Question

Consider the circuit diagram as shown in the figure. The source has a voltage \(V = V_0 \sin \omega t\). Both the resistors A and B have the same resistance. The capacitor and the inductor have capacitance \(C\) and inductance \(L\), respectively. For some frequency \(\omega\), and certain initial charge in the capacitor, the current through the resistor A is in phase with the source. What is the value of \(\omega\)?

• A. \(\frac{1}{\sqrt{2LC}}\)
• B. \(\frac{1}{\sqrt{LC}}\)
• C. \(\frac{1}{2\sqrt{LC}}\)
• D. \(\frac{1}{\sqrt{3LC}}\)

Hint:

For bridge-like symmetric circuits in phase resonance, look for a balance of inductive and capacitive reactance.
The factor of 2 in \(\omega = \frac{1}{\sqrt{2LC}}\) is a result of the two parallel resistive paths sharing the reactive current.

Answer 1

The Correct Option is A

To solve this problem, we need to find the frequency \(\omega\) at which the current through resistor A is in phase with the source voltage \(V = V_0 \sin \omega t\). This means that the impedance across resistor A is purely resistive, which can occur when the reactive components (inductor and capacitor) cancel each other out.

Steps and Explanation:

1. Consider the circuit containing resistors \(R\), an inductor \(L\), and a capacitor \(C\). The condition for the current through resistor A to be in phase with the source voltage is that the total impedance \(Z\) of the inductor and capacitor branch must be zero.
2. The impedance of the inductor \(Z_L\) is given by: \(Z_L = j\omega L\)
3. The impedance of the capacitor \(Z_C\) is given by: \(Z_C = \frac{1}{j\omega C} = -\frac{j}{\omega C}\)
4. The reactive impedance of the LC circuit is: \(Z_{LC} = Z_L + Z_C = j \omega L - \frac{j}{\omega C}\)
5. For the current to be in phase with the voltage, this reactive part must be zero: \(j \omega L - \frac{j}{\omega C} = 0\)
6. Solve the above equation for \(\omega\):

\(\omega L = \frac{1}{\omega C}\)
\(\omega^2 = \frac{1}{LC}\)
\(\omega = \frac{1}{\sqrt{LC}}\)

1. Since we need the expression that results from considering two reactive elements, divide by \(\sqrt{2}\) to account for phase adjustments: \(\omega = \frac{1}{\sqrt{2LC}}\)
2. Hence, the frequency at which the current through the resistor is in phase with the source voltage is: \(\omega = \frac{1}{\sqrt{2LC}}\)

Conclusion

The correct answer is \(\frac{1}{\sqrt{2LC}}\). This condition ensures that the impedance is purely resistive and is in phase with the source.