To determine why the brightness of the bulb decreases, we need to analyze how different changes in the circuit impact the current flowing through the bulb and thus its brightness.
When a coil (inductor) is connected in series with a bulb and an AC source, the impedance of the circuit affects the current flow. Impedance \( Z \) in an AC circuit with inductance \( L \) is given by:
Z = \sqrt{R^2 + (X_L - X_C)^2}
Where:
- X_L = 2\pi f L is the inductive reactance.
- X_C = \dfrac{1}{2\pi f C} is the capacitive reactance (if a capacitor is present).
- R is the resistance of the bulb.
- f is the frequency of the AC source.
Let's evaluate each option:
- Frequency of the AC source is decreased:
- Decreasing the frequency \( f \) reduces the inductive reactance \( X_L \). This generally decreases the impedance \( Z \), allowing more current to flow, which would increase, not decrease, the brightness.
- Number of turns in the coil is reduced:
- Reducing the number of turns in the coil reduces the self-inductance \( L \). This reduces the inductive reactance \( X_L \), similarly decreasing impedance and increasing brightness.
- A capacitance of reactance \( X_c = X_L \), is included in the same circuit:
- This introduces resonance, minimizing the total impedance. Current and brightness would increase.
- An iron rod is inserted in the coil:
- Inserting an iron rod in the coil increases the self-inductance \( L \), which in turn increases the inductive reactance \( X_L \). This increases the impedance \( Z \), reducing the current through the bulb and thereby decreasing its brightness. Hence, this is the correct option.
Therefore, the brightness of the bulb decreases when an iron rod is inserted in the coil.