Obtain the power factor for the circuit in the figure. Specify the power factor as leading or

lagging. Round your answer to four decimal places.

− j 2 || (

4 Ω j6 Ω

-j2 Ω -j2 Ω

j 6 − j 2) = − j 2 || j 4 = = − j 4(− j 2)( j 4)

j 2

Z T = 4 − j 4 = 5.66 −45

pf = cos(−45) = 0.7071 (leading)

Assume that the values of voltage and current are the peak values.

Oscilloscope measurements indicate that the voltage across a load and the

current through it are, respectively, 166 V and 3 A. Determine the

apparent power.

47 63

= VI = (166 )(3 ) = 249 S 1

2

1

247 −63 −16

S = 249 VA

Assume that the value of current is the RMS value.

Obtain the complex power delivered by the source in the circuit shown below. Please

express the answer in rectangular form. The answer should be given in terms of VA and

rounded to three decimal places. Round your answers to three decimal places.

− j 3 3 =

90 3 Ω

7 Ω j7 Ω

-j3 Ω 3 Ω30° A

90

I o

Vo 3 Ω8.5 + j5.5 Ω30° A

= 1.5 − j 1.5(3) (− j 3)

3 − j 3

7 + j 7 + = 8.5 + j 5.5(− j 3) 3

The circuit is reduced to that shown below.

I o = 90 = 8.5 + j 5.5

11.5 + j 5.530

10.124 × 90 (32.91 + 30 − 25.56)

12.748

= 71.479 A37.35

V o = 3 I o = 214.437 V37.35

S = V o I s *

= 214.437 9037.35 −30

S = 19299.299 = ( 19140.923 + j 2467.385) VA 7.35

In the op amp circuit shown below, v s

= 2 cos 103 t V. Find the average power delivered to

the 65‐‐‐‐kΩ resistor. Round the final answer to two decimal places if necessary.

C = 1 nF

1,000 kΩ

1 nF 65 kΩ

= = − j 1,000 kΩ1

jωC

− j

103 10

−9

At the non‐inverting terminal,

= V o = = 2 − V o 0°

1,000

V o

− j 1,000

2

1 + j

2

2−45°

V o ( t ) = cos (103 t − 45°)

2

2

P = = W V

2rms

R

2

2

1

2

2 1

65 103

P ≈ 15.38 µW

Find the complex power, the average power and the reactive power if

V = 60 V rms and Z = 90 Ω.

(Round your answer to 2 decimal digits if needed.)

45° 30°

S = = = 40 = 34.64 + j (20) VA | V |

2

Z *

(60)2

90 −30°30°

Average power = 34.64 W

Reactive power = 20 VAR

Assume that the values of voltage and current are the peak values.

Find the complex power delivered by v s

to the network shown.

Let Please express the answer as a phasor with positive

magnitude. For both the final answer and intermediate calculations, round magnitudes

to six decimal places and phases to two decimal places, if necessary.

v s

= 160 cos2400 t V.

30 µF =

vs

40 Ω 30 µF

i x

90 mH

50 Ω

5 i x

160 V

Io 40 Ω -j13.888889 V

o

I x

j216

50 Ω

5Ix

= − j 13.8888891

j ω C

1

j 2400 × 30 × 10−6

90 mH j ω L = j 2400 × 90 × 10−3

= j 216

We apply nodal analysis to the circuit shown below.

+ + = 0 where I x = V o − 160

40 − j 13.888889

V o − 0

j 216

V o − 5 I x

50

V o

j 216

1

40 − j 13.888889 = = 0.02231 + j 0.007747

40 + j 13.888889

1600 + 192.901238

= 0.023622 19.15

and

− = j 0.000463 V o

5I x

50

which leads to

(0.02231 + j 0.007747 − j 0.00463 + 0.02 + j 0.000463) V o =

= 160 0.023622 19.15

= 160 0.023622 19.15

or

(0.04231 + j 0.00358) V o = 0.042462 V o = 3.779524.84 19.15

Thus, V o = 89.009467 V = ( 86.247735 + j 22.000304) V14.31

I o = 160 − V o

40 − j 13.888889

= 160 (0.02231 + j 0.007747) − 89.009467 (0.023622 )14.31 19.15

= 3.5696 + j 1.23952 − 2.102582 33.46

= 3.5696 + j 1.23952 − ( 1.754123 + j 1.159268)

= 1.815477 + j 0.080252 = 1.81725 A2.53

S = V s (I o )* = 0.5 160 1.81725

1

20 −2.53

= 145.38 VA−2.53