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I am a complete beginner in circuit analysis and while reading the book „Circuit analysis for dummies” I’ve come across the following circuit:

enter image description here

What I don't understand is the "Vx" labelling assigned to voltage across resistor R1 and not R1 + R2 as it would occur from the source transformation applied to the Is current source (later in the text it is explicitely pointed out that Va = Vx). Here is what happens after the source transformation:

enter image description here Now, according to what wikipedia says (https://en.wikipedia.org/wiki/Dependent_source) the transformed circuit is a voltage controlled current source and the voltage across both the R1 and R2 resistors in series is a factor (Vx) to the current source (gVx), but the book says that only voltage across R1 should be the factor. Could someone tell me where I am wrong?

(Update) OK, all answers and comments so far tell me that I performed the source transformation too eagerly thus losing the original Vx. As far as I know, the first picture in my question shows the so called hybrid model of the circuit. How would the standard one (i.e. with a BJT transistor instead of the dependent source) look like? How would be the R2 resistor connected in such circuit?

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    \$\begingroup\$ A few hints: (1) The things are called resistors not registers. (2) The controlling quantitiy for the dependent current source is \$g v_x\$ and it is simply a given constant \$g\$ times the voltage marked as \$v_x\$ in the original circuit. There is no \$R_1\$ or \$R_2\$ involved so far. (3) If you do a source transformation (Norton source (\$i_s\$, \$R_1\$) --> Thevenin source (\$v_s\$, \$R_1\$) make sure that you leave the definiton for \$v_x\$ unchanged. I.e. you have to split your combined \$R_1 + R_2\$ to mark where \$v_x\$ is measured. \$\endgroup\$ – Curd Mar 20 '18 at 13:52
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    \$\begingroup\$ R1 is not an identifiable entity in the transformed circuit. \$\endgroup\$ – Chu Mar 20 '18 at 14:49
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You combined R1 and R2 to a single resistor after source transformation. Split it back to R1 and R2. Because \$ V_x\$ is the voltage across R1. If you combine it with R2, the voltage across them is not \$ V_x \$ anymore. It would be \$ V_x + V_{ab} \$

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    \$\begingroup\$ Right. To be exact: \$v_x\$ is the voltage across R1 only in the original circuit. After source transformation (without combining R1 and R2) \$v_x\$ is the voltage measured from the common node between R1 and R2 to the GND node. \$\endgroup\$ – Curd Mar 20 '18 at 14:57
  • \$\begingroup\$ In the real circuit with the actual BJT transistor what would be the role of the R2 resistor? Will it be placed in the collector-emitter part of the circuit? \$\endgroup\$ – krzysiekb Mar 20 '18 at 19:49
  • \$\begingroup\$ That would be another different question. \$\endgroup\$ – Meenie Leis Mar 20 '18 at 20:00
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    \$\begingroup\$ R1 and R2 cannot be 'split back'. In the Thevenin equivalent of Is/R1/R2, R1 has zero volts across it; in the original circuit R1 has IsR1 volts across it. The equivalent circuit is not a simple re-arrangement of the original components. \$\endgroup\$ – Chu Mar 20 '18 at 22:20

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