Question

What is the natural frequency (wo) of a circuit that has an inductor of 0.048 henery's, and a capacitor of 0.0004 farads?

Answer 1

The natural frequency (ω0) of an LC circuit with an inductor of 0.048 henries and a capacitor of 0.0004 farads is approximately 228.0 rad/s.

Step-by-step explanation:

The natural frequency (ω0) of a circuit containing an inductor (L) of 0.048 henries and a capacitor (C) of 0.0004 farads can be determined using the formula for the resonant frequency of an LC circuit, which is ω0 = 1/√(LC). Plugging the given values into the formula, we calculate the natural frequency as follows:

ω0 = 1/√(0.048 H × 0.0004 F) = 1/√(0.0000192 H·F)

To find the natural frequency, we take the square root of the product of the inductance and capacitance values and then take the reciprocal of that number:

ω0 = 1/√(0.0000192) = 1/0.004382 = 228.0 rad/s (approximately).

Therefore, the resonant frequency of the LC circuit is about 228.0 rad/s.

Answer 2

The natural frequency will be 228.11 rad/sec

Step-by-step explanation:

We have given Inductance L = 0.048 Henry

And the capacitance C = 0.0004 farad

We have to find natural frequency

When only inductor and capacitor is present in the circuit then it is known as LC circuit and Natural frequency of LC circuit is given by
`\omega _0=(1)/(√(LC))=(1)/(√(0.048* 0.0004))=228.21\ rad/sec`

So the natural frequency will be 228.21