A 115- resistor, a 0.237-µF capacitor, and a 5.56-mH inductor are connected in s

Question

A 115- resistor, a 0.237-µF capacitor, and a 5.56-mH inductor are connected in series to a generator whose voltage is 29.0 V. The current in the circuit is 0.130 A. Because of the shape of the current-versus-frequency graph (shown), there are two possible values for the frequency that correspond to this current. Obtain these two values. Give (a) the smaller value, and (b) the larger value. Note: The ac current and voltage are rms values and power is an average value unless indicated otherwise.

I got up to the part where I now have the quadratic equation and am solving for x
. I got 1.3177x^2 - 45302.5x - 10^9.

Is this correct? I am having trouble getting this to work and i need both x values so I can plug them into the other equation to get the two frquencies!

Explanation / Answer

Z = V/I

Z = 29.0/0.130

(R2 - (L - 1/(C))) = 29.0/0.130

(R2 - ((2f)L - 1/((2f)C))) = 29.0/0.130

NOTE: this equation has tow different answers:

>>>>> (R2 - ((2f)L - 1/((2f)C))) = 29.0/0.130

and

>>>>> (R2 - ((2f)L - 1/((2f)C))) = -29.0/0.130

These two different quadratic equation have four frequencies but only two of them are positive:

Positive frequencies are:

You can see here: (mostly recommende :) )

http://www.wolframalpha.com/input/?i=%28%28115^2+%2B+%282*3.14*x*5.56*10^-3+-+1%2F%282*3.14*x*0.237*10^-6%29%29^2+%29^0.5+%29+%3D+29.0%2F0.130

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