Question

In an AC circuit, R = 2$\pi^2\Omega$, L = 0.02$\pi$H and is powered by an AC voltage of frequency 50 Hz. Find impedance of the circuit ?

Hint:

Whenever resistance equals reactance ($R = X$), the impedance forms an isosceles right triangle, so $Z$ will always be $R\sqrt{2}$ (or $X\sqrt{2}$). Also, note the phase angle would be exactly $45^\circ$ ($\pi/4$).

Answer 1

Step 1: Understanding the Concept:
The problem describes an RL series AC circuit. The total opposition to alternating current in such a circuit is called impedance (\(Z\)). Impedance is the vector sum of resistance (\(R\)) and inductive reactance (\(X_L\)).
Step 2: Key Formula or Approach:
1. Calculate the inductive reactance: \(X_L = 2\pi f L\)
2. Calculate the total impedance: \(Z = \sqrt{R^2 + X_L^2}\)
Where \(f\) is frequency and \(L\) is inductance.
Step 3: Detailed Explanation:
Given values from the problem text:
Resistance, \(R = 2\pi^2 \, \Omega\)
Inductance, \(L = 0.02\pi \text{ H}\)
Frequency, \(f = 50 \text{ Hz}\)
First, find the inductive reactance (\(X_L\)):
\[ X_L = 2\pi f L \]
\[ X_L = 2 \cdot \pi \cdot 50 \cdot (0.02\pi) \]
Multiply the numerical parts and \(\pi\) parts separately:
\[ X_L = (2 \cdot 50 \cdot 0.02) \cdot (\pi \cdot \pi) \]
\[ X_L = (100 \cdot 0.02) \cdot \pi^2 \]
\[ X_L = 2\pi^2 \, \Omega \]
Notice that the inductive reactance happens to be exactly equal to the resistance (\(X_L = R\)).
Now, calculate the impedance (\(Z\)):
\[ Z = \sqrt{R^2 + X_L^2} \]
Substitute \(R\) and \(X_L\):
\[ Z = \sqrt{(2\pi^2)^2 + (2\pi^2)^2} \]
\[ Z = \sqrt{2 \cdot (2\pi^2)^2} \]
Take the square root:
\[ Z = (2\pi^2) \cdot \sqrt{2} \]
\[ Z = 2\sqrt{2}\pi^2 \, \Omega \]
Step 4: Final Answer:
The impedance of the circuit is \(2\sqrt{2}\pi^2 \, \Omega\).