Question

Find the currents (in A) flowing in the circuit in the figure. (Indicate the direction with the sign of your answer. Due to the nature of this problem, do not use rounded intermediate values in your calculations, including answers submitted in WebAssign.) E, = 24.0 V b w 0.102 RS R 5.02 E, = 48.0 V 17 R e 1 0.502 462 >0.2002 R 822 E4 = 36.0 V Eg = 6.0V 13 i 0.05 2 h A A 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 maximum size of 1 MB.) Choose File No file chosen

Answer 1

E24V b 71 H m d Ri Billie 17 52 R5 22.2 E-46V =0.512 R2 a е. 46.2 Iz P3 {II 74 = 0.212 82522 Ej=6V 73 E4 = 36V 0.05.12 مه For loops, we can use Kirchhoff's loop rule For junction as using junction rule, Ti + 12 Iz For loop abdea, IRI+I, & + I, Rs - Iq Rz - Izr2 = Ei-Eq. >> I, CRI + xy +R5) – 12 ( R2+Y2) = £,- Ea. >> I. (5 +0.1 +22) – 12(46+0.5) = 24 - 48 => 27.1 I 46:5 12 = -24 2 . CS Scanned with CamScanner

For loop regja, Iz 82 + I2 R2 + I3 84 + I3 3 + I3 R₂ = E2-E4 + €3 - I2 CO2 + R2)+ I3 (4+13 + R3) - Ex- Es + Ez >> Iz C 0.5+46) + 13 (0.2+0.05+82) = 48-36 +6 - 46.5 Iz +2.25 I3 =18 > from @ 46:5 I2 + 82.25 ( I+ In) 18, → 46.5 I2 + 82.25 I, +82.25 I2 = 18 82.25 I, + 128.75 I2 = 18 Iz 18-82.25 Ii 128.75 put value of Is from ③ in ① > 27.11 46.5 18-82:25 Ii = -24. 128.75 27.1 Ii 6.5 + 29.706 I, = -24 13 ( 82.255) > = -24 +6.5 56.8058 Il 0.30806 A 1, => CAns) CS Scanned with CamScanner

From equation ® , I, - (8 82.25 % (-0.30806) 128.75 0.3366 A (Ans) From equation o Caus) I3 0.02854 A CS Scanned with CamScanner