Question

The figure below shows a multiloop circult. The voltages and resistances in the circuit are: V, - 14,0 V, V, - 36.0 V, V, - 54,0 VR - 120.00, R2 - 140.00, R, - 180.00. Use these values to answer the questions below. (Use the following as necessary: 1. Is, and .) B aminy intit D a) write the equation for the loop ABHEPGA in the order that you go around the loop O-I(140.00) - 36,0 V + 14.0 V-1(120.00) - 0 +1 (140.00) 36.0V +14.0V-1120.00) O +13 (140,00) 54.0 V-14.0 V-1 (140.00) - 0 O -(180.00) -S4.0 V +14,0 V -1,140,00) - 0 (b) write the equation for the loop BCKDEHB in the order that you go around the loon + (140.00) 36.0 V +14.0 V-(120.00) - 0 O +1 (180.0 m) -54.0 V + 36.0V + (14000) - O +13 (140.00) 54,0 V -14.0 V - (140.00 - 0 O-I, (18000) -54,0 V +36.0 V-1, (140.00) - O (C) Write the equation for the ABCDEFGA in the order that you go around the loop +1, (140.00) +36.0V-54.0V -> (120.00 - 0 O-13(1800 M -54.0 V +14.0 V-14000) - 0 +1, (150.00) -54.0 V + 14,0 V - I, (120.00) - O +1, (140,00) +36.0V +140 V, 120.000

Answer 1

Q Given that voltages Vi= 14.0 V, V = 360 V, V3 -54.0V Resistens R, = 120.0r, R= 140.0r and Rg = 180.00 ond CKT Apply KVL svi Ro sum Vo = voltoge drop R cicross ith element G H + _< V take cunent as positive in direction of loop E (a) 08 At loop 20 ABHEFGA loop 0 Li R, + B R + V, -V, = 0 12 (140.02) +i (120.01) + 36.00 – 14.00 =0 OR -iz (140.01) - 36.0V +14.00 - (120.00) This is correct option. (b) Apply KVL at lock BCKDEHB loop ③ -BR2-ig Rg + Vy -V2 = 0 i Rz - V₂ +V2 tl2 R₂ =0 iz (180.02) -54.00 + 36.0V +121140. or) So This is consect option

us (O Athly KUL at loop ABCKD EFGA loop 3 lir, -iz R₂ +V₂ =V, = 0 i Rz U₂ +Voir 1180.02) -54.00 +14.00 (120.02) R MALAL =o This iſ the Conect option