Physics Problems & Solutions: AC Circuit

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Electromagnetism - RLC Circuit

In the RLC circuit shown, the applied voltage is ε(t) = ε_mcos ωt For a constant ε_m, at what angular frequency ω does the current have its maximum steady-state amplitude after the transients have died out?

A. 1/RC
B. 2L/R
C. 1/√(LC)
D. √[(1/LC) − (R/2L)²]
E. √[(1/RC)² − (L/R)²]

(GR9277 #81)

Solution:

I_max when X_L = X_C
→ ωL = 1 / ωC

ω = 1/√(LC)

Answer: C

Electromagnetism - Impedance

An alternating current electrical generator has a fixed internal impedance R_g + jX_g and is used to supply power to a passive load that has an impedance R_g + jX₁, where j =√(−1). R_g ≠ 0 and X_g ≠ 0. For a maximum power transfer between the generator and the load, X₁ should be equal to

A. 0
B. X_g
C. −X_g
D. R_g
E. −R_g

(GR8677 #64)

Solution:

Maximum power transfer theorem (Impedance matching):

Maximum power is transferred from a source to a load when

Load resistance = the internal resistance of the source, or
Load impedance = complex conjugate source impedance (Z_l = Z_s^*)

Z_s = R_g + jX_g with j =√(−1)

Z_s^*= R_g − jX_g

For max power transferred:
Z_l= Z_s^*
X₁= −X_g

Answer: C

Electromagnetism - RLC Circuit

An AC circuit consists of the elements shown above, with R = 10,000 ohms, L = 25 millihenrie and C an adjustable capacitance. The AC voltage generator supplies a signal with amplitude of 40 volts and angular frequency of 1,000 radians per second. For what value of C is the amplitude of the current maximized?

A. 4 nF
B. 40 nF
C. 4 μF
D. 40 μF
E. 400 μF

(GR0177 #38)

Solution:

I_max when X_L = X_C
→ ωL = 1 / ωC

C = 1 / ω²L
= 1 / [(10³)²× 25 × 10⁻³]
= 1 / (25 × 10³)
= 40 × 10⁻⁶
= 40 μF

Answer: D

Electromagnetism - High-Pass Filters

Which two of the following circuits are high-pass filters

A. I and II
B. I and III
C. I and IV
D. II and III
E. II and IV

(GR0177 #39)

Solution:

High-pass filter: V_out ≈ V_in for frequency, ω → ∞

Inductive reactance, X_L = ωL
Capacitive reactance, X_C = 1 / ωC

For ω → ∞
X_L → ∞
X_C → 0

Voltage divider:
$V_{\text{out}}=\left( \frac{R_2}{R_1+R_2}\right ) V_{\text{in}}$

Case I.
$V_{\text{out}}=\left( \frac{R}{R+X_L}\right) V_{\text{in}}=\left( \frac{R}{R+\infty}\right ) V_{\text{in}}=0$
→ Low-pass filter

Case II.
$V_{\text{out}}=\left( \frac{X_L}{R+X_L}\right) V_{\text{in}}=\left( \frac{\infty}{R+\infty}\right ) V_{\text{in}}=V_{\text{in}}$
Note: R ≪ ∞
→ High-pass filter

Case III.
$V_{\text{out}}=\left( \frac{R}{R+X_C}\right) V_{\text{in}}=\left( \frac{R}{R+0}\right ) V_{\text{in}}=V_{\text{in}}$
→ High-pass filter

Case IV.
$V_{\text{out}}=\left( \frac{X_C}{R+X_C}\right) V_{\text{in}}=\left( \frac{0}{R+0}\right ) V_{\text{in}}=0$
→ Low-pass filter

Answer: D