Question

If L=0.2 henry C=0.0012muF.What is the Maximum value of resistance in an oscillator
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Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

The maximum value of the resistance, R, in an oscillator measured is $2.58\times 10^{4}\Omega$.

Explanation:

Given,

The inductance in an oscillator, L = $0.2 henry$

The capacitance in an oscillator, C = $0.0012\mu F$ = $1.2\times 10^{-9}F$

The maximum value of the resistance in an oscillator, R =?

As we know,

• The maximum value of the resistance in an oscillator can be calculated by the formula given below:
• $R^{2} = \frac{4L}{C}$
• $R = \sqrt{\frac{4L}{C} }$

After putting the given values in the equation, we get:

• $R = \sqrt{\frac{4\times 0.2}{1.2\times 10^{-9} } }$
• $R = \sqrt{\frac{0.8\times10^{9} }{1.2 } }$
• $R = \sqrt{\frac{8\times10^{8} }{1.2 } }$
• $R =\sqrt{6.66\times 10^{8} }$
• $R =2.58\times 10^{4}\Omega$

Hence, the maximum value of the resistance in an oscillator, R = $2.58\times 10^{4}\Omega$.