Find the value of inductance in a series inverter circuit having the frequency of 5 kHz and a capacitance μF. If the inverter is operating under resonance condition. The value of inductance is given by:
40 mH
To solve this problem, we need to calculate the inductance value in a series inverter circuit operating under resonance condition. Given parameters are:
- Resonant frequency, \( f = 5 \, \text{kHz} = 5000 \, \text{Hz} \)
- Capacitance, \( C = \text{1 μF} = 1 \times 10^{-6} \, \text{F} \)
In a series LC circuit operating at resonance, the resonant frequency is given by:
\(f = \frac{1}{2 \pi \sqrt{LC}}\)
Rearranging for \( L \), we get:
\(L = \frac{1}{(2 \pi f)^2 C}\)
Substituting the given values:
\(L = \frac{1}{(2 \pi \times 5000)^2 \times 1 \times 10^{-6}}\)
Simplifying further:
\(= \frac{1}{(2 \times 3.14159 \times 5000)^2 \times 10^{-6}}\)
\(= \frac{1}{(31415.9)^2 \times 10^{-6}}\)
\(= \frac{1}{986960441 \times 10^{-6}}\)
\(= \frac{1}{986.96}\)
\(= 0.001013 \, \text{H} = 40.13 \, \text{mH}\)
Therefore, the value of inductance in the inverter circuit is approximately 40 mH.
Thus, the correct answer is 40 mH.
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Inverters are having following properties:
A. VSI uses force commutation
B. VSI has negligible impedance
C. VSI has large source impedances
D. They work like choppers
Choose the correct answer from the options given below:
