Solve the current ia(t) in the following circuit: 50 20 20 cos(5t + 45°)V 3H 50 Solve the current i,(t) in the following circuit: 50 is(t) 20 3H 50 3 sin(2t) A The following circuit in problem 2(c) is the combination of the circuits in part 2(a) and part 2(b). Use the superposition principal to find i(t). [To elaborate, write down the equation for i¿(t) based on the results found for part 2(a) and part 2(b)] 20 cos(5t + 45°)V( 3H 3 sin(2t) A 1/3 ll

Delmar's Standard Textbook Of Electricity
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Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter24: Resistive-inductive-capacitive Parallel Circuits
Section: Chapter Questions
Problem 1RQ: An AC circuit contains a 24 resistor, a 15.9-mH inductor, and a 13.3F capacitor connected in...
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Solve the current ia(t) in the following circuit:
50
20
20 cos(5t + 45°)V
3H
Solve the current i,(t) in the following circuit:
50
is(t)
20
3H
)3 sin(2t) A
The following circuit in problem 2(c) is the combination of the circuits in part 2(a) and part 2(b).
Use the superposition principal to find i(t). [To elaborate, write down the equation for i¿(t)
based on the results found for part 2(a) and part 2(b)]
20 cos(5t + 45°)V(
3H
3 sin(2t) A
ll
Transcribed Image Text:Solve the current ia(t) in the following circuit: 50 20 20 cos(5t + 45°)V 3H Solve the current i,(t) in the following circuit: 50 is(t) 20 3H )3 sin(2t) A The following circuit in problem 2(c) is the combination of the circuits in part 2(a) and part 2(b). Use the superposition principal to find i(t). [To elaborate, write down the equation for i¿(t) based on the results found for part 2(a) and part 2(b)] 20 cos(5t + 45°)V( 3H 3 sin(2t) A ll
with w= zrads 1, Chouse iii
Broblem z
a). voltage sourie: 206430, Capaciton -}ja, Inductore 1878e.
(V- 20c40)15 + vl15;+ Vil(7-0.bj) =0
talt)
Vi Š t s t o.14+ o.0lj) = 4L45% Vi (D.34-0.057;)= 4645°
Vi-0.34<350.48° = 4L45° Vi= 11.76L-305.48°
Ia= 11.762-30s.48° /1-0.6j) = 1766-305.48/7.03635.1=トレ7ムー660.88°
= 67-300.558° talt)= 167 c0s(5t+59.4で)A
b) current source: 36-90°, Capacitor: -15je, Inductm: bjh.
Rth = 552, Vth#32-90° x5=15-90°.
VA/F + VA/bj +(VA>1544900)/(7-15;)=0.
16-ト分)= D.137+ e.oz9j= a.14211.95°
VA(+ to.137to.029;)= 182-90° x o.14Ll1.9560
VA(ţ + to.37 to.0293) = 2-14-101.95°
VA (0.337-0.138j) = z./2-101.95° VA 0.3642=22.2692=;
2.12-10190 0=57592-79.681° I6%=(M-け2-10°)/(7-65)
%3D
ード
Tbit)
O32-70
%3D
%3D
VA
50
5n
Transcribed Image Text:with w= zrads 1, Chouse iii Broblem z a). voltage sourie: 206430, Capaciton -}ja, Inductore 1878e. (V- 20c40)15 + vl15;+ Vil(7-0.bj) =0 talt) Vi Š t s t o.14+ o.0lj) = 4L45% Vi (D.34-0.057;)= 4645° Vi-0.34<350.48° = 4L45° Vi= 11.76L-305.48° Ia= 11.762-30s.48° /1-0.6j) = 1766-305.48/7.03635.1=トレ7ムー660.88° = 67-300.558° talt)= 167 c0s(5t+59.4で)A b) current source: 36-90°, Capacitor: -15je, Inductm: bjh. Rth = 552, Vth#32-90° x5=15-90°. VA/F + VA/bj +(VA>1544900)/(7-15;)=0. 16-ト分)= D.137+ e.oz9j= a.14211.95° VA(+ to.137to.029;)= 182-90° x o.14Ll1.9560 VA(ţ + to.37 to.0293) = 2-14-101.95° VA (0.337-0.138j) = z./2-101.95° VA 0.3642=22.2692=; 2.12-10190 0=57592-79.681° I6%=(M-け2-10°)/(7-65) %3D ード Tbit) O32-70 %3D %3D VA 50 5n
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