An AC voltage source of variable angular freq | vedclass

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(B) The impedance $Z$ of an $RC$ series circuit is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.Subs

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the bulb glows brighter

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(B) The impedance $Z$ of an $RC$ series circuit is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.Substituting $X_C$,we get $Z = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}$.As the angular frequency $\omega$ increases,the capacitive reactance $X_C = \frac{1}{\omega C}$ decreases.Since $Z = \sqrt{R^2 + X_C^2}$,a decrease in $X_C$ leads to a decrease in the total impedance $Z$ of the circuit.The current in the circuit is given by $I = \frac{V_0}{Z}$. Since $V_0$ is constant and $Z$ decreases,the current $I$ increases.The brightness of the bulb is proportional to the power dissipated,$P = I^2 R$. As the current $I$ increases,the power dissipated increases,and the bulb glows brighter.

An $AC$ voltage source of variable angular frequency $\omega$ and fixed amplitude $V_0$ is connected in series with a capacitance $C$ and an electric bulb of resistance $R$ (inductance zero). When $\omega$ is increased,

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