Question 31: AC fundamentals

Question

An AC voltage source of variable angular frequency $$\omega$$ and fixed amplitude $$\mathrm{V}_0$$ is connected in series with a capacitance $$\mathrm{C}$$ and an electric bulb of resistance $$\mathrm{R}$$ (inductance zero). When $$\omega$$ is decreased

Accepted Answer

Let's analyze the circuit step by step:
The capacitive reactance is given by the formula:
$$X_C = \frac{1}{\omega C}$$
As the angular frequency $$\omega$$ is decreased, the capacitive reactance $$X_C$$ increases. This is because they are inversely related.
The total impedance of the circuit is:
$$Z = \sqrt{R^2 + X_C^2} = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}$$
With a lower $$\omega$$, the term $$\left(\frac{1}{\omega C}\right)^2$$ increases, leading to a larger overall impedance.
The current in the circuit is:
$$I = \frac{V_0}{Z}$$
Since an increased impedance results in a lower current, the bulb receives less current.
The brightness of the electric bulb (a resistive load) depends on the power dissipated:
$$P = I^2R$$
With a decrease in current, the power decreases, causing the bulb to glow dimmer.
Thus, when $$\omega$$ is decreased, the bulb glows dimmer, corresponding to Option D.

Answer

the bulb switches off

Answer

total impedance of the circuit is unchanged

Answer

the bulb glows brighter

Answer

the bulb glows dimmer

AC fundamentals

An AC voltage source of variable angular frequency $\omega and fixed amplitude \mathrm{V}_0 is connected in series with a capacitance \mathrm{C} and an electric bulb of resistance \mathrm{R} (inductance zero). When \omega$ is decreased

Choose the correct answer: