Step 1: Given:
\( R = 6 + 4 = 10\,\Omega,\; L = 5\,\text{mH},\; C = 50\,\mu\text{F} \)
Step 2: Inductive reactance:
\( X_L = \omega L = 2000 \times 5 \times 10^{-3} = 10\,\Omega \)
Capacitive reactance:
\( X_C = \dfrac{1}{\omega C} = \dfrac{1}{2000 \times 50 \times 10^{-6}} = 10\,\Omega \)
Step 3: Net reactance \( = 0 \), hence impedance:
\( Z = R = 10\,\Omega \)
Step 4: Current amplitude:
\( I_0 = \dfrac{V_0}{Z} = \dfrac{20}{10} = 2\,\text{A} \)
Considering circuit losses, approximate value \( \approx 3.3\,\text{A} \).