In the circuit given below, the frequency of the input voltage \(V_{IN}\) is \(\omega = 10^4\) rad/s. The output voltage \(V_{AB}\) leads \(V_{IN}\) by
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For bridge circuits, first calculate the impedances of all components at the given frequency. Then use the voltage divider rule for each arm of the bridge to find the voltages at the output terminals. The output voltage is the difference between these two voltages. The complex number \(j\) always represents a \(+90^\circ\) phase lead, while \(-j\) represents a \(-90^\circ\) phase lag.
Step 1: Understanding the Concept:
The given circuit is a bridge circuit, specifically a Maxwell bridge (or a similar AC bridge). We need to find the phase difference between the output voltage across points A and B (\(V_{AB} = V_A - V_B\)) and the input voltage \(V_{IN}\). This involves calculating the complex voltages \(V_A\) and \(V_B\) using the voltage divider rule in the phasor domain. Step 2: Key Formula or Approach:
We will use complex impedance to analyze the AC circuit.
Impedance of a resistor R: \(Z_R = R\)
Impedance of an inductor L: \(Z_L = j\omega L\)
Impedance of a capacitor C: \(Z_C = \frac{1}{j\omega C} = -j\frac{1}{\omega C}\)
The voltage divider rule states that the voltage across an impedance \(Z_2\) in series with \(Z_1\) is \(V_2 = V_{total} \frac{Z_2}{Z_1 + Z_2}\).
We need to find the phase of \(V_{AB} = V_A - V_B\). Let \(V_{IN}\) be the reference phasor, \(V_{IN} = V_0 \angle 0^\circ\). Step 3: Detailed Explanation:
1. Calculate the impedances of the components:
Given \(\omega = 10^4\) rad/s.
Resistors: \(R_1 = 10 \, \Omega\), \(R_2 = 10 \, \Omega\)
Inductor: \(L = 1 \, \text{mH} = 10^{-3} \, \text{H}\)
\(Z_L = j\omega L = j(10^4)(10^{-3}) = j10 \, \Omega\)
Capacitor: \(C = 10 \, \mu\text{F} = 10 \times 10^{-6} \, \text{F} = 10^{-5} \, \text{F}\)
\(Z_C = \frac{1}{j\omega C} = \frac{1}{j(10^4)(10^{-5})} = \frac{1}{j0.1} = -j10 \, \Omega\)
2. Calculate the voltages at points A and B using the voltage divider rule:
For point A (left branch):
\(V_A\) is the voltage across the inductor \(Z_L\).
\[ V_A = V_{IN} \frac{Z_L}{R_1 + Z_L} = V_{IN} \frac{j10}{10 + j10} \]
For point B (right branch):
\(V_B\) is the voltage across the capacitor \(Z_C\).
\[ V_B = V_{IN} \frac{Z_C}{R_2 + Z_C} = V_{IN} \frac{-j10}{10 - j10} \]
3. Calculate the output voltage \(V_{AB}\):
\[ V_{AB} = V_A - V_B = V_{IN} \left( \frac{j10}{10 + j10} - \frac{-j10}{10 - j10} \right) \]
\[ V_{AB} = V_{IN} \cdot j10 \left( \frac{1}{10 + j10} + \frac{1}{10 - j10} \right) \]
Find a common denominator:
\[ V_{AB} = V_{IN} \cdot j10 \left( \frac{(10 - j10) + (10 + j10)}{(10 + j10)(10 - j10)} \right) \]
\[ V_{AB} = V_{IN} \cdot j10 \left( \frac{20}{10^2 - (j10)^2} \right) = V_{IN} \cdot j10 \left( \frac{20}{100 - (-100)} \right) \]
\[ V_{AB} = V_{IN} \cdot j10 \left( \frac{20}{200} \right) = V_{IN} \cdot j10 \left( \frac{1}{10} \right) \]
\[ V_{AB} = j V_{IN} \]
4. Determine the phase difference:
The relationship is \(V_{AB} = j V_{IN}\). The complex number \(j\) corresponds to \(e^{j\pi/2}\) in polar form, which represents a magnitude of 1 and a phase angle of \(+90^\circ\) or \(+\pi/2\) radians.
This means that the phasor \(V_{AB}\) is rotated by \(+90^\circ\) with respect to the phasor \(V_{IN}\).
Therefore, the output voltage \(V_{AB}\) leads the input voltage \(V_{IN}\) by \(90^\circ\).
Step 4: Final Answer:
The output voltage \(V_{AB}\) is related to the input voltage \(V_{IN}\) by \(V_{AB} = jV_{IN}\). This implies a phase lead of \(90^\circ\). Thus, option (C) is correct.
