- #1
An1MuS
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As seen from a and b.
My teacher did the norton's equivalent in the class, and now for studying purposes i was trying to get to the Thévenin's one. However it seems my equations are somehow wrong, and i can't figure out why.
The Eq resistance is 8||2, which is 1.6 ohms.
I tried to apply Nodal analysis first
so,
for node V1: [tex]\frac{20-V_1}{3}=\frac{V_1-V_3}{2}[/tex]
for node V2: [tex]\frac{V_1-V_3}{2}=6+\frac{V_2-0}{5}=[/tex]
The relationship between two voltages is also known, which is [tex]V_2=V_3+10[/tex]
which gives V1=2 ; V2=0 ; V3=-10.
V1-V2 should indeed be 2, because from my teachers resolution, Vab = 2V. I don't get is how can V2 be 0? Then there can't be another voltage drop when it reaches the ground (0V)
Also, then i tried mesh analysis:
For mesh 1[tex]20+10=3I_1+2I_1+5(I_1+I_2)[/tex]
For mesh 2 [tex]0=5(I_2-I_1)+v[/tex]
[tex]I_2=6[/tex]
Where v is the voltage drop across the current source.
It gives I1 = 6 and v (which is equal to the V2 of the nodal analysis) = 0 again.
My teacher did the norton's equivalent in the class, and now for studying purposes i was trying to get to the Thévenin's one. However it seems my equations are somehow wrong, and i can't figure out why.
The Eq resistance is 8||2, which is 1.6 ohms.
I tried to apply Nodal analysis first
so,
for node V1: [tex]\frac{20-V_1}{3}=\frac{V_1-V_3}{2}[/tex]
for node V2: [tex]\frac{V_1-V_3}{2}=6+\frac{V_2-0}{5}=[/tex]
The relationship between two voltages is also known, which is [tex]V_2=V_3+10[/tex]
which gives V1=2 ; V2=0 ; V3=-10.
V1-V2 should indeed be 2, because from my teachers resolution, Vab = 2V. I don't get is how can V2 be 0? Then there can't be another voltage drop when it reaches the ground (0V)
Also, then i tried mesh analysis:
For mesh 1[tex]20+10=3I_1+2I_1+5(I_1+I_2)[/tex]
For mesh 2 [tex]0=5(I_2-I_1)+v[/tex]
[tex]I_2=6[/tex]
Where v is the voltage drop across the current source.
It gives I1 = 6 and v (which is equal to the V2 of the nodal analysis) = 0 again.