Question

PLEASE ANSWER WITH CLEAR PROCESS AND DETAILS

Student Instructions: 1) Procedures: Using the following circuit models, modify the value of the voltage source ur I number as follows: Define the value of the left side voltage source as V1 and the right side voltage source as V2, where Vi= the sum of your first 3 digt of your D number and V2 = the sum of the rest of your ID number. (Example:If the ID. number is 021831, Vi=3V and V2=12V).. 2) Solve the circuit using the mesh-current method for problem 1 to findi.. j. and is 3) Solve the circuit using the mesh-current method for problem 2 to calculate current at the 2 Ohms resistance. 4) You have until next monday to complete this activity. 5) Total of 20 points each part value (10 points each case). 6) Your results will be summited in a PDF files showing the complete procedures and calculations. Use Blackboard platform to send the file(s). Problem OR **O2-0.0

V1= 8v

V2= 16

Answer 1

given circuit NESV A) 16. um um farsito-is menog va=l6v. Со tloop3 SL .) 16 V-GV it to- 'is takenog he voltage drop tot is takenas the voltage drop. Yhoop z Apply KVL for loop and loop2 by removing cardent source :- ar as in na u - Vg2 - Vz+V, = 0 . Bug 2. BIGV -3[ia-in] -2 [ic-ib] -1678 =0 -zia tsin -dictin = 8 zia - sig thic - - Apply KCl af Node ai i a ichib enlesing current at node = leaving cuttent of Node Karths Tá sia 6+5 = icky 15 = ic-ia)- Apply KUL for loof3 :- Vio - Vga - Vzx=0 loin -2 [ib-ic] -3 [ib-ia] = ė bia-15 is thic = o- After solving en by lareing calculator we get la= -4.4A ib= -0.8A ic=0.6A

given circuiti wa u in A loop3 doe loop di un flora Apply kul for Loops: lozib -vy-dio - Vao ao cuib - Qib - do [ib-ia] =0 -26 16 tloia 30-0 Apply KUL fet 100P2 : 210 t16-10-2 0 Dis +16 - loic -e [ic-ja] =0 2 in +16-10ic - Pictoria =0 dat dib- l2ic =-16-② Apply kufor loop! - 42o - V2 +8=0 -20[ia-ib] -2 [ia-ic] +8=0 -acia thoib teica-8- After solving equations. , L we get Ia=177102A custent through aon resistance Ib=1.3617A is Ic- Ia:16553 -1.77102 Ic=1:6553 A l IQ-y = 0.06498 Al