# Simple circuit analysis with dependent source

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

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:

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?

• 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. – Curd Mar 20 '18 at 13:52
• R1 is not an identifiable entity in the transformed circuit. – Chu Mar 20 '18 at 14:49

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 $V_x + V_{ab}$

• 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. – Curd Mar 20 '18 at 14:57
• 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? – krzysiekb Mar 20 '18 at 19:49
• That would be another different question. – Meenie Leis Mar 20 '18 at 20:00
• 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. – Chu Mar 20 '18 at 22:20