I am well aware that Thevenin voltage is equal to the open circuit voltage at the end of the circuit.
The circuit information are: \$𝐕_𝐴\$=50∠0 V, \$𝐈_𝐵\$=5∠0 A, \$𝑅_1\$=10 Ω, \$𝑅_2=10\$ Ω, and \$𝑍_𝐿=10i\$ Ω.
The circuit diagram is this:
My question is why is thevenin voltage equal to source voltage here. There should be a voltage drop cross R1 right? Can anyone help with a clear information on this?
From KCL at the top-middle node: $$I_2 = I_1 + I_B$$
From KVL at the leftmost loop: $$V_A -V_{R_1} - V_{R_2} = 0$$ $$V_A - I_1 \cdot R_1 - I_2 \cdot R_2 = 0$$ $$V_A - I_1 \cdot R_1 - (I_1 + I_B) \cdot R_2 = 0$$
Solving for \$I_1\$, after substituting the rest of the variables, yields: $$I_1=0[A]$$
Therefore, there is no voltage drop across R1, and the Thevenin voltage is equal to \$V_A\$.
My question is why is thevenin voltage equal to source voltage here. There should be a voltage drop cross R1 right? Can anyone help with a clear information on this?
In this particular example, there is no current passing through \$R_1\$ because of the presence of \$I_B\$ and \$R_2\$. Coincidentally, they produce the same 50 volts terminal voltage as \$V_A\$. Thus, no current flows through \$R_1\$ and, the Thevenin voltage is wholly determined by \$I_B\$ and \$R_2\$ (which happens to be coincidentally the same value as \$V_A\$).
In general, for this topology, \$V_{Th}\ne V_S\$.
The equality exists for these values only. Change any of \$V_S\$, \$R_2\$, or \$I_B\$ will effect an inequality instead.
Change any two such that \$V_S=R_2 I_B\$ then the equality will remain.
