Equivalent Resistance between A and B

Question

I am having difficulties finding the equivalent resistance between points A and B. While a solution is appreciated, I'm rather looking for an explanation/redrawing and understanding of the circuit. Thank you.

Answer 1

^{simulate this circuit – Schematic created using CircuitLab}

From this rearrangement, you can see that there are 4 nodes. One way to solve this is nodal analysis. You put a unit current source between a and b, and set B to 0, giving you three node equations:

$$A:\frac{(V_A-V_3)}{2}+\frac{(V_A-V_4)}{1}+\frac{(V_A-0)}{5}-1=0$$

$$3: \frac{(V_3-V_A)}{2}+\frac{(V_3-V_4)}{2}+\frac{(V_3-0)}{3}=0$$

$$4: \frac{(V_4-V_A)}{1}+\frac{(V_4-V_3)}{2}+\frac{(V_4-0)}{4}=0$$

You then solve for A to get the voltage. Since you know the current going into the system (1 Amp), you can use:

$$\frac{V}{I}=R$$

Since I=1, the resistance will actually just be the voltage at A.

Answer 2

The main thing here is to use delta-wye conversions to transform the resistors into a shape that can be analyzed.

^{simulate this circuit – Schematic created using CircuitLab}

Notice the ∏ shape formed by R3, R7, and R8. This can be transformed to a delta shape by using \$Ra = \frac{R7*R3}{R3+R7+R8}\$, \$Rb = \frac{R8*R3}{R3+R7+R8}\$, and \$Rc = \frac{R7*R8}{R3+R7+R8}\$. This gives you:

^{simulate this circuit}

Next, combine the series resistors R6 and Rc, giving:

^{simulate this circuit}

Figuring out the resistance should be fairly straightforward from here, yielding a final resistance of \$1.64\Omega\$

Answer 3

This hasn't a trivial solution which uses the common serial and parallel connection formulas. General circuit analysis methods must be used. An example which has only 2 unknowns to be solved:

Let node A to be your GND. Input voltage +1V to node B (minus to GND)

Unknown 1 = V1 = voltage of the joint of 1 Ohm and 4 Ohm resistors

Unknown 2 = V2 = voltage of the joint of 2 Ohm and 3 Ohm resistors.

Write normal nodal voltage analysis equations for V1 and V2 and solve them.

Calculate voltages over 3 and 4 Ohm resistors. (= one volt - V1 and one volt - V2)

Calculate the currents through 3,4 and 5 Ohm resistors.

Get the sum (=I) of the currents

Calculate the total resistance = 1V/I

Answer 4

The best way to solve this type of circuits is to apply the FACTs and, in this particular case, the extra-element theorem or EET forged by Dr. Middlebrook. The key to solving this type of exercise is to simplify and reorganize the circuit as correctly suggested by BeB00. A quick sim with a 1-A dc current source and a dc bias calculation shows the resistance value we want and obtained from both schematics: \$R_{AB}=1.639\;\Omega\$.

Now, to determine this value swiftly without effort but also without writing a single line of algebra, you can realize that \$R_1\$ is in || with the resistance you want from terminals A and B. Therefore, you can temporarily remove this resistance and will later place it back in || with the expression obtained without it to form the final result. This is the first step. Then, to determine the transfer function - because an impedance or here a resistance is a transfer function liking a response to a stimulus - I will install a current source \$I_T\$ (the stimulus) forcing a voltage response across its terminals, \$V_T\$. Obviously, the ratio of \$V_T\$ by \$I_T\$ is the resistance we want:

Then, to apply the EET, we have to identify an element whose presence troubles us to determine the transfer function. To me, it's \$R_6\$ and I select it as the extra element (you could select any component). I then have the choice to either replace it by a short circuit or open circuit it to calculate the reference value, \$R_{ref}\$. I choose to calculate the reference value when \$R_6\$ is open circuited. Redraw the circuit without \$R_1\$ and, in your head, what resistance do you "see" from the left-side connecting points?

You see - like if you would connect an ohm-meter in your head - the series-parallel combination of the resistors: \$R_{ref}=(R_2+R_3)||(R_4+R_5)\$.

Now, to apply the EET, you have to turn the excitation off. The excitation in this resistance determination exercise is the current source. Turn it to 0 A or open-circuit it. What resistance \$R_d\$ do you "see" between \$R_6\$ connecting terminals? \$R_d=(R_2+R_4)||(R_3+R_5)\$.

Then, you have to null the response in order to determine \$R_n\$. Nulling the response means that despite a current generator \$I_T\$ placed across \$R_6\$ terminals, the response (the voltage \$V_T\$ between A and B) is 0 V. A 0-V condition across a current source is a degenerate case and the current source can be replaced by a short circuit. Now, if you update the schematic as below in which the current source is shorted, what resistance \$R_n\$ do you "see" between \$R_6\$ connecting terminals? \$R_n=(R_2||R_3)+(R_4||R_5)\$.

This is it, it's done! Apply the EET formula as follows

\$R_{AB}=R_1||(R_{ref}\frac{1+\frac{R_n}{R_6}}{1+\frac{R_d}{R_6}})=R_1 || ((R_2+R_3)||(R_4+R_5)\times\frac{1+\frac{(R_2||R_3)+(R_4||R_5)}{R_6}}{1+\frac{(R_2+R_4)||(R_3+R_5)}{R_6}}))\$

and if you capture the formula in Mathcad, you have the exact value as given by SPICE:

I do not expect you to master the technique overnight but you can see how appealing these FACTs are: you can solve by inspection very often (no algebra) and you can individually correct the intermediate steps in case you identify a mistake. Really cool and I encourage EE students to acquire this skill which dates back to H. Bode book from 1945 : ) It is of invaluable help to determine transfer functions of any order. A tutorial on FACTs is available here.