There are several ways to determine the resistance offered from terminals A and B. However, I recommend using the Fast Analytical Circuits Techniques or FACTs based on the Extra-Element Theorem or EET (see https://en.wikipedia.org/wiki/Extra_element_theorem). You will see how easy it is to determine what you need just by going through a few sketches and without writing a single line of algebra. We must first select who is this "extra" element. Usually, you chose the one that bothers you in your analysis. Here, for me, it is the \$4\;\Omega\$ resistor in the left side of your circuit but you could select any other one. We can decide to remove this element (temporarily set it to an infinite value) or make it 0 \$\Omega\$ for the analysis. Let's make it zero and replace it by a short circuit. What resistance do you "see" from terminals A and B if the \$4\;\Omega\$ resistor is replaced by a short circuit? You see: \$R_{ref}=R_1||R_2+R_3||R_4\$.
Generally speaking, if you want to determine the resistance seen from A and B, you install a test generator \$I_T\$ feeding A and B which will develop a voltage \$V_T\$ across its terminals. The resistance you want or the transfer function you look for is the ratio of a response signal, \$V_T\$, by the excitation signal or stimulus, \$I_T\$. The next part of the exercise will consist of reducing the stimulus to 0 (\$I_T\$ is 0 A) and calculating the resistance offered by the \$4\;\Omega\$ resistor terminals when temporarily removed from the circuit. We will call it \$R_d\$. If we remove the \$4\;\Omega\$ resistor from the circuit and "look" at the resistance offered by its terminals while \$I_T\$ is 0 A (current source is open-circuited), then we find \$R_d=(R_1+R_2)||(R_3+R_4)\$. Now, the final part. We will run the same exercise but this time nulling the response, implying that \$V_T\$ is zero while we look at the resistance offered by the \$4\;\Omega\$ resistor terminals. This is a degenerate case and a current source whose voltage across its terminals is 0 V can be replaced by a short circuit. In this case, the resistance \$R_n\$ observed from the \$4\;\Omega\$ resistor while a short is installed between A and B is equal to \$R_n=R_1||R_3+R_2||R_4\$. This is it, we have everything to apply the EET and find that the resistance offered by terminals A and B is equal to:
\$R_{AB}=R_{ref}\frac{1+\frac{R_5}{R_n}}{1+\frac{R_5}{R_d}}=(R_1||R_2+R_3||R_4)\frac{1+\frac{R_5}{R_1||R_3+R_2||R_4}}{1+\frac{R_5}{(R_1+R_2)||(R_3+R_4)}}\$
If you apply the numerical values of your drawing, you find \$R_{AB}=2.07143\;\Omega\$
If have captured a quick SPICE simulation showing the few sketches you have to go through. No equations at all, just inspection!
If you now chose to have the reference circuit being the \$4\;\Omega\$ resistor to be infinite-valued, simply remove it from the circuit and determine what resistance you "see" from terminals A and B when that resistance is remove:\$R_{refinf}=(R_1+R_3)||(R_2+R_4)\$. The EET can now be reformulated considering this new reference value:
\$R_{AB}=R_{refinf}\frac{1+\frac{R_n}{R_5}}{1+\frac{R_d}{R_5}}=(R_1+R_3)||(R_2+R_4)\frac{1+\frac{R_1||R_3+R_2||R_4}{R_5}}{1+\frac{(R_1+R_2)||(R_3+R_4)}{R_5}}\$
If you do the maths, results are exactly similar. Both expressions are so-called low-entropy expressions.
This is the beauty of the FACTs! Very often, you can determine the transfer functions of a complicated passive circuit just by inspection, without writing a single line of algebra. If you are interested - and as a student I encourage you to acquire this skill - have a look at
http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf
and the examples solved in the book
http://cbasso.pagesperso-orange.fr/Downloads/Book/List%20of%20FACTs%20examples.pdf
My guess is no. Because both of those points are connected to a ground symbol by blank wires, they both have potential of 0V.
Your guess is right, but your reasoning is incorrect.
In any idealized circuit like this, the potential is equal at all points on the wire, but we still allow that current can flow. Because the wire is modeled to have 0 resistance, it doesn't require any potential difference across it to establish a current.
The reason there is no current in this wire is because of the cut-set form of KCL. If you were to break that wire, there'd be no other connection between the two halves of the circuit. Therefore no complete circuit can form using that wire, and so we know no current is flowing through it even when it isn't broken.
Best Answer
When the transistor is turned on in that manner it can be switched into saturation if the base current is sufficient. The collector-emitter emitter voltage will drop to about 0.2 V. This voltage will be applied to the LED.
Figure 1. Current through various colours of LED as a function of forward voltage. Image source IV curves.
Figure 1 shows that none of the LEDs from infrared to ultra-violet will pass any significant current at 0.2 V. There just isn't enough voltage to get the charge carriers to jump the P-N junction.
Figure 2. A water check-valve analogy. Image source: What is an LED?.
If you look at the check-valve in the figure above, it should be clear that the spring normally keeps the ball in position and prevents back-flow. When “forward-biased” the ball shut-off can be moved against the spring but it will take some initial pressure to move the ball. This results in a pressure drop across the valve: the pressure downstream will be less than the inlet pressure.
In a similar manner the PN junction in an LED causes a voltage drop. For a red LED it is about 1.5 V to 2.0 V. You need to exceed the Vf to get enough current to flow and light the LED.
The links are to articles by me and may help you further.