An AC voltage source of variable angular freq | vedclass

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

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The bulb glows brighter

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(B) The impedance $Z$ of the $RC$ series circuit is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.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 rms current in the circuit is given by $I_{rms} = \frac{V_{rms}}{Z}$. As $Z$ decreases,$I_{rms}$ increases.The power dissipated by the bulb is $P = I_{rms}^2 R$. Since $I_{rms}$ increases,the power $P$ increases,which means 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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