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 (103t + 68.3o) A.The nature of the elements would be
R-Ccauses a positive phase shift in voltage
vo (t) = ?
10 sin (t + 30°) = 10 cos (t - 60°)
Vo = ?
The circuit is as shown in fig.
i1( t) = ?
The circuit is as shown in fig
i2(t) = ?
Ix = ?
Vx = ?
Let Vo be the voltage across current source
Vo(20 + j10) - (20 + j40) Vx = j600
Determine the complex power for hte given values in question.
P = 269 W, Q = 150 VAR (capacitive)
S = P-jQ = 269-j150 VA
Determine the complex power for hte given valuesin question.
Q = 2000 VAR, pf =09. (leading)
pf = cos θ = 0.9 ⇒ θ = 25.84°
Q = S sin θ ⇒
Determine the complex power for hte given values in question.
S = 60 VA, Q = 45 VAR (inductive)
Q = S sin θ ⇒
Determine the complex power for hte given values in question.
Vrms = 220 V, P = 1 kW, |Z| = 40Ω (inductive)
= 0.8264 or θ = 34.26°,
Determine the complex power for hte given values in question
Vrms = 21∠20°V, Vrms = 21∠20°V, Irms = 8.5∠-50°A
S = Vrms I*rms = (21∠20°)(8.5∠50°)
= 61+j167.7VA
Determine the complex power for hte given values in question.
Vrms = 120∠30°V, Z = 40 + j80Ω
= 72 + j144 VA
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
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
Z= 30 + j(0.5)(2π)(50) = 30 + j157,
Apparent power = 275.6 VA
In the circuit shown in fig. power factor is
= 4 - j6 = 7.21∠ - 56.31°, pf = cos 56.31° = 0.555 leading
The power factor seen by the voltage source is
I1 = 1∠36.9°
pf = cos 36.9° = 0.8 leading
The average power supplied by the dependent source is
(2∠ - 90°)4.8 = -Ix (4.8 + j1.92) + 0.6Ix(8)
Ix = 5∠0°, Va = 0.6 x 5 x 8 = 24∠0°,
In the circuit of fig. the maximum power absorbed by ZL is
The value of the load impedance, that would absorbs the maximum average power is
Use Code STAYHOME200 and get INR 200 additional OFF
|
Use Coupon Code |
![]() |
|
![]() |
|
![]() |
|
![]() |
|
|
|
|
|
|