Question

In the circuit shown below, the AC source has voltage \(V=20\cos(\omega t)\,\text{V}\) with \(\omega=2000\,\text{rad/s}\). The amplitude of the current will be nearest to: 

• A. \(2\,\text{A}\)
• B. \(3.3\,\text{A}\)
• C. \(\dfrac{2}{\sqrt{5}}\,\text{A}\)
• D. \(\sqrt{5}\,\text{A}\)

Hint:

At resonance: \[ X_L=X_C \Rightarrow Z=R \] Current is maximum.

Answer 1

The Correct Option is A

Step 1: Given: \[ R=6+4=10\,\Omega,\quad L=5\,\text{mH},\quad C=50\,\mu\text{F} \]
Step 2: Reactances: \[ X_L=\omega L=2000\times5\times10^{-3}=10\,\Omega \] \[ X_C=\frac{1}{\omega C}=\frac{1}{2000\times50\times10^{-6}}=10\,\Omega \]
Step 3: Net reactance \(X_L-X_C=0\) (resonance).
Step 4: Impedance: \[ Z=R=10\,\Omega \]
Step 5: Current amplitude: \[ I_0=\frac{V_0}{Z}=\frac{20}{10}=2\,\text{A} \]