Question

In a series \(RL\) circuit shown in the figure, the current through the resistor lags the AC source voltage by \(45^{\circ}\). Let \(R = 100\pi\) \(\Omega\) and \(L = 2\) H.

The frequency of the source is ____ Hz. (rounded off to the nearest integer)

Hint:

The phase angle of a series RL circuit is given by tan(phi) = wL/R; a 45 degree lag means wL = R.

Answer 1

Correct Answer: 25

Step 1: Understand the phase relationship in a series RL circuit.
In a series \(RL\) circuit driven by a sinusoidal source, the resistor is a purely resistive element with no phase shift and the inductor is a purely reactive element. Because current is common to both elements in a series circuit, the phase angle by which the current lags the source voltage is entirely decided by the ratio of the inductive reactance to the resistance. This phase angle \(\phi\) is given by
\[ \tan\phi = \frac{X_L}{R} = \frac{\omega L}{R} \]
where \(\omega\) is the angular frequency of the source, \(L\) is the inductance, and \(R\) is the resistance.

Step 2: Write down the given information.
We are told the current lags the voltage by
\[ \phi = 45^{\circ} \]
and we are given
\[ R = 100\pi \ \Omega, \qquad L = 2 \text{ H} \]

Step 3: Apply the phase angle formula.
Since \(\tan 45^{\circ} = 1\), the condition becomes
\[ \frac{\omega L}{R} = 1 \]
This tells us that at this particular frequency the inductive reactance exactly equals the resistance, which is why the phase shift comes out to exactly \(45^{\circ}\), halfway between a pure resistor at \(0^{\circ}\) and a pure inductor at \(90^{\circ}\).

Step 4: Solve for the angular frequency.
\[ \omega = \frac{R}{L} = \frac{100\pi}{2} = 50\pi \text{ rad/s} \]

Step 5: Convert angular frequency to ordinary frequency.
Angular frequency and ordinary frequency are related by
\[ \omega = 2\pi f \]
so
\[ f = \frac{\omega}{2\pi} = \frac{50\pi}{2\pi} \]

Step 6: Simplify.
The factor \(\pi\) cancels out top and bottom, leaving
\[ f = \frac{50}{2} = 25 \text{ Hz} \]

Final Answer:
The frequency of the source is
\[ \boxed{25 \text{ Hz}} \]