Consider an AC bridge shown in the figure with $R=300\Omega$, $R1=1000 \Omega$, $R2=500 \Omega$, $L=30mH$, and a detector D. At the bridge balance condition, the frequency of the excitation source $V_s$ is _________ kHz (rounded off to two decimal places).
To find the frequency at which the AC bridge is balanced, we use the balance condition of an AC bridge:
Complex impedance in opposite arms are equal.
Given:
\(R = 300 \, \Omega\), \(R_1 = 1000 \, \Omega\), \(R_2 = 500 \, \Omega\), \(L = 30 \, \text{mH}\)
The balance condition for the bridge is:
\(\dfrac{R_1}{R_2} = \dfrac{R}{j\omega L}\)
Where \(\omega = 2\pi f\)
\(f = \dfrac{R}{2\pi L} \cdot \dfrac{1}{1 + \dfrac{jR_2}{R_1}}\)
Simplifying, we get:
\(f = \dfrac{R \cdot R_1}{2\pi L \cdot R_2}\)
Plug in values:
\(f = \dfrac{300 \times 1000}{2\pi \times 30 \times 10^{-3} \times 500}\)
\(f = \dfrac{300000}{94.248}\)
\(f \approx 1591.55 \, \text{Hz} = 1.59 \, \text{kHz}\)
The frequency is \(1.59 \, \text{kHz}\), which is within the range of \(1.55 \, \text{kHz}\) to \(1.65 \, \text{kHz}\).
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