Question

A light bulb and an open coil inductor are connected in series across an ac source of variable frequency. How will the glow of the bulb be affected when the frequency of the source is decreased ? Justify your answer. Assume that other factors remain unchanged.

Hint:

An inductor inherently acts to aggressively block high frequencies while letting low frequencies pass easily; radically reducing the AC frequency always allows the electrical current to aggressively pass much more freely.
Ensure you always logically step through the chain reaction: Frequency $\rightarrow$ Reactance $\rightarrow$ Impedance $\rightarrow$ Current $\rightarrow$ Power $\rightarrow$ Glow.

Answer 1

Concept:
• The total current strictly flowing through a series RL (resistor-inductor) circuit is intricately controlled by the overarching impedance, which beautifully combines both fixed resistance and variable reactance.

• Inductive reactance is not a static constant; it is fundamentally a dynamic quantity directly and linearly proportional to the physical operating frequency of the applied alternating current.

• Manipulating the source frequency inevitably alters the reactance, which heavily manipulates the current, ultimately dictating the power entirely dissipated by the light bulb.

Step 1: Relate Frequency to Inductive Reactance
We strictly analyze the specific scenario where the AC voltage source's operating frequency $f$ is intentionally and manually dialed down.
The inductive reactance $X_L$ of a coil is mathematically defined precisely by the foundational AC equation:
\[ X_L = \omega L = 2\pi f L \]
This fundamental mathematical relation unequivocally shows that the inductive reactance is directly and strictly proportional to the applied source frequency ($X_L \propto f$).
Therefore, deliberately decreasing the source frequency actively causes the corresponding inductive reactance $X_L$ to cleanly and linearly decrease.

Step 2: Relate Reactance to Circuit Impedance
The total working impedance $Z$ for this series configuration is given exactly by the standard magnitude formula:
\[ Z = \sqrt{R_{bulb}^2 + X_L^2} \]
Since the physical resistance $R_{bulb}$ of the light bulb essentially remains solidly constant, a considerably lower inductive reactance $X_L$ logically leads directly to a markedly reduced total electrical impedance $Z$ for the entire series combination.

Step 3: Conclude the Effect on Bulb Brightness
According to Ohm's law strictly adapted for AC circuits, the RMS current strictly depends inversely on the impedance:
\[ I_{rms} = \frac{V_{rms}}{Z} \]
Driven by the assumed constant supply voltage, this newly lowered total impedance graciously permits a significantly higher operational RMS current $I_{rms}$ to successfully course freely through the circuit.
Since the current has vigorously increased, the thermal power seamlessly dissipated within the bulb's resistive tungsten filament ($P = I_{rms}^2 R$) aggressively surges higher.
Consequently, emitting much more thermal and radiant energy, the visual brightness and overall glow of the light bulb will undeniably and visibly increase.