circuit analysishomework
I need to solve exercise 57. I need to determine the voltage between points A and B in diagrams (a) and (b).
What I think so far:
Is it correct?
Thanks for your help.
You don't need meshes. Just calculate everything one source at a time, while keeping the other sources zeroed out (i.e. a voltage source is replaced with a wire, and a current source is replaced with an open (cut) circuit). Then sum the results from each source.
So in this case, the voltage source would see nothing but four resistors in series, and both current sources would see (different sets of) 2 series resistors connected in parallel with 2 other series resistors. Hopefully this helps you see how to "zero" the sources.
I'm not familiar with MIT's Circuit Sandbox, but it shouldn't matter where you put the ground since you just need to check the current flowing in different places - so just pick a spot and call it ground!
Let's say the model for your LED is \$V_{fwd}=1.56\:\textrm{V}+20\:\Omega\cdot I\$ and the schematic is as follows:
simulate this circuit – Schematic created using CircuitLab
Then I compute:
$$I_B=\frac{5\:\textrm{V}-V_{LED}-V_{BE}}{R_2+\left(\beta+1\right)\cdot R_1}\approx 150\:\mu\textrm{A}$$
From that (and \$\beta\approx 200\$) I would get that \$V_B\approx 4.5\:\textrm{V}\$ and \$V_E\$ about \$800\:\textrm{mV}\$ less, or about \$V_E\approx 3.7\:\textrm{V}\$. The LED drop isn't yet certain but let's say it's another \$2.4\:\textrm{V}\$ for now. This means \$1.3\:\textrm{V}\$ across \$R_1=47\:\Omega\$. Or about \$28\:\textrm{mA}\$.
Which isn't far from what you got. Given that lower current and the assumed model I laid out at the top, I'd refine my estimate of \$V_{LED}\approx 2.1\:\textrm{V}\$. This would add another \$300\:\textrm{mV}\$ across \$R_1\$ and add another \$6\:\textrm{mA}\$ to make it \$34\:\textrm{mA}\$. Putting that back in, I'd then get \$V_{LED}=2.2\:\textrm{V}\$, losing \$100\:\textrm{mV}\$ across \$R_1\$, dropping the current to \$32\:\textrm{mA}\$ as a still further refinement. (I could do that several times more to nail it down.) But my model was just made up from your single set of values for it in your question and I've no idea if it is very accurate. The main point here is that you should NOT be surprised.
But the above circuit isn't the way most people try and do this. And it's not how you'll see web sites laying it out, either.
Best Answer
If we apply the first law of Kirchhoff we can tell that the current in both resistors (in both circuits) is the same, as you said. Thus, the voltage fall on the 10 ohm resistors is 1V and the other is 2V. Now, you have to apply the second law to determine V between A and B.
We need to add al voltages (including the source) and equal them to 0. The only thing to have in mind is the "direction" of that fall. In the first, is is against the direction so you'll need to substract, whereas in the second you'll need to add.
Vab1 = +1.5 V
Vab2 = -7.5 V