# Am I solving this basic Kirchoff's current law and Ohm's law problem correctly?

After moving onto some of the practice textbook exercises in the section covering Kirchhoff's current law and Ohm's law, I came across a problem which evades my best attempts at solving it.

Using Ohm's law, it was relatively easy to determine a equation involving voltage drops around the right loop:$$(1000\,\Omega)I_C+(1000\,\Omega)I_E=V_2\\(1000\,\Omega)(I_C+I_E)=V_2$$Similarly, it was simple to write an equation for the node where $I_B,I_E$ meet:$$I_B+I_C+150I_B=I_E\\151I_B+I_C=I_E$$... but given just $I_B=100\,\mu A=10^{-4}\,A$ it seems impossible to determine a numerical value for $I_C,I_E$.

Nevertheless it's straightforward to solve for $I_C,I_E$ in terms of $I_B,V_2$:I_C+I_E=\frac1{1000}V_2\\-I_C+I_E=151I_B\\\implies\begin{align*}2I_E&=151I_B+\frac1{1000}V_2\\2I_C&=-151I_B+\frac1{1000}V_2\end{align*}

Substituting $I_B=10^{-4}\,A$ yields $I_C=\frac1{20000}(10V_2-151),I_E=\frac1{20000}(10V_2+151)$

Am I approaching the problem correctly? Am I justified in ignoring the unlabeled voltage source bridging the two loops along with the labelled elements in the left one (i.e. $V_1,R_1,R_2$)? Am I treating the CCCS properly in writing my nodal equation?

In particular, a current source will look to a circuit like "whatever resistance" it needs to be in order for the correct current to flow. So you can't apply Ohm's law to the right hand circuit - instead you have to assume that the current source is doing its job, and that a current of 150 $I_B$ is flowing in the collector.
Incidentally - as drawn this circuit has a fatal flaw: there is no return path for the base current... Normally you would have a common ground between the left and right circuits, where the base current could return (and which would allow the emitter current to be $151 I_B$. So really - as drawn, this circuit doesn't quite work...
• what about the CCCS? what do I do about the current source supplying $150I_B$? – oldrinb Sep 5 '14 at 3:39
• Since $V_2$ is not given, you can ignore it. That's how transistors work. They provide "whatever resistance is needed" to make the simple equation work. At least in the DC model they are assuming here. We have to assume that $V_2$ is big enough that the voltage drops across the load resistors in the emitter and collector are less than the supply voltage. Absent information to the contrary that is a reasonable assumption. $V_2$, through the 1k resistors, is providing the current. – Floris Sep 5 '14 at 3:44
• the simple equation being $I_B+I_C=I_E$? what other equation do I use to solve for $I_C, I_E$ in terms of $I_B$? the loop equation? – oldrinb Sep 5 '14 at 3:47
• You also have $I_C = 150 I_B$. But see my updated answer - normally $I_E=I_B+I_C$ but that is not quite possible in this configuration since there is no return path for the base current... – Floris Sep 5 '14 at 3:49