LC/LRC/AC Circuits

Equations and constants
I = (V/R)*(1-e^(-t/tau)) (charging LR circuit)
tau = L/R (time constant)
X_L = 2pifL ( impedance of inductor, inductive reactance )
X_C = 1/2pifC ( impedance of capacitor, capacitive reactance )
Z = sqrt(R^2 + (X_L -X_C )^2) ( total impedance of LRC circuit )
cos(phi) = R/Z. ( phase angle )
tan(phi) = (X_L -X_C )/R ( phase angle )
f_0 = 1/(2pi*sqrt(LC)) ( resonance )

A resistor behaves such that the voltage and the current are in phase.
An inductor behaves such that the current lags the voltage by pi/2.
A capacitor behaves such that the current leads the voltage by pi/2.

Resistance is the rating of a resistor.
Impedance is the term used to describe the sum resistance of the circuit.
Reactance is a term that refers just to the inductive properties.

A phasor diagram is a pseudovector diagram. It can be used to show the relation between voltage and current lag/lead.

Resonance is obtained using an LC circuit when X_C = X_L . The impedance is purely resistive and cos(phi) = 1.
Explain what is meant by impedance matching

Impedance matching is the process of making two connected devices' impedance output equal to the input. This maximizes power transfer.

Problem 1
It takes 7.20ms for the current in an LR circuit to reach 80% of its max value. Determine a) the time constant of the circuit and b) the inductance of the circuit if R = 250Ohms.

a) Since we're given the percentage, we have I as 0.8I_max. The equation I = (V/R)*(1-e^(-t/tau)) simplifies to 0.8 = 1-e^(-t/tau). Solving for tau, we arrive at 4.4736*10^(-3)s.
b) The inductance of the circuit can be found from the equation tau = L/R. L is then 1.1184H.

Problem 2
After how many time constants does the current in Fig 21-29 reach within a) 10% b) 1% and c) 0.1% of its max value?

a) I is 0.90I_max. We solve the charging equation, I = I_max*(1-e^(-t/tau)) for t/tau which is 2.302 time constants.
b) Repeat part a)'s step, except now I = 0.99I_max. t/tau = 4.605 time constants
c) Using the same reasoning as above, I = 0.999I_max, t/tau 6.908 time constants.

Problem 3
At what frequency will a 160*10^(-3)H inductor have a reactance of 1.5*10^3ohms?

Frequency is related to reactance and inductance by the equation X_L = 2pifL. Solving for f, X_L /(L*2pi) = f. Making the substitutions, we arrive at f = 1492.36Hz.

Problem 4
A 9.20*10^(-6)F capacitor is measured to have a reactance of 250ohms. At what frequency is it being driven?

Frequency is related to reactance and capacitance by the equation X_C = 1/(2pifC). Solving for f, we have f = 1/(2pi*X_C *C). Making the substitutions, we arrive at f = 69.21Hz.

Problem 5
Calculate the impedance of, and rms current in, a 160*10^(-3)H radio coil connected to a 240-V (rms) 10.0kHz ac line. Ignore resistance.

To calculate the impedance, we need to know the inductance of the reactance. Because that was given, we use the equation X_L = 2pifL to arrive at X_L = 1005.12ohms. Next, we use the equation V = IX_L to solve for the rms current. Just substitute the rms value of V to solve for the rms value of I, 0.238777A.

Problem 6
An inductance coil operates at 240V and 60Hz. It draws 12.8A. What is the coil's inductance?

Solve for the impedance first, V/I = X_L : 18.75ohms. The inductance is related to the impedance by X_L = 2pifL, solving for L: X_L /2pif = L. Make the substitutions, L = 0.0497H.

Problem 7
a) What is the impedance of a well-insulated 0.030*10^(-6)F connected to a 2.0kV (rms) 700Hz line? b) What will be the peak value of the current?

a) The impedance for a capacitor is related to capacitance and frequency by X_C = 1/2pifC. Solving for X_C , we arrive at 7580.24ohms.
b) The peak value of the current can be found by I = V/X_C . Use the peak value of voltage found by 2*10^3/.707 in this equation. I is then 0.3732A.

Problem 8
A 30*10^3ohm resistor is in series with a 0.50-H inductor and an ac source. Calculate the impedance of the circuit if the source frequency is a) 60Hz and b) 3.0*10^4Hz.
The impedance of a circuit is given by Z = sqrt(R^2 +(X_L -X_C )^2). With no capacitor in the circuit, the impedance can be simplified to the square root of the sum of the squares of the resistance and the inductive reactance.

a) X_L is equal to 2pifL, 188.46ohms. The total impedance is then 30000.59195ohms.
b) X_L becomes 94230ohms in this case. The total impedance is then 98890.30741ohms.

Problem 9
For a 120-V_rms 60-Hz voltage, a current of 70*10^(-3)A passing through the body for 1.0s could be lethal. What would be the impedance of the body for this to occur?

Impedance is related to voltage and current quite nicely in the equation Z = V/I. Making the substitutions, Z = 1714.2857ohms.

Problem 10
What is the longest wavelength photon that could produce a proton-antiproton pair? Each has a mass of 1.67*10^(-27)kg.

We find the total initial energy by using E = mc^2 => 3.006*10^(-10)J. This will be equal to the energy of the photon emitted since the proton and antiproton annihilate each other. E = hc/ ?, ? = hc/E. ? = 6.61*10^(-16)m.

Problem 11
a) What is the rms current in an RC circuit if R = 28.8*10^3ohm, C = 0.80*10^(-6)F, and the rms applied voltage is 120V at 60Hz? b) What is the phase angle between voltage and current? c) What is the power dissipated by the circuit? d) What are the voltmeter readings across R and C?

a) Solve for the impedance first of all, Z = sqrt(R^2 + (-X_C )^2), X_C is 3316.35ohms. Z = 28990ohms. The rms current can be found by I = V/Z, 0.00414A.
b) The phase angle between voltage and current can be found by arctan(-X_C /R), -6.57degrees.
c) The power dissipated in the circuit can be found by P = IVcos(phi) => 0.49354W.

Problem 12
a) What is the rms current in an RL circuit when a 60Hz 120V_rms ac voltage is applied, where R = 1.8*10^3ohms and L = 900*10^(-3)H? b) What is the phase angle between voltage and current? c) How much power is dissipated? d) What are the rms voltage readings across R and L?

a) First find the impedance, Z = sqrt(R^2 + X_L ^2), X_L = 339.228 ohms. Z = 1831.7ohms. The rms current is then given by V_rms /Z = I_rms , I_rms = 0.0655A
b) The phase angle is given by tan(phi) = (X_L -X_C )/R, phi = 10.675 degrees.
c) The power dissipated is given by P = IVcos(phi), 7.724W.
d) Since the components are in series, use ohm's law for the voltage across R, V = IR, V = 117.90V. The voltage across L is given by V = IX_L , V = 22.22V.

Problem 13
What is the total impedance, phase angle, and rms current in an LRC circuit connected to a 10.0*10^3Hz, 300V (rms) source if L = 22*10^(-3)H, R = 8.7*10^3ohms, and C = 5000*10^(-12)F?

To find the total impedance, we first find X_C and X_L . To find X_C , we use the equation X_C = 1/2pifC => 3183.7ohms. X_L is given by 2pifL, 1382.04ohms. The total impedance is given by Z = sqrt(R^2+(X_L -X_C )^2), Z is then 8884.59ohms.
The phase angle is given by phi = arctan((X_L -X_C )/R) => -0.2042 radians = -11.7 degrees.
The rms current is given by V_rms /Z = I_rms . I_rms = 0.033766A

Problem 14
What is the resistance of a coil if its impedance is 35ohms and its reactance is 30 ohms?

Total impedance is given by Z = sqrt(R^2+reactance^2). R is then 18.03ohms.

Problem 15
A voltage V = 4.8sin754t is applied to an LRC circuit. If L = 3.0*10^(-3)H, R = 1.40*10^3ohms and C = 3.0*10^(-6)F, how much power is dissipated in the circuit?

The power dissipated in the circuit is found by P = IVcos(phi). The frequency of the circuit can be found by 2pif = 754, 120.025Hz. This can be used to find the impedance of the circuit. X_L is 2.262ohms. X_C is 442.086649ohms. Z is then 1467.46ohms. The angle phi between voltage and current can be found by arctan((X_L -X_C )/R) => -0.3044rads = -17.44degrees. P = IVcos(phi) can be rewritten as P = V^2/Z*cos(phi). V for this equation is 4.8*.707. P = 0.007487W.

Problem 16
An LRC circuit has L = 4.8mH and R = 4.4ohms. a) What value must C have to produce resonance at 3600Hz? b) What will be the maximum current at resonance if the peak external voltage is 50 V?

a) Solve for the capacitance in the equation f 0 = 1/2pi * sqrt(1/LC), C = 4.07*10^(-7)F.
b) The maximum current at resonance can be found with the equation V = IR, I = 11.36A.

Problem 17
A 230mH coil whose resistance is 18.5ohms is connected to a capacitor C and a 3360Hz source voltage. If the current and voltage are to be in phase, what value must C have?

If the current and voltage in phase, then the angle must be 0. This means that X_L = X_C , 2pifL = 1/2pifC. C = 1/(4pi^2f_0^2*L), C = 9.75883*10^(-9)F.