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23.12 Rlc series ac circuits  (Page 3/9)

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Resonance in AC circuits is analogous to mechanical resonance, where resonance is defined to be a forced oscillation—in this case, forced by the voltage source—at the natural frequency of the system. The receiver in a radio is an RLC circuit that oscillates best at its f 0 size 12{f rSub { size 8{0} } } {} . A variable capacitor is often used to adjust f 0 size 12{f rSub { size 8{0} } } {} to receive a desired frequency and to reject others. [link] is a graph of current as a function of frequency, illustrating a resonant peak in I rms size 12{I rSub { size 8{"rms"} } } {} at f 0 size 12{f rSub { size 8{0} } } {} . The two curves are for two different circuits, which differ only in the amount of resistance in them. The peak is lower and broader for the higher-resistance circuit. Thus the higher-resistance circuit does not resonate as strongly and would not be as selective in a radio receiver, for example.

The figure describes a graph of current I versus frequency f. Current I r m s is plotted along Y axis and frequency f is plotted along X axis. Two curves are shown. The upper curve is for small resistance and lower curve is for large resistance. Both the curves have a smooth rise and a fall. The peaks are marked for frequency f zero. The curve for smaller resistance has a higher value of peak than the curve for large resistance.
A graph of current versus frequency for two RLC series circuits differing only in the amount of resistance. Both have a resonance at f 0 size 12{f rSub { size 8{0} } } {} , but that for the higher resistance is lower and broader. The driving AC voltage source has a fixed amplitude V 0 size 12{V rSub { size 8{0} } } {} .

Calculating resonant frequency and current

For the same RLC series circuit having a 40.0 Ω resistor, a 3.00 mH inductor, and a 5.00 μF capacitor: (a) Find the resonant frequency. (b) Calculate I rms size 12{I rSub { size 8{"rms"} } } {} at resonance if V rms size 12{V rSub { size 8{"rms"} } } {} is 120 V.

Strategy

The resonant frequency is found by using the expression in f 0 = 1 2π LC size 12{f rSub { size 8{0} } = { {1} over {2π sqrt { ital "LC"} } } } {} . The current at that frequency is the same as if the resistor alone were in the circuit.

Solution for (a)

Entering the given values for L and C into the expression given for f 0 size 12{f rSub { size 8{0} } } {} in f 0 = 1 2π LC size 12{f rSub { size 8{0} } = { {1} over {2π sqrt { ital "LC"} } } } {} yields

f 0 = 1 2π LC = 1 2π ( 3 . 00 × 10 − 3 H ) ( 5 . 00 × 10 − 6 F ) = 1 . 30 kHz . alignl { stack { size 12{f rSub { size 8{0} } = { {1} over {2π sqrt { ital "LC"} } } } {} #" "= { {1} over {2π sqrt { \( 3 "." "00" times "10" rSup { size 8{ - 3} } " H" \) \( 5 "." "00" times "10" rSup { size 8{ - 6} } " F" \) } } } =1 "." "30"" kHz" {} } } {}

Discussion for (a)

We see that the resonant frequency is between 60.0 Hz and 10.0 kHz, the two frequencies chosen in earlier examples. This was to be expected, since the capacitor dominated at the low frequency and the inductor dominated at the high frequency. Their effects are the same at this intermediate frequency.

Solution for (b)

The current is given by Ohm’s law. At resonance, the two reactances are equal and cancel, so that the impedance equals the resistance alone. Thus,

I rms = V rms Z = 120 V 40 . 0 Ω = 3 . 00 A. size 12{I rSub { size 8{"rms"} } = { {V rSub { size 8{"rms"} } } over {Z} } = { {"120"" V"} over {"40" "." "0 " %OMEGA } } =3 "." "00"" A"} {}

Discussion for (b)

At resonance, the current is greater than at the higher and lower frequencies considered for the same circuit in the preceding example.

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Power in RLC Series ac circuits

If current varies with frequency in an RLC circuit, then the power delivered to it also varies with frequency. But the average power is not simply current times voltage, as it is in purely resistive circuits. As was seen in [link] , voltage and current are out of phase in an RLC circuit. There is a phase angle     ϕ size 12{ϕ} {} between the source voltage V size 12{V} {} and the current I size 12{I} {} , which can be found from

cos ϕ = R Z . size 12{"cos"ϕ= { {R} over {Z} } } {}

For example, at the resonant frequency or in a purely resistive circuit Z = R size 12{Z=R} {} , so that cos ϕ = 1 size 12{"cos"ϕ=1} {} . This implies that ϕ = 0 º size 12{ϕ=0 rSup { size 8{ circ } } } {} and that voltage and current are in phase, as expected for resistors. At other frequencies, average power is less than at resonance. This is both because voltage and current are out of phase and because I rms size 12{I rSub { size 8{"rms"} } } {} is lower. The fact that source voltage and current are out of phase affects the power delivered to the circuit. It can be shown that the average power is

Questions & Answers

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If auger is pair are the roots of equation x2+5x-3=0
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Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
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find product (-6m+6) ( 3m²+4m-3)
SIMRAN Reply
-42m²+60m-18
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-24m+3+3mÁ^2
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Bajemah
-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
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12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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