Let's break down the circuit and the given values: **Given:** - Amplitude of AC generator: \( \varepsilon_{max} = 35\,V \) - Inductance: \( L = 30\,mH = 0.03\,H \) - Capacitance: \( C = 9\,\mu F = 9 \times 10^{-6}\,F \) - Voltage across inductor: \( V_{L, max} = 11\,V \) - Voltage across capacitor: \( V_{C, max} = 6\,V \) - Need to find the resistance \( R \). ### Step 1: Find the Current Amplitude For an RLC circuit, the voltage across each element may be found using: \[ V_L = I_{max} \cdot \omega L \] \[ V_C = I_{max} \cdot \frac{1}{\omega C} \] Let \( I_{max} \) be the amplitude of the current. From the two equations, \[ I_{max} = \frac{V_{L,max}}{\omega L} = \omega C V_{C,max} \] Equating the two, \[ \frac{V_{L,max}}{\omega L} = \omega C V_{C,max} \] \[ V_{L,max} = \omega^2 L C V_{C,max} \] \[ \omega^2 = \frac{V_{L,max}}{L C V_{C,max}} \] Plug in the values: \[ \omega^2 = \frac{11}{0.03 \times 9 \times 10^{-6} \times 6} \] \[ = \frac{11}{0.03 \times 54 \times 10^{-6}} \] \[ = \frac{11}{1.62 \times 10^{-6}} \] \[ = 6.79 \times 10^6 \] \[ \omega = \sqrt{6.79 \times 10^6} \approx 2.61 \times 10^3 \text{ rad/s} \] ### Step 2: Find the Current Amplitude \[ I_{max} = \frac{V_{L,max}}{\omega L} = \frac{11}{2.61 \times 10^3 \times 0.03} \] \[ = \frac{11}{78.3} \] \[ \approx 0.140\,A \] ### Step 3: Find the Resistance \( R \) The total voltage amplitude: \[ \varepsilon_{max}^2 = (I_{max} R)^2 + (V_{L,max} - V_{C,max})^2 \] \[ 35^2 = (0.140 R)^2 + (11 - 6)^2 \] \[ 1225 = (0.140 R)^2 + 25 \] \[ 1200 = (0.140 R)^2 \] \[ 0.140 R = \sqrt{1200} \] \[ 0.140 R = 34.64 \] \[ R = \frac{34.64}{0.140} \] \[ R \approx 247.4\,\Omega \] --- ## **Final Answer:** \[ \boxed{R \approx 247 \,\Omega} \] Let me know if you'd like to see the answer formatted in LaTeX!