Question

In the circuit shown below, the ac source has voltage \(V = 30cos(ωt)\) volt with \(ω = 2000\) rad/s. What will be the amplitude of the current?

• A. \(\sqrt{5}\) A
• B. \(\frac{3}{\sqrt{5}}\) A
• C. \(3\) A
• D. \(3.3\) A

Hint:

Find X_L and X_C. They turn out to be equal, so the impedance equals the total resistance of 10 ohm.

Answer 1

The Correct Option is C

Step 1: Read the circuit
The ac source is \(V=30\cos\omega t\) with \(\omega=2000\) rad/s. In series: a \(6\ \Omega\) resistor, a coil of \(5\) mH with \(4\ \Omega\) resistance, and a \(50\ \mu\)F capacitor.

Step 2: Total resistance
\(R=6+4=10\ \Omega\).

Step 3: Reactances
\[ X_L=\omega L=2000\times5\times10^{-3}=10\ \Omega \] \[ X_C=\frac{1}{\omega C}=\frac{1}{2000\times50\times10^{-6}}=10\ \Omega \] Since \(X_L=X_C\), the circuit is at resonance.

Step 4: Impedance and current
\[ Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{10^2+0}=10\ \Omega \] \[ I_0=\frac{V_0}{Z}=\frac{30}{10}=3\ \text{A} \]

Step 5: Check the options
\(\sqrt5\) A and \(\frac{3}{\sqrt5}\) A would need a nonzero net reactance. \(3.3\) A would need an impedance of about 9 ohm, which is smaller than the total resistance 10 ohm, and that is impossible. Only 3 A fits.

Final Answer:
The current amplitude is 3 A, option (C). \[ \boxed{3\ \text{A}} \]