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)
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The phase angle of a series RL circuit is given by tan(phi) = wL/R; a 45 degree lag means wL = R.
Step 1: Write the circuit impedance in complex form.
For a series $RL$ branch, the total impedance seen by the source is $Z = R + j\omega L$, a complex number with a real part from the resistor and an imaginary part from the inductor.
Step 2: Recall what the phase angle of the current means.
The source voltage is the reference, so its phase is $0^{\circ}$. The current is $I = V/Z$, so the phase of the current relative to the voltage is minus the phase angle of $Z$. Since the current lags the voltage by $45^{\circ}$, the impedance $Z$ itself must have a phase angle of $+45^{\circ}$.
Step 3: Write the phase angle of Z.
The angle of $Z=R+j\omega L$ is
\[ \angle Z = \tan^{-1}\!\left(\frac{\omega L}{R}\right) \]
Setting this equal to $45^{\circ}$ gives $\dfrac{\omega L}{R}=\tan 45^{\circ}=1$, so the reactive part equals the resistive part in size: $\omega L = R$.
Step 4: Substitute the numbers.
With $R=100\pi\ \Omega$ and $L=2$ H,
\[ \omega(2) = 100\pi \]
\[ \omega = 50\pi \text{ rad/s} \]
Step 5: Convert to frequency in Hz.
Using $f=\omega/(2\pi)$,
\[ f = \frac{50\pi}{2\pi} = 25 \text{ Hz} \]
\[ \boxed{25 \text{ Hz}} \]
