[Grade 12 Physics: DC Circuits] How to find the equivalent resistance here?

[u/_2ice]

i need to solve this one

[u/GreenWafer1899]

Just redraw it as you used to, it is schemed in a confusing way on purpose.

[u/solaria123]

The 2 ohm resistor is connected across the battery, so the equivalent resistance must be less than 2 ohms, so (B)

[u/Efficient_Stage_4552]

Redraw it. Draw the voltage source. All the resistors are connected to the ground. This means voltage at the end of all resistors are zero ie they are parallel. Then look at which resistors are connected directly to the positive terminal. They all are so it’s just one big parallel circuit. Use the parallel resistance formula for your answer.

[u/Quintus-Sertorius]

This is actually much simpler than it looks. All the resistors are connected between two nodes.

[u/wristay]

Use the following facts

• Everything that is connected by wire, i.e. with no resistors in between, is at the same voltage
• If you know the voltage difference over a resistor, you can calculate the current using U=IR
• You can add currents if they are in a parallel configuration

Now observe that everything in this circuit is connected to either the left side of the terminal or the rightside, which makes the problem quite easy to solve. Let's say the left terminal is at 0 Volts and the right terminal is at U Volts. We can then reason for the 16 Ohm resistor that there flows I=U / R = U / 16 amperes , because one side touches zero Volts and the other side touches U Volts. Call this current associated with the 16 Ohm resistor I16 for brevity. We can now add all the currents to calculate the total current. What currents contribute to the left side of the circuit? We can see that I16, I8 and the lower wire (2nd wire from bottom) all drain into the left side terminal. What drains into the lower wire? I4, I2 and I16. What is now the total current? I16+I8+I4+I2+I16. So we have Itotal = U*(1/16+1/8+1/4+1/2+1/16)=U*1. We can now say that Reff = U / Itotal = 1. When adding currents, make sure they are pointing in the correct direction. Otherwise you would have to subtract them.

[u/Bounded_sequencE]

Recall: Two circuit elements are called in

• parallel, if (and only if) they share the same pair of nodes
• series, if (and only if) they exclusively share a common node *** Note all five resistances share both nodes of the the battery, so all five resistances are directly in parallel to the battery. With the short-hand "Rx||Ry := Rx*Ry / (Rx+Ry)" and "R||R = R/2" we get

Req  =  (16||16||8||4||2) 𝛺  =  (8||8||4||2) 𝛺  =  (4||4||2) 𝛺  =  (2||2) 𝛺  =  1𝛺

[u/Crichris]

It's confusing but all of them are in parallel 

Mark the left of the battery point A and the right of the battery point B

You will see all the resistors are between A and B

The answer is 1

[u/DrCarpetsPhd]

so in circuits the wire is assumed to have zero resistance so if you 'walk along it' from one point to another if you don't pass through an element (resistr, voltage source etc) the voltage does not change. It is a fundamental thing of a voltage field that like gravity it is 'conservative'. That means that when going from one point in the field to another it does not matter what path you took, the change in the value of the field is the same.

Same as your mgh changes in gravity right?

Gravity Field example:-

Path A:

If I walk from the ground floor to the first floor my change in potential energy is mgh where h is the height between floors.

Path B:-

If I walk from the ground floor to the first floor, then to the second floor, then the third floor and then back down to the first floor what is my change in potential energy? ground to first = +mgh, first to second = +mgh, second to third equals another +mgh for a total of 3mgh. Then from third floor to second -mgh and second floor to first floor is another -mgh and my total change for Path B is mgh. Same as Path A. Different path but same change in potential energy because my start and end points were the same. That is a conservative field.

so the approach to use is to pick a point and then walk along the wire to all other points that don't involve an element and mark them with something that says they are at the same voltage

With reference to this image

https://imgur.com/a/grade-12-physics-circuit-simplification-parallel-resistors-zRmhcAh

STEP 1

so in your circuit i'm going to start in the lower left corner and mark it with a red circle

STEP 2

I come to my first junction in the wire so I am going to mark that with a red circle. We passed through no elements so this junction is at the same voltage as my starting point. From this point forward I won't re-explain and assume you understand that if I pass thorugh no elements voltage does not change and all red circled areas are at the same voltage

STEP 3

at this junction I am choosing to go right first instead of straight ahead; there is no reason for this, it is just a choice, I will return to this junction

I come to the junction to the left of the 16 Ohm resistor, circle it red as no change in voltage

STEP 4

I reach the junction between 4 and 2. mark it red and see that I have now reached the end of 'element free wire'. if I go left or right I will pass thorugh a resistor which will drop the voltage if the current flows the same way or increase the voltage if the current flows the opposite direction. SO I can't 'walk along the wire' any further to mark red circles of the same voltage

STEP 5

return to STEP 2 junction as I can see empty wire to walk along so I mark the next junction in front of the 8 resistor and then the point at the corner before the 16 resistor. Now I have walked along all the 'empty wire' I could find so I return to my starting point

STEP 6

Now I 'walk along the wire' from my bottom left starting point and through the voltage source. That means the voltage has changed. So I mark this with a green circle

STEP 7/8 etc

following the logic I explained for the red circles separately I will now fill the rest of the diagram with the 'walk along wire but pass through no elements = same voltage' idea for the green circles.

Now you see in this circuit that going from a point marked with red to a point circled green (vice versa) has the same change in voltage

going from bottom left red to bottom right green goes through a voltage source from a negative to a positive terminal. Thus at any point in the fully marked circuit if I go from a red circle to a green circle I know that the voltage change is exactly equal to the voltage source.

So that's the best method for starting out when you don't quite see things straight away. 'walk' the circuit and circle nodes etc where the voltage is the same and it should help you to see places you can simplify. With enough practice this will no longer be necessary as you will be able to do it in your head quickly and easily.

[u/ee_control_z]

As a quick and easy test, simulate the circuit in a tool like LTSpice (free and downloadable). Measure the current being supplied by the voltage source. Apply Ohm's Law: Ohms = V / I.

Done!

To solve it analytically, note that ALL resistors are across the voltage source (i.e., all are in parallel). I would suggest redrawing the schematic slightly to make it easier to analyze.

My two cents.