Test 8.1 Complex Power

Complex power S contains two components, which are

Peak absorbed power and average exchanged power
Peak absorbed power and peak exchanged power
Average absorbed power and peak exchanged power
Average absorbed power and average exchanged power

Test 8.2 Complex Power

For a purely capacitive load, the average absorbed power is

P_av = I_2rms/ωC
P_av = 0
P_av = I_2rmsωC
P_av = ½ I_2rms/ωC

Test 8.3 Power Factor

For a purely resistive load, the power factor is

pf = 0
pf = 0.5
pf = ∞
pf = 1

Test 8.4 Power Factor

For a purely capacitive load, the power factor is

pf = 0
pf = 0.5
pf = ∞
pf = 1

Test 8.5 Maximum Power Transfer

A source circuit with impedance Z_s = R_s + jX_s is connected to a load Z_L = R_L + jX_L. To effect maximum transfer of power to the load, the load should be designed such that

R_L + jX_L = R_s + jX_s
|R_L + jX_L| = |R_s + jX_s|
R_L + jX_L = R_s - jX_s
R_L + jX_L = |R_s + jX_s|

Test 8.6 Average Value

What is the average vale of ν(t)?

V_av = 6 V
V_av = 2 V
V_av = 3 V
V_av = 1.5 V

Test 8.7 RMS Value

What is the rms value of ν(t)?

V_rms = 2 V
V_rms = 2.45 V
V_rms = 1.22 V
V_rms = 0.61 V

Test 8.8 Power Factor

The voltage and current at the input to a load circuit are given by υ(t) = 100 cos(377t - 30°) V, i(t) = 5 cos(377t + 30°) A. The load power factor is

pf = 2 leading
pf = 1 lagging
pf = 0.5 lagging
pf = 0.5 leading

Test 8.9 Average Power

The voltage and current at the input to a load circuit are given by V_rms = 100∠75° V, I_rms = 2∠15° A. The average absorbed power is

P_av = 100 W
P_av = 200 W
P_av = 50 W
P_av = 0

Test 8.10 Average Power

In the circuit, i_s(t) = 0.2 sin 10⁵t A, R = 20 Ω, L = 0.05 mH, and C = 2 μF. Determine the average power dissipated in R.

P_av = 0
P_av = 0.8 W
P_av = 0.2 W
P_av = 0.4 W

Test 8.11 Power Factor

In the circuit, i_s(t) = 0.2 sin 10⁵t A, R = 20 Ω, L = 0.05 mH, and C = 2 μF. The power factor of the source is

pf = 0.2
pf = 1
pf = 0
pf = 0.5

Test 8.12 Op-Amp Circuit

In the circuit, υ_in(t) = 10 cos(1000t) V, R = 10 kΩ, R_L = 1 kΩ, and C = 1 μF. The average power dissipated in R_L is:

P_av = 0.2 W
P_av = 0.1 W
P_av = 0.2 mW
P_av = 0.5 mW

Test 8.13 Op-Amp Circuit

In the circuit, υ_in(t) = 10 cos(1000t) V, R = 10 kΩ, R_L = 2 kΩ, and C = 1 μF. The average power dissipated in R_L is:

P_av = 0.2 W
P_av = 0.1 W
P_av = 0.25 mW
P_av = 0.5 W

Test 8.14 Maximum Power Transfer

Choose the load Z_L so that the power dissipated in it is a maximum.

Z_L = (0.8 - j1.6) Ω
Z_L = (1.6 - j0.8) Ω
Z_L = (0.8 + j1.6) Ω
Z_L = (1.6 - j3.2) Ω

Test 8.15 Maximum Power Transfer

In the phasor-domain circuit, R = 2 kΩ and Z_L= j2 kΩ. Choose the load impedance so that the power dissipated in it is a maximum.

Z_L = (2 + j2) kΩ
Z_L = 2(1 - j1) kΩ
Z_L = 0.5(1 + j1) kΩ
Z_L = (1 - j1) kΩ