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QUESTION: 1

In a two element series circuit, the applied voltage and the resulting current are v(t) = 60 + 66sin (10 t) V, i(t) = 2.3sin (10^{3}t + 68.3^{o}) A.The nature of the elements would be

Solution:

R-Ccauses a positive phase shift in voltage

QUESTION: 2

v_{o }(t) = ?

Solution:

10 sin (t + 30°) = 10 cos (t - 60°)

QUESTION: 3

V_{o} = ?

Solution:

QUESTION: 4

The circuit is as shown in fig.

i_{1}( t) = ?

Solution:

QUESTION: 5

The circuit is as shown in fig

i_{2}(t) = ?

Solution:

QUESTION: 6

I_{x} = ?

Solution:

QUESTION: 7

V_{x} = ?

Solution:

Let V_{o} be the voltage across current source

V_{o}(20 + j10) - (20 + j40) V_{x} = j600

QUESTION: 8

Determine the complex power for hte given values in question.

P = 269 W, Q = 150 VAR (capacitive)

Solution:

S = P-jQ = 269-j150 VA

QUESTION: 9

Determine the complex power for hte given valuesin question.

Q = 2000 VAR, pf =09. (leading)

Solution:

pf = cos θ = 0.9 ⇒ θ = 25.84°

Q = S sin θ ⇒

QUESTION: 10

Determine the complex power for hte given values in question.

S = 60 VA, Q = 45 VAR (inductive)

Solution:

Q = S sin θ ⇒

QUESTION: 11

Determine the complex power for hte given values in question.

V_{rms} = 220 V, P = 1 kW, |Z| = 40Ω (inductive)

Solution:

= 0.8264 or θ = 34.26°,

QUESTION: 12

Determine the complex power for hte given values in question

V_{rms} = 21∠20°V, V_{rms} = 21∠20°V, I_{rms} = 8.5∠-50°A

Solution:

S = V_{rms} I*_{rms} = (21∠20°)(8.5∠50°)

= 61+j167.7VA

QUESTION: 13

Determine the complex power for hte given values in question.

V_{rms} = 120∠30°V, Z = 40 + j80Ω

Solution:

= 72 + j144 VA

QUESTION: 14

In a two element series circuit, the applied voltage and the resulting current are v(t) = 60 + 66 sin (1000t) V, i(t) = 2.3sin (1000t + 68.3) 3 A. The nature of the elements would be

Solution:

QUESTION: 15

A relay coil is connected to a 210 V, 50 Hz supply. If it has resistance of 30Ω and an inductance of 0.5 H, the apparent power is

Solution:

Z= 30 + j(0.5)(2π)(50) = 30 + j157,

Apparent power = 275.6 VA

QUESTION: 16

In the circuit shown in fig. power factor is

Solution:

= 4 - j6 = 7.21∠ - 56.31°, pf = cos 56.31° = 0.555 leading

QUESTION: 17

The power factor seen by the voltage source is

Solution:

I_{1} = 1∠36.9°

pf = cos 36.9° = 0.8 leading

QUESTION: 18

The average power supplied by the dependent source is

Solution:

(2∠ - 90°)4.8 = -I_{x} (4.8 + j1.92) + 0.6I_{x}(8)

I_{x} = 5∠0°, V_{a} = 0.6 x 5 x 8 = 24∠0°,

QUESTION: 19

In the circuit of fig. the maximum power absorbed by Z_{L} is

Solution:

QUESTION: 20

The value of the load impedance, that would absorbs the maximum average power is

Solution:

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