hi,
attached are my working. Pls verify if my answer were correct.
first convert norton part to thevenin.
thus the current-ctrl voltage source is 70i & 20Ω in series with the circuit.
to compute V\(_{ab}\),
-10 + 10i + 20i - 70i=0
i= -0.25A
i(20)-(70i)= -5+17.5= 12.5V =V\(_{ab}\)
next to compute I\(_{sc}\)
short the ab terminal,
I\(_{sc}\)= i - 3.5i
v=0,
(-10+v)/10 = -i
i= 1
thus, I\(_{sc}\)= 1 - 3.5 = -2.5A
hence R\(_{th}\)= 12.5/-2.5 = -5Ω
i m wondering if i got them right.
attached are my working. Pls verify if my answer were correct.
first convert norton part to thevenin.
thus the current-ctrl voltage source is 70i & 20Ω in series with the circuit.
to compute V\(_{ab}\),
-10 + 10i + 20i - 70i=0
i= -0.25A
i(20)-(70i)= -5+17.5= 12.5V =V\(_{ab}\)
next to compute I\(_{sc}\)
short the ab terminal,
I\(_{sc}\)= i - 3.5i
v=0,
(-10+v)/10 = -i
i= 1
thus, I\(_{sc}\)= 1 - 3.5 = -2.5A
hence R\(_{th}\)= 12.5/-2.5 = -5Ω
i m wondering if i got them right.
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