Let's analyze the problem step by step.
Given:
• An AC voltage with angular frequency ω and fixed amplitude V0V_0 is connected in series with a capacitance CC and an electric bulb of resistance RR, with inductance being zero.
• The question asks what happens when the angular frequency ω\omega is increased.
Step 1: Impedance of the circuit
For a series combination of resistance RR and capacitance CC, the total impedance ZZ of the circuit is given by:
Z=R2+(1ωC)2Z = \sqrt{R^2 + \left( \frac{1}{\omega C} \right)^2}
where:
• RR is the resistance of the bulb,
• 1ωC\frac{1}{\omega C} is the capacitive reactance,
• ω\omega is the angular frequency of the AC supply.
Step 2: Effect of increasing ω\omega
As the angular frequency ω\omega increases, the capacitive reactance 1ωC\frac{1}{\omega C} decreases because it is inversely proportional to ω\omega. Therefore, as ω\omega increases, the total impedance ZZ of the circuit decreases.
Z=R2+(1ωC)2Z = \sqrt{R^2 + \left( \frac{1}{\omega C} \right)^2}
Since 1ωC\frac{1}{\omega C} becomes smaller as ω\omega increases, the impedance ZZ decreases.
Step 3: Current in the circuit
The current II in the circuit is given by Ohm's law:
I=V0ZI = \frac{V_0}{Z}
Since the impedance ZZ decreases as ω\omega increases, the current II increases.
Step 4: Power delivered to the bulb
The power delivered to the bulb is given by:
P=I2RP = I^2 R
Since the current II increases as ω\omega increases, the power PP delivered to the bulb also increases. This makes the bulb glow brighter.
Final Answer:
When ω\omega is increased, the bulb glows brighter.
Thus, the correct answer is (B) The bulb glows brighter.
