Question

I AJ Howing in the circuit in the figure. (Indicate the direction with the sign of your answer. Due to the nature of this prot not use rounded intermediate values in your calculations-including answers submitted in WebAssign.) E-240 V 0.106 R 5.00 E-48.0V R 220 R 0502 380 4 30.2002 R 66 0 36.0 V 2. = 6.0V w 10.050 h А 12 13 A Explicitly show how you follow the steps in the Problem Solving Strategies for Series and Parallel Resistors. (Submit a file with a maxi 1 MB) Choose File no file selected +

Answer 1

Criven, circuit :- Matrix : Oslo 24 im 222 38.5 66.2 27.1 -38.5 0 24V 48V no -24 T 0.501 38.1 - 661 0.2012 36 V 0.05r 6V we have 3 loops! using Kirchoff's law ! 0 loop! - 6 + 66 13 -48 +0.5 I2 +38 I2 +0.2I, +30 - 0 66.213 38.5 I, teens -24 0 from wor 6 - 24 V + 0.11, + 221, -3812 -0.5 IZ +5I, +48=0 27.11, 38.512 +24 = 0 and we have Is It Iz egin 0,0 & ③. uring we get, I, - - 0.29498 A Iz = +0.41574 A Auswer Iz z to. 120756A

Uvpote, if the answer was helpful. Have a nice and safe day.