- **Note:** If the current is indeed \(100~\text{A}\), the resistance would be \(R = 2 \times 10^{-5}~\Omega\), and the capacitance remains the same as above. Please clarify if that's the case.

# Solution to Series LCR at Resonance ## Given Data - Resonance current, \( I = 0.01 \) A (10 mA) - Voltage, \( V = 2 \) mV \( = 2 \times 10^{-3} \) V - Frequency, \( f = 200 \) kHz \( = 2 \times 10^5 \) Hz - Inductance, \( L = 50 \) H ## Objective - (a) Find the circuit resistance, \( R \) - (b) Find the circuit capacitance, \( C \) --- ## Step 1: At Resonance At resonance in a series LCR circuit: \[ Z = R \quad \text{(since } X_L = X_C) \] The current equation is: \[ I = \frac{V}{R} \] --- ## Step 2: Calculate Resistance \( R \) Using the formula for resistance: \[ R = \frac{V}{I} \] Substituting the values: \[ R = \frac{2 \times 10^{-3}}{0.01} = 0.2~\Omega \] --- ## Step 3: Calculate Capacitance \( C \) At resonance, the relation is: \[ \omega^2 = \frac{1}{LC} \] Where: \[ \omega = 2\pi f = 2\pi \times 2 \times 10^5 = 4\pi \times 10^5~\text{rad/s} \] Rearranging for \( C \): \[ C = \frac{1}{L \omega^2} \] Calculating \( \omega^2 \): \[ \omega^2 = (4\pi \times 10^5)^2 = 16\pi^2 \times 10^{10} \] Now substituting in: \[ C = \frac{1}{50 \times 16\pi^2 \times 10^{10}} \] Calculating \( 16\pi^2 \): \[ \pi^2 \approx 9.87 \quad \Rightarrow \quad 16\pi^2 \approx 158.0 \] Thus, \[ C = \frac{1}{50 \times 158 \times 10^{10}} = \frac{1}{7900 \times 10^{10}} = \frac{1}{7.9 \times 10^{13}} \] Calculating \( C \): \[ C \approx 1.27 \times 10^{-14}~\text{F} = 12.7~\text{fF (femtofarads)} \] --- ## Final Answers ### (a) Circuit Resistance \[ \boxed{R = 0.2~\Omega} \] ### (b) Circuit Capacitance \[ \boxed{C \approx 1.27 \times 10^{-14}~\text{F} = 12.7~\text{fF}} \] --- ## Summary Table | Quantity | Value | |--------------|-----------------------| | Resistance | \(0.2~\Omega\) | | Capacitance | \(1.27 \times 10^{-14}~\text{F}\) | --- **Note:** If the initial current value of \(100~\text{A}\) is confirmed, then the resistance calculation would be adjusted accordingly, but the capacitance would remain unchanged. Please verify the initial conditions if necessary.