Step 1: Understanding the Concept:
Resonance in a series \(LCR\) circuit is a special electrical state that occurs when the circuit is driven at a frequency where the inductive reactance (\(X_L\)) and the capacitive reactance (\(X_C\)) are equal in magnitude.
Inductive reactance increases with frequency (\(X_L = \omega L\)), while capacitive reactance decreases as frequency increases (\(X_C = 1/\omega C\)).
At the resonant frequency, these two reactive components perfectly cancel each other out because their voltage phases are \(180^{\circ}\) apart.
As a result, the total impedance (\(Z\)) of the circuit is reduced to its absolute minimum value, which is simply equal to the resistance \(R\).
At this point, the current amplitude in the circuit reaches its peak value.
The frequency at which this happens is called the resonant frequency, and it depends solely on the values of the inductor (\(L\)) and the capacitor (\(C\)).
Step 2: Key Formula or Approach:
To find the resonant angular frequency \(\omega_r\), we set the reactances equal:
\[ X_L = X_C \implies \omega_r L = \frac{1}{\omega_r C} \]
Rearranging this equation to solve for \(\omega_r\):
\[ \omega_r^2 = \frac{1}{LC} \]
\[ \omega_r = \frac{1}{\sqrt{LC}} \]
Step 3: Detailed Explanation:
First, we must ensure all given metrics are converted into standard SI units:
- Inductance \(L = 2\) H (already in standard units).
- Capacitance \(C = 32 \mu F = 32 \times 10^{-6}\) F.
- Resistance \(R = 10 \Omega\) (Standard units, though not required for the frequency calculation).
Now, let's calculate the product of \(L\) and \(C\) inside the square root:
\[ LC = 2 \times (32 \times 10^{-6}) = 64 \times 10^{-6} \text{ s}^2 \]
Next, calculate the square root of this value:
\[ \sqrt{LC} = \sqrt{64 \times 10^{-6}} = \sqrt{64} \times \sqrt{10^{-6}} \]
\[ \sqrt{LC} = 8 \times 10^{-3} \text{ s} \]
Finally, determine the resonant angular frequency by taking the reciprocal:
\[ \omega_r = \frac{1}{8 \times 10^{-3}} = \frac{1000}{8} \]
\[ \omega_r = 125 \text{ rad/s} \]
The angular frequency is \(125\) radians per second. If the question had asked for the cyclic frequency \(f\) in Hertz, we would have divided this result by \(2\pi\).
Note that the resistor value of \(10\Omega\) only affects the "sharpness" (Quality Factor) and the height of the current peak at resonance, but it does not shift the resonant frequency itself.
Step 4: Final Answer:
The resonant angular frequency of the given \(LCR\) circuit is \(125\) rad/s.