A series RLC circuit is driven by a generator with an emf amplitude of 80 V and a current amplitude of 1.25 A. The current leads the emf by 0.65 rad. What are the impedance and the resistance of the circuit?

Solution:

- Concepts:

AC circuits - Reasoning:

We have a series RLC circuit and a AC generator, generating a sinusoidal voltage. - Details of the calculation:

In general: V = IZ

Z = R + iωL + 1/(iωC) = R + i(ωL - 1/(ωC)) = (R^{2}+ (ωL - 1/(ωC))^{2})^{½}exp(iφ)

φ = tan^{-1}((ωL - 1/(ωC))/R)

All that is needed for this problem:

Z = R + iX.

Given: V = 80 V exp(iωt), I = 1.25 A exp(i(ωt + 0.65)),

Z = V/I = 64 Ω exp(-i0.65), R = 64 Ω cos(0.65) = 50.95 Ω.

In the circuit below a 30 V (peak) AC source at 60 Hz is connected to a 90 Ω resistor, a 50 μF capacitor, and a 60 mH inductor in series. Find the phase of the current through the circuit with respect to the voltage across the source.

Solution:

- Concepts:

AC circuits - Reasoning:

We are asked to analyze an AC circuit. - Details of the calculation:

V = IZ.

Z = R + iωL + 1/(iωC) = [90 + i*22.6 – i*53)] Ω.

= [90 - i*30.4] Ω = 95 exp(iφ) Ω with φ = -0.326.

I = V/Z = (30 V)/(95 exp(-i0.326)) Ω = 0.316 A exp(i0.326).

The peak value of the current is 0.316 A.

φ_{I}– φ_{V}= 0.326. I(t) leads V(t).

For the circuits shown below, V_{in} = V_{0}exp(iωt).

In terms of R L and C
find V_{out} and identify which type of filter each circuit
represents.

Solution:

- Concepts:

Filter circuits - Reasoning:

We are asked to analyze 3 different AC filter circuits. - Details of the calculation:

(a) I = V_{in}/(Z_{C}+ Z_{R}), V_{out}= IZ_{C}= V_{in}Z_{C}/(Z_{C}+ Z_{R}).

Z_{C}/(Z_{C}+ Z_{R}) = 1/(1 + iRωC) = (1 - iRωC)/(1 + ω^{2}R^{2}C^{2}) = [1/(1 + ω^{2}R^{2}C^{2})^{½}]e^{-iφ}.

tanφ = RωC.

V_{out}= V_{0}[1/(1 + ω^{2}R^{2}C^{2})^{½}]e^{i(ωt-φ)}.

V_{out}--> 0 as ω --> ∞, V_{out}--> V_{0}as ω --> 0, we have a low-pass filter.

(b) I = V_{in}/(Z_{L}+ Z_{R}), V_{out}= IZ_{L}= V_{in}Z_{L}/(Z_{L}+ Z_{R}).

Z_{L}/(Z_{L}+ Z_{R}) = ωLexp(iπ/2)/(R + iωL) = exp(iφ)ωL/(R^{2}+ (ωL)^{2})^{½}.

V_{out}= V_{0}[ωL/(R^{2}+ (ωL)^{2})^{½}]e^{i(ωt+φ)}.

V_{out}--> 0 as ω --> 0, V_{out}--> V_{0}as ω --> ∞, we have a high-pass filter.

(c) I = V_{in}/(Z_{c}+ Z_{L}+ Z_{R}), V_{out}= IZ_{R}= V_{in}Z_{R}/(Z_{C}+ Z_{L}+ Z_{R}).

Z_{R}/(Z_{C}+ Z_{L}+ Z_{R}) = exp(iφ) R/(R^{2}+ (ωL - 1/(ωC))^{2})^{½}.

V_{out}= V_{0}[R/(R^{2}+ (ωL - 1/(ωC))^{2})^{½}]e^{i(ωt+φ)}.^{ }V_{out}= V_{0}when ω = 1/(LC)^{½}.

V_{out}--> 0 as ω --> 0 and as as ω --> ∞, we have a band-pass filter.

Find the currents in each arm of the parallel LRC circuit
with V_{AC} = Re(V_{0}e^{iωt}). Find the total current
and the average power drawn from the generator.

Solution:

- Concepts:

AC circuits - Reasoning:

We have a parallel RLC circuit and a AC generator, generating a sinusoidal voltage. - Details of the calculation:

Let V = V_{0}e^{iωt}.

I_{L}= Re(V/Z_{L}) = Re(V/iωL) = Re((V_{0}/(ωL))e^{i(ωt - π/2)}) = (V_{0}/(ωL))cos(ωt - π/2)

= (V_{0}/(ωL))sin(ωt).

I_{C}= Re(V/Z_{c}) = Re(ViωC) = Re(V_{0}ωC e^{i(ωt + π/2}) = V_{0}ωC cos(ωt + π/2)

= -V_{0}ωC sin(ωt).

I_{R}= Re(V/R) = Re((V_{0}/R)e^{iωt}) = (V_{0}/R)cos(ωt).

I_{tot}= V_{0}[(1/(ωL) - ωC)sin(ωt) + (1/R)cos(ωt)].

P_{avg}= ½V_{0}^{2}/R, there is no energy dissipated in the reactive part.

For the circuit shown below, determine the value of the
current output through the AC generator, in terms of the symbols given, for two
limiting conditions.

(a) The frequency of the AC generator approaches zero.

(b) The frequency of the AC generator approaches infinity.

Solution:

- Concepts:

AC circuits - Reasoning:

We have a two-terminal network and a AC generator, generating a sinusoidal voltage. - Details of the calculation:

Assume V(t) and I(t), are proportional to exp(iωt). Assume idealized circuit elements. Define the impedance Z as Z = V/I. Then

Z(capacitance) = Z_{C}= 1/(iωC),

Z(inductance) = iωL,

Z(resistance) = Z_{R }= R.

(a) As ω approaches zero the inductance becomes a short circuit and the capacitors become open circuit elements. Then the circuit is just a simple loop with 3 resistors in series. I = V_{0}exp(iωt)/(R_{1}+R_{2}+R_{4}).

(b) As ω approaches infinity the capacitors become a short circuits and the inductor becomes an open circuit element. The circuit then contains 2 loops in parallel, one with resistance R_{4}and one with the resistors R_{1}R_{2}and R_{3}in series. The current through the AC generator is I = V_{0}exp(iωt)(1/(R_{1}+R_{2}+R_{3}) + 1/R_{4}).

On the input of the RLC filter shown below the periodic
voltage oscillating as U(t) = A sin^{4}(ωt) is applied. Calculate the
output voltage after all transients have decayed if the elements R, L, C have
been chosen such that 4ω^{2}LC = 1 and ωRC = 2.

Solution:

- Concepts:

AC circuits, Fourier series, the principle of superposition - Reasoning:

U = ZI for a sinusoidally varying voltage. - Details of the calculation:

sin^{4}(x) = ⅛cos(4x) – ½cos(2x) + 3/8.

Use complex notation, the real part of the expression describes the physical quantity.

U(t) = (A/8)exp(i4ωt) + (A/2)exp(i2ωt) + 3A/8 = U_{1}+ U_{2}+ U_{3}.

Use the principle of superposition.

For each of the 3 terms in the Fourier series: U_{i}= IZ_{i}, U_{i}' = IZ_{i}', U_{i}' = U_{i}Z_{i}'/Z_{i}.

Here U_{i}is the input and U_{i}' the output voltage.

Z_{i}/Z_{i}' = (i(ω_{i}L - 1/(ω_{i}C)) + R)/(i(ω_{i}L - 1/(ω_{i}C))) = 1 - iω_{i}RC/(ω_{i}^{2}LC - 1).

Z_{i}/Z_{i}' = [1 + (ω_{i}RC/(ω_{i}^{2}LC - 1))^{2}]^{½}exp(-φ), with tanφ = ω_{i}RC/(ω_{i}^{2}LC - 1).

Z_{i}'/Z_{i}= [1 + (ω_{i}RC/(ω_{i}^{2}LC - 1))^{2}]^{-½}exp(φ).

For ω_{i}= 0, Z_{i}'/Z_{i}= 1.

For ω_{i}= 2ω, Z_{i}'/Z_{i}= i(4ω^{2}LC - 1)/(i(4ω^{2}LC - 1) + 2ωRC) = 0.

For ω_{i}= 4ω, Z_{i}'/Z_{i}= [1 + (4ωRC/(16ω^{2}LC - 1))^{2}]^{-½}exp(φ) = [1 + (8/3)^{2}]^{-½}exp(φ) = 0.3511exp(φ).

U' = 3A/8 + (A/8)exp(i(4ωt + φ) = 3A/8 – 0.3511(A/8)cos(4ωt + φ).

The filter passes only the 4ω component (plus a constant offset).