# Problem Overview Given the circuit with: - Source voltage: \( v(t) = 60\cos(300t + 30^\circ) \) V - Inductor: \( L = 60\,\text{mH} \) - Resistor: \( R = 100\,\Omega \) - Capacitor: \( C = ? \) We are asked to **find the value of capacitance \( C \) such that the current \( i(t) \) is in phase with \( v(t) \)**. This means the **total impedance is purely real** (i.e., the imaginary part is zero). --- # Step 1: Write the Impedance Expression Total impedance in series: \[ Z = R + j\omega L + \frac{1}{j\omega C} \] \[ Z = R + j\omega L - j\frac{1}{\omega C} \] \[ Z = R + j\left(\omega L - \frac{1}{\omega C}\right) \] --- # Step 2: Condition for Zero Phase (Impedance Real) For \( i(t) \) and \( v(t) \) **to be in phase**: \[ \text{Im}(Z) = 0 \implies \omega L - \frac{1}{\omega C} = 0 \] \[ \omega L = \frac{1}{\omega C} \] \[ C = \frac{1}{\omega^2 L} \] --- # Step 3: Plug in the Values Given: - \( \omega = 300 \,\text{rad/s} \) - \( L = 60 \,\text{mH} = 60 \times 10^{-3} \,\text{H} \) \[ C = \frac{1}{(300)^2 \times 60 \times 10^{-3}} \] First, calculate \( (300)^2 \): \[ (300)^2 = 90,000 \] So, \[ C = \frac{1}{90,000 \times 60 \times 10^{-3}} \] \[ C = \frac{1}{90,000 \times 0.06} \] \[ C = \frac{1}{5,400} \] \[ C \approx 1.85 \times 10^{-4} \,\text{F} = 185\,\mu\text{F} \] --- # Step 4: Final Answer ## Required Capacitance \[ \boxed{ C = 185\,\mu\text{F} } \] --- ## Summary Table | Parameter | Value | |-----------|----------------| | \( R \) | \( 100\,\Omega \) | | \( L \) | \( 60\,\text{mH} \) | | \( C \) | \( 185\,\mu\text{F} \) (to make \( i(t) \) in phase with \( v(t) \)) | | \( \omega \) | \( 300\,\text{rad/s} \) | --- ## Circuit Diagram *(Not provided, but if drawing: source in series with R, L, and C)* **Alt text:** Series circuit with AC voltage source (60cos(300t+30°) V), 100 Ω resistor, 60 mH inductor, and unknown capacitor C. --- ## Additional Notes - With this \( C \), the circuit is **resonant** at the source frequency, so the impedance is purely resistive (\( Z = R \)). - The current amplitude is then \( I = \frac{V}{R} \), and current and voltage are in phase.