Hi there!
Does someone know why it is not possible to obtain the network equivalent resistance as seen from the terminal a-b by short circuiting the voltage sources and doing network reduction in this case? Could it have something to do with the dependent voltage source \(6I_x\)
By Network Reduction:
\( R' = 4 || 6 = 2.4 \Omega \)
Applying KVL on the open circuit a-b
\( -20 + 4I_x - 6I_x + 6I_x = 0 \)
\(
I_x = 5 (A)
V_{oc} = 30 V
\)
Applying KVL on the short circuit a-b
\(
I_x = 0
-20 + 4I = 0
I_{sc} = 5 A
\)
By Thevenin's Theorem:
\( R' = \frac{V_{oc}}{I_{sc}} = 6 \Omega\)
Does someone know why it is not possible to obtain the network equivalent resistance as seen from the terminal a-b by short circuiting the voltage sources and doing network reduction in this case? Could it have something to do with the dependent voltage source \(6I_x\)
By Network Reduction:
\( R' = 4 || 6 = 2.4 \Omega \)
Applying KVL on the open circuit a-b
\( -20 + 4I_x - 6I_x + 6I_x = 0 \)
\(
I_x = 5 (A)
V_{oc} = 30 V
\)
Applying KVL on the short circuit a-b
\(
I_x = 0
-20 + 4I = 0
I_{sc} = 5 A
\)
By Thevenin's Theorem:
\( R' = \frac{V_{oc}}{I_{sc}} = 6 \Omega\)