Determine the voltage drop from the top terminal to the bottom terminal, vab, in the right hand branch and, vcd, in the left hand branch of the circuit. Determine each voltage drop based on the elements in the corresponding branch.

Hello your question is incomplete attached below is the missing part of the question

answer ;

voltage drop in the Vcd branch = 30 V

Voltage drop in the middle branch = 40v - 30v = 10 volts

voltage drop in AB = 60 + ( -600 * 0.05 ) = 60 - 30 = 30 volts

Explanation:

Determine voltage drop from top terminal to bottom terminal ( Vab ) in the right hand branch and Vcd in left hand branch

40v and 50mA are in series hence;  Ix = 50mA

also Vcd = 30V

CD is parallel to AB hence; Vcd = Vab = 30 V

Vab = ∝*Ix + 60 v

 30v  = ∝ ( 50mA ) + 60

therefore ∝ = -600

voltage drop in the Vcd branch = 30 V

Voltage drop in the middle branch = 40v - 30v = 10 volts

voltage drop in AB = 60 + ( -600 * 0.05 ) = 60 - 30 = 30 volts

answer:

ask me the quest

explanation: