Question

Two identical incandescent light bulbs are connected as shown in the figure. When the circuit is an AC voltage source of frequency \(f\), which of the following observations will be correct?

• A. both bulbs will glow alternatively
• B. both bulbs will glow with same brightness provided frequency \(f = \frac{1}{2\pi\sqrt{LC}}\)
• C. bulb \(b_1\) will light up initially and goes off, bulb \(b_2\) will be ON constantly
• D. bulb \(b_1\) will blink and bulb \(b_2\) will be ON constantly

Hint:

In parallel RL and RC branches, equal current occurs at resonance condition \(\omega L = \frac{1}{\omega C}\Rightarrow f=\frac{1}{2\pi\sqrt{LC}}\).

Answer 1

The Correct Option is B

Step 1: Identify circuit branches.
The figure shows two parallel branches connected to AC source:
Branch 1 contains \(R\) and \(C\) with bulb \(b_1\).
Branch 2 contains \(R\) and \(L\) with bulb \(b_2\).
Step 2: Condition for equal brightness.
Brightness depends on current through each bulb.
Currents will be equal when impedances of two branches are equal.
Step 3: Impedance of RC and RL circuits.
\[ Z_{RC} = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} \] \[ Z_{RL} = \sqrt{R^2 + (\omega L)^2} \]
Step 4: Set the reactive parts equal for equality.
\[ \omega L = \frac{1}{\omega C} \Rightarrow \omega^2 = \frac{1}{LC} \]
Step 5: Convert \(\omega\) to frequency.
\[ \omega = 2\pi f \Rightarrow 2\pi f = \frac{1}{\sqrt{LC}} \Rightarrow f = \frac{1}{2\pi\sqrt{LC}} \]
Step 6: Conclusion.
At this frequency, both branches carry equal current, so both bulbs glow with same brightness.
Final Answer:
\[ \boxed{\text{(B) both bulbs glow equally when } f=\frac{1}{2\pi\sqrt{LC}}} \]