R=175 ohms C=12.5 mircroF L=8 mH All conected to AC source having variable frequ

Question

R=175 ohms

C=12.5 mircroF

L=8 mH

All conected to AC source having variable frequency and voltage 25 Volts.

(

a) At What angular Frequency will impepedance be lowest?

(b) With this frequency, what is Max current through Inductor?

(c) Find potential difference across AC source, capacitor, resistor , and Inductor at instant current is equal to one half its greatest value?

Explanation / Answer

A)impedance = sqrt(R^2 + (Lw-1/Cw)^2)...it si min. at resonance => w= sqrt(1/LC) so w = 3.16*10^3 s- B)max current = max voltage/Lw = 25sqrt2 /25.3 = 1.397 amp C) let V = 25sqrt2 sinwt ... => i = 25sqrt2/175 sinwt = .143sqrt2 sinwt current is half its max value when sinwt = .5 => coswt = sqrt3/2 so V' = 25 sqrt2/2 = 16.76 V V across L = L di/dt = 8*10^-3 *.143 sqrt2 *w coswt= 4.42 V V across C = -4.42 as at resonance voltags across L and C cancel each other V across R = 16.76 V

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