- Q He Xc Llll Ih W M For The Circuit In The Below Figure Xl 4 5 Kohm Xc 5 Kohm R1 2 2 Kohm R2 8 2 Kohm R 1 (12.04 KiB) Viewed 34 times
- Q He Xc Llll Ih W M For The Circuit In The Below Figure Xl 4 5 Kohm Xc 5 Kohm R1 2 2 Kohm R2 8 2 Kohm R 2 (67.96 KiB) Viewed 34 times
- Q He Xc Llll Ih W M For The Circuit In The Below Figure Xl 4 5 Kohm Xc 5 Kohm R1 2 2 Kohm R2 8 2 Kohm R 3 (61.35 KiB) Viewed 34 times
- Q He Xc Llll Ih W M For The Circuit In The Below Figure Xl 4 5 Kohm Xc 5 Kohm R1 2 2 Kohm R2 8 2 Kohm R 4 (18.36 KiB) Viewed 34 times
For the circuit in the below figure, XL = 4.5 kOhm, XC = 5 kOhm, R1 = 2.2 kOhm, R2 = 8.2 kOhm, R3 = 2.7 kOhm, 1V1 | = 24 V phase 0 degrees, and V2] = 18 V phase 90. Use the superposition theorem to determine which of the following is true? Round up your answer to exactly two decimal points. Warning: selecting a wrong answer takes marks away. None of the answers |Zt_V11 = 3.66 kOhm (total impedance for V1) Angle of Zt_V1 = -40.59 degrees (total impedance = for V1) 0 1Zt_v11 = 2.26 kOhm (total impedance for V1) 1Zt_V1 | = 6.45 kOhm (total impedance for V1) Angle of Zt_V1 = -52.13 degrees (total impedance for V1) Angle of Zt_V1 = -52.14 degrees (total impedance for V1)
For the circuit in the below figure, XL1 = 15 Ohm, XL2 = 20 Ohm, XC = 90 Ohm, R = 20 Ohm, Vs] = 20 V phase 160 degrees, and is = 1.6 A phase 160 degrees. Use Mesh analysis to determine which of the following is true? Round up your answer to exactly two decimal points. Warning: selecting a wrong answer takes marks away. 0 (+j5) 11 + (-j20 ) 13 = (-27-j15.71 ) --> (Equation for Loop-1) (-20) 11 + ( 20 + j110) 13 = ( 49.25 + j135.32 ) --> (Equation for Loop-3) 0 +>( (+j5 ) 11 + (-130 ) 13 = (-7.18 +j64.84 ) --> (Equation for Loop-1) 0 (+j5 ) 11 + (-130 ) 13 = (-14.57 +j12.36 ) --> > (Equation for Loop-1) 0 > (-j30 ) 11 + ( 51 + j55 ) 13 = ( 15.39-j42.29 ) --> (Equation for Loop-3) None of the answers 0 (-130 ) 11 + ( 11 + j75 ) 13 = ( 8.86-j1.56 ) --> (Equation for Loop-3)
R w XLI (13 000 XL2 m Xc E Vs 2. Is