The brightness of the bulb depends on the power dissipated ($P = I^2 R$), which is determined by the current ($I$) flowing through the series circuit. The current in an RL series circuit is given by $I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + X_L^2}}$, where $X_L = \omega L = 2\pi f L$ is the inductive reactance .

To decrease the brightness, the current $I$ must decrease, which requires the impedance $Z$ (and thus $X_L$ or $L$) to increase.

1. Inserting an iron rod: The self-inductance of a solenoid is given by $L = \mu_r \mu_0 n^2 Al$. Inserting an iron rod (a ferromagnetic material with high relative permeability $\mu_r$) dramatically increases the self-inductance $L$ . This increases $X_L$ and the total impedance $Z$, thereby reducing the current and the bulb's brightness.

2. Analysis of other options: Reducing turns ($N$) decreases $L$ ($L \propto N^2$), decreasing impedance and increasing brightness. Adding capacitance where $X_C = X_L$ causes resonance ($Z = R$, minimum impedance), which maximizes current and brightness.

• Decreasing frequency ($f$) decreases $X_L$ ($X_L \propto f$), decreasing impedance and increasing brightness.