LC circuit inductance help.?
Calculate the inductance of an LC circuit if the resonance frequency is 190 Hz. The capacitance is equal to 2.3e-5 F. Show your work. I know the formula is L = (1/C) (1 / w_0)^2 L = (1/2.3E-5)(1/190)^2 But when i plug it in im getting 1.204 which is the wrong answer. The answer should be in mHz so i tried to convert but im still getting it wrong! Any help on where im going wrong?
Capacitance and inductance of a circuit?
fo=102.048Hz , (As f1: XL=13.2Ω , XC= 10.1Ω) ωL=13.2 1/ωC=10.1 ω=1/10.1C L/10.1C=13.2 L= 133.32C f=1/(2π√LC) 102.048= 1/(2π√133.32C*C) √133.32C^2= 1/(2π*102.048) 11.5464C= 1/(2π*102.048) C= 1/(2π*102.048*11.5464) C= 0.00013507282453703619921558922145648 C= 135.073[μF] ,Part A answer. L=133.32C = 133.32*135.073*10^-6= 18.0079*10^-3= 18[mH] ,Part B answer. (f1= 116.66Hz)
Find the (a) minimum and (b) maximum inductance of the tank circuit?
The tank circuit is resonant when the capacitive-reactance of the circuit equals the inductive-reactance. XL = 2 * PI * F * L XL is the inductive-reactance in ohms, PI is 3.14159, F is frequency in Hertz, L is inductance in Henries Xc = 1/(2 * PI * F * C) Xc is capacitive-reactance in ohms, PI and F like above, C is capacitance in farads Since the tank circuit is resonant when XL = Xc 2 * PI * F * L = 1/(2 * PI * F * C) Solve for L when F=3.8 MHz (remember in Hertz) and 9.1 MHz, you know PI and C
A circuit consisting of a resistance of 12 ohms and an inductance of 0,2 Henry in series is connected to a 250V, 50Hz supply. What current will flow in the circuit?
First, calculate the inductive reactance of 0.2 H at 50 Hz, which is given by 2piFL and expressed in ohms:XL = 2piFL = 2 x 3.1416 x 50 Hz x .2 H = 62.83 ohmsThen add the above value to the series resistance of 12 ohms:62.83 ohms + 12 ohms = 74.83 ohmsUsing Ohm's’ Law, calculate the current through 74.8 ohms when applying a potential of 250 volts:E/R = I250 volts / 74.83 ohms = 3.34 amperes
What is the relations between inductance and frequency in a.c circuit?
What is the relations between inductance and frequency in a.c circuit?XL = 2Ï€fLXL is inductive reactance expressed in Ohms.f is frequency expressed in Hertz.L is inductance expressed in Henrys.You can rearrange the equation to see how two variables affect and determine the third value.Generally speaking the higher the frequency, the greater the reactance if the inductance stays the same.
What is stray inductance?
Stray inductance is unintended and unwanted inductance in a circuit.Inductance does not exist only within inductors . In fact, any wires or component leads that have current flowing through them create magnetic fields. When these magnetic fields are created, they can produce an inductive effect. Thus, wires or components leads can act as inductors if they are long enough. Such effects are often present within circuits (for example, between conductive runs of wire traces or components with long leads such as capacitors), even though they are not intended. This unintended inductance is referred to as stray inductance, and it can result in a disruption of normal current flow within a circuit.for more please go to the link :What is Stray Inductance?
If the capacitance and inductance in an RLC series circuit are each doubled, but the resistance remained same, what happens to the frequency?
If f is the the frequency of RlC circuit before doubling the values of L and C, then it would be f/2 after doubling the values of L and C.By the way theoratically the value of R doesnt affect the resonant frequency. However, it affects the bandwidth of the circuit.
Calculate the inductance of an LC circuit that oscillates at 152 Hz when the capacitance is 7.7 µF. Answer in?
The circuit oscillates at the resonate frequency. The resonate frequency is that frequency at which the inductive reactance equals the capacitive reactance.; Capacitive reactance = 1/(2pi fc) = 1/[(955)*(.0000077)] = 136 Ohms 136 = 2pi fL L = 136 /955) = 142.4mH
How to find inductance and peak voltage? help :o?
The period of the LC circiut is 2*π*√[L*C] 5*10^-3 = 2*π•√[L*200*10^-6] L = 25*10^-6/(4*π²*200*10^-6 = 0.30 H = 300 mH The inductive impedance is 2*π*f*L = 2*π*L/T = 388 ohm The peak voltage is 388*0.025 = 9.68 V