**Frequency of an L-R-C Circuit**

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## Question “Frequency of an L-R-C Circuit”

An L-R-C circuit has an inductance of 0.430 , a capacitance of 2.45×10−5 , and a resistance of R.

(A) What is the angular frequency of the circuit when ?Expressyour answer in radians per second.

(B) What value must *R* have to give a decrease in angular frequency of 6.00 compared to the value calculated in Part A?Express your answer in ohms.

## Answer

### Answer 1

Given:

#####
Question Details

A L-R-C circuit has an Inductance of 0.430, a Capitance of 2.45×10-5, and a Resistance of R.

(A) What is ‘s angular frequency? Give your answer in radians per minute.

(B) What value must be to provide a decrease of angular frequency at 6.00 as compared to Part A?

These are the voltage equations for each of the circuit elements:

(3)

We see in equation (3) that is the initial voltage of the capacitor, which is a constant.

I will assume that current flows in clockwise direction. This would mean that voltage drops are occurring in the clockwise direction.

direction I will add the voltage rises in counterclockwise direction. Kirchhoff’s voltage law

The sum of all these voltage rises equals zero. The sum will be calculated starting at the top of resistor

. Consider the derivative of this equation.

and you get:

. This equation can be divided by

and the differential can be written.

Equation for the series circuit as:

(4)

This is the auxilliary equation that solves this differential equation.

.

This auxilliary equation can be solved using the quadratic method:

.

When :

as can be written.

(5)

inequation (5) is the radian frequencythat we are looking for :

(6)

and tag24.

has an angular frequency of (6) that is:

.

B) We can solve equation 6 for

(I am actually using).

instead of the rounded off value —

) .

We get the following:

. Take the square root from both sides and multiply it by

.

we get :

.

Substitute HTMLMediaTag23 for in this equation and you get:

rounding to six significant figures we have :

.

### Answer 2

(A) We assume that the resistor is of zero resistance so it doesn’t exist in this section. It’s a LC circuit using the equation

(B) Now, the angular frequency should be reduced to 6% of LC value

So _{0}=289.61

Now to Find R

So R=90.4 O

## Conclusion

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