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

Step 1: Understanding the Question:
This AC circuit question asks us to find the angular frequency \(\omega\) such that the current through resistor A is in phase with the source voltage.

Step 2: Key Formula or Approach:
We use AC nodal analysis with complex impedances:

• Resistor impedance: \(Z_R = R\)

• Inductor impedance: \(Z_L = i \omega L\)

• Capacitor impedance: \(Z_C = \frac{1}{i \omega C}\)

• For the current in resistor A to be in phase with the source, the ratio of the node voltage across A to the source voltage must be purely real.

Step 3: Detailed Explanation:

• Let the bottom-left node connected to the source be the reference node (0 V).

• The top-left node has potential \(V_1 = V\).

• Let the top-right node be \(V_2\), and the bottom-right node be \(V_3\).

• Resistor A is connected between \(V_1\) and \(V_2\). Resistor B is connected between \(V_1\) and \(V_3\).

• Applying Kirchhoff's Current Law (KCL) at node 2:
\[ \frac{V_2 - V_1}{R} + i\omega C(V_2 - V_3) = 0 \implies V_2(1 + i\omega RC) - i\omega RC V_3 = V_1 \quad \text{--- (Eq. 1)} \]
• Applying KCL at node 3:
\[ \frac{V_3 - V_1}{R} + i\omega C(V_3 - V_2) + \frac{V_3}{i\omega L} = 0 \implies V_3\left(1 + i\omega RC - i\frac{R}{\omega L}\right) - i\omega RC V_2 = V_1 \quad \text{--- (Eq. 2)} \]
• Solving Eq. 1 for \(V_3\).:
\[ i\omega RC V_3 = V_2(1 + i\omega RC) - V_1 \]
• Substituting this into Eq. 2 and organizing terms yields:
\[ V_2 \left[ \left(1 + \frac{R^2 C}{L}\right) + i\left(2\omega RC - \frac{R}{\omega L}\right) \right] = V_1 \left[ 1 + i\left(2\omega RC - \frac{R}{\omega L}\right) \right] \]
• For the voltage \(V_2\) (and thus the current through A) to be in phase with \(V_1\), the imaginary terms in the bracket must equal zero:
\[ 2\omega RC - \frac{R}{\omega L} = 0 \] \[ 2\omega^2 L C = 1 \implies \omega = \frac{1}{\sqrt{2LC}} \]

Step 4: Final Answer:
The value of \(\omega\) is \(\frac{1}{\sqrt{2LC}}\).