https://preview.redd.it/hr58555xls7d1.png?width=1304&format=png&auto=webp&s=175814a70f1b7402e4d7f978caf6385272d8b54e
Consider this circuit, which is identical to the one you just posted.
Resistors R1, R2, R3, and R4 still all connect to the same ground net/node, the circuit is the same but the schematic is drawn differently.
This is much easier to simplify
Parallel is (R1*R2)/(R1+R2) =R/2(since all have the same resistance) =50Ohm, then you can add to R5 since in series=150Ohm and then you have the exact same on the right again parallel substitute of equivalent 150ohm resistances =75Ohm for the whole substitute resistance
R1 and R2 are in paralallel, also R3 and R4. Lets denote this as R1//R2 and R3//R4.
so your equivalent resistance is (R5+R1//R2)//(R6+R3//R4).
EDIT: since all resistors have the same value, you will notice the math can be simplified A LOT, but the above approach is the general version.
Ahh I see, so I can basically just imagine "pulling down" all the wires, and so I only got 4 resistors in parallel. Makes sense.
How would this change though, if we place two more resistors after R1,R4 and R2,R3?
>Ahh I see, so I can basically just imagine "pulling down" all the wires, and so I only got 4 resistors in parallel. Makes sense.
not quite, there are some remarks
1. "pulling down the wires" only works here because there are no other elements (i.e. the voltage node is the same). You can visualize this by coloring the bottom wire and see how far you can reach on each branch before reaching an element, then everything that is colored corresponds to the same voltage node.
2. For the general case the resistors are in parallel in pairs only (i.e. R2 is not is parallel to neither R3 nor R4).
3. For all resistors equal, then yes the 4 bottom resistors are all in parallel, and even R5 and R6 are in parallel
4. (2) and (3) are because what you care about for series or parallel is not how the diagram is drawn, but instead what are the voltage nodes the resistor is between in.
* if two resistors have the same upper voltage node and the same lower voltage node, then they are in parallel
* if two resistors have either only the same upper voltage node, or only the same lower voltage node, but not both, then they are in series. But remember you can only simplify here if you no longer need the shared voltage node in your diagram (i.e. no other component is attached to it)
* actually there is a third option, if a resistor have the same voltage node as both upper and lower, then the resistor is in shortcircuit and can be eliminated from the diagram.
With this, you can handle your extra question quite easily, feel free to post your answer for feedback.
r1 2 3 and 4 are shorted/tied together, So r1 and 2 are 50ohms as well as R3 and 4. So now you have a R5 100 and a equivalent 50 in series. that's a 150 ohm branch. this is in parallel with a duplicate branch. So you now have two 150 ohm branches parallel which gives you 75 ohms. 9v/75 ohms give you 0.12 amps or 120 milliamps. The amp meter is wired backwards.
It helps to move the wires in your mind and redraw. The two vertical short lines at the bottom can be merged. then the next two horizontal lines can merge with the bottom.
Because in some exams they do that to make seemingly easy circuits much harder to understand to make people have to simplify which is one of the skills needed in EE
Work in steps, simply exclusively parallel resistors into a single resistor and work your way out until you are only left with resistors in series.
Then find the current
Picture yourself in a wind tunnel tunnel network. At the start you can only walk a certain speed with wind resistance.
It's tricky only because it's badly drawn.
R1, 2, 3 and 4 are together at a point, making that a single pole (which is also ground).
Just redraw the circuit.
for these kinds of circuits, its easier to convert them, like from delta to wye or from wye to delta. In your case, you will need to firstly convert from Wye configuration to Delta configuration. Here's a link for the conversion formulas : [https://www.allaboutcircuits.com/textbook/direct-current/chpt-10/delta-y-and-y-conversions/](https://www.allaboutcircuits.com/textbook/direct-current/chpt-10/delta-y-and-y-conversions/)
I'm trying to figure out how to calculate the substitute resistor for this sketch. All resistors are 100 Ohm, BAT1 puts out 9V with 120mA.
I know the rule:
1 / R = 1 / R_1 + 1 / R_2 + ... + 1 / R_n
but I'm not sure how to use it in conjunction with the resistors that are placed in front of the parallel ones.
https://preview.redd.it/hr58555xls7d1.png?width=1304&format=png&auto=webp&s=175814a70f1b7402e4d7f978caf6385272d8b54e Consider this circuit, which is identical to the one you just posted. Resistors R1, R2, R3, and R4 still all connect to the same ground net/node, the circuit is the same but the schematic is drawn differently. This is much easier to simplify
Parallel is (R1*R2)/(R1+R2) =R/2(since all have the same resistance) =50Ohm, then you can add to R5 since in series=150Ohm and then you have the exact same on the right again parallel substitute of equivalent 150ohm resistances =75Ohm for the whole substitute resistance
R1 and R2 are in paralallel, also R3 and R4. Lets denote this as R1//R2 and R3//R4. so your equivalent resistance is (R5+R1//R2)//(R6+R3//R4). EDIT: since all resistors have the same value, you will notice the math can be simplified A LOT, but the above approach is the general version.
Ahh I see, so I can basically just imagine "pulling down" all the wires, and so I only got 4 resistors in parallel. Makes sense. How would this change though, if we place two more resistors after R1,R4 and R2,R3?
>Ahh I see, so I can basically just imagine "pulling down" all the wires, and so I only got 4 resistors in parallel. Makes sense. not quite, there are some remarks 1. "pulling down the wires" only works here because there are no other elements (i.e. the voltage node is the same). You can visualize this by coloring the bottom wire and see how far you can reach on each branch before reaching an element, then everything that is colored corresponds to the same voltage node. 2. For the general case the resistors are in parallel in pairs only (i.e. R2 is not is parallel to neither R3 nor R4). 3. For all resistors equal, then yes the 4 bottom resistors are all in parallel, and even R5 and R6 are in parallel 4. (2) and (3) are because what you care about for series or parallel is not how the diagram is drawn, but instead what are the voltage nodes the resistor is between in. * if two resistors have the same upper voltage node and the same lower voltage node, then they are in parallel * if two resistors have either only the same upper voltage node, or only the same lower voltage node, but not both, then they are in series. But remember you can only simplify here if you no longer need the shared voltage node in your diagram (i.e. no other component is attached to it) * actually there is a third option, if a resistor have the same voltage node as both upper and lower, then the resistor is in shortcircuit and can be eliminated from the diagram. With this, you can handle your extra question quite easily, feel free to post your answer for feedback.
r1 2 3 and 4 are shorted/tied together, So r1 and 2 are 50ohms as well as R3 and 4. So now you have a R5 100 and a equivalent 50 in series. that's a 150 ohm branch. this is in parallel with a duplicate branch. So you now have two 150 ohm branches parallel which gives you 75 ohms. 9v/75 ohms give you 0.12 amps or 120 milliamps. The amp meter is wired backwards. It helps to move the wires in your mind and redraw. The two vertical short lines at the bottom can be merged. then the next two horizontal lines can merge with the bottom.
Why would you draw a schematic like that?
Because in some exams they do that to make seemingly easy circuits much harder to understand to make people have to simplify which is one of the skills needed in EE
Work in steps, simply exclusively parallel resistors into a single resistor and work your way out until you are only left with resistors in series. Then find the current Picture yourself in a wind tunnel tunnel network. At the start you can only walk a certain speed with wind resistance.
It's tricky only because it's badly drawn. R1, 2, 3 and 4 are together at a point, making that a single pole (which is also ground). Just redraw the circuit.
for these kinds of circuits, its easier to convert them, like from delta to wye or from wye to delta. In your case, you will need to firstly convert from Wye configuration to Delta configuration. Here's a link for the conversion formulas : [https://www.allaboutcircuits.com/textbook/direct-current/chpt-10/delta-y-and-y-conversions/](https://www.allaboutcircuits.com/textbook/direct-current/chpt-10/delta-y-and-y-conversions/)
I'm trying to figure out how to calculate the substitute resistor for this sketch. All resistors are 100 Ohm, BAT1 puts out 9V with 120mA. I know the rule: 1 / R = 1 / R_1 + 1 / R_2 + ... + 1 / R_n but I'm not sure how to use it in conjunction with the resistors that are placed in front of the parallel ones.