Question

R2 R1 0.8K 0.7K L3 L2 L1 1.2mH 1.5mH 1mH C2 C1 2.2nF 2.4nF 10 360 us2 Usl 14 < 00

Frequency is 60Hz

• Use mesh analysis and Phasor approach to find both Phasors and the corresponding time domain representation of the current through and voltage across each circuit element
• Verify the correctness of the results by means of Tellegen’s theorem (Power Balance)
• Draw the phasor diagram with respect to either extraordinary node

Answer 1

Cus He giren as 6oHz = 23x6o = 1207 Yad) CLe fquny thun, SReaetances of euments: . (L21 J.5mM 377 xImH L21.2mi jo3775 2-2nF 3 XL= JwLz =J377X ! -Sm H = jo .565S2 G n G us2 -il-lo5X lo X wc 377X24X1-9 = j2 206 Xlo'n Xes w 142D 372.2 X l0"9 Now for song uing merh gpply KVL in loop-:- Rt Rt ixi^xq-im.)m, + (a+i Me-ive-»} = {I4 4036) (R frco +8oo t jo 377+jo.yg 2y -j/tosXIo-j2 206x/»6 -(Boojo.us24 -j2-206%lo)T2 = 1y - l0236 8-335 - (22059 99-69-89.I9) 444 (33/0 999.512-89-97) -8.3354.84 - (I- Rt JXi^-JX^) o -us +I2 (Boot joy 5 24 p j2-206 X jo-5655 2 2205999.127289.3792) - (29b5999.93 L-81-9732=o36° soling '33 l0 999-51 L-89-1) NOw g.9g) (10ac)I, (22459F693 L-89 977 S999.693 -89 - (220. 220 59927 2-89.9792) vaasr 8.33SL-44.8y as to3gy-18947 -oos 92)J (220 59I36.93 -89.979 8-335L-444 + 00L36+ 79899 o b9) I5y22 Z-58.66 5y22 L-58.66 uA

Now, T2 OL36 + T ( 220 5999 6 93 L-89 .JF92 I2 = 2205999 127 2-89-2792) I2 9-753 XID -80 74 2 Amp J.753-80 742' Hence avount () (5-422 L-5866° uA) - (9753 -807u2un) 5.1493 75-9yuA Ref.) #finding cweunt and volhage accos the each efemen:- (Take Us, as Vo ltag Time domain. phcsor cwvant (uA) Element SNo Time clomain Rep. phusor Rep (I-2-A VRI 3354XJ Sin(377t-ss) 99-S5-7 |3954 58-68) -5422 X SE Sin(377t 1. Z 5.422-58:66 R1 5422S8.66UA 5422J2 sinl377-E -536) 2. Sin (s7731-3 31.34 5.492 J2 sinf377t| U= 5-99VI = 5992 sinl37785 C1 |71= 5-y222-5866 u 58.6)uA 99.8ni- lls 5-u93sin{397t| VR3 =3-629591 D36J2 sin (371t 75.gu).uA (T-I2) = 5-/M93E594 +7594)m UA |(72-32)= D 5-4935-94°MA V2233765-94 MVolk D514 33JEsin(377 t +75.94 5. 2.332 Sin A(h591att8) ,90-h/-787-11=3 1-29/2 05.1493275 94° UAP514 93 J2 sin (3?7t +75.940UA Volls Sin37t-1to6 Volh I2-9.53-go.742g7532 sin(377t-8a72 9.258 tauz uVolh 5-515J Sin (377|- f teaze .258) V5.515 M

in l00p-- F powo Balance ( Tellegen's Theorem) Dolng (shouldl be egual Ho) MAnaanas36 VR+Ve+ Vi-2) o 3) ) -66)A 1347594)uA (3- 79542-58 '66 mV + &.045231.3y uV 7 5-992-148-66)( 5 422 2-S8 Yolh 4 mrナ 2-332165-9y uvt.282- hosキフ9-]+ 22-58) (-1046) S-u 22 3-2478ム152-7 5.8 Xlo-s 99.85 85-7804t)(668 19750). +/5.1493 -68-06 でE1673199 erlenterty top2 k) (G-613891-302) Xiu-5 Hmee proreedt This can be Paaurng phasar diayun.. VL3 Us=4Lo Ver