Are you confused, that how to deduce parallel & series combination of resistors in a complicated circuit or network? Then read it to clear your doubts.

### Where do you get confused?

Your teacher must have told you that when potential difference across two resistors is same, consider that they are in parallel. Also when same current is flowing through two resistors, consider that they are in series.

Well, this is OK. But when it comes to complicated circuits containing resistors, student gets confused and doesn’t understand which resistors are in parallel or in series. However, sometimes, some students get the idea of this, when the circuit is neatly drawn and contains only vertically or horizontally drawn resistors.

But when a circuit contains some resistors, which are drawn in say 45º angle, or something like it, then its becomes completely out of the mind for a student.

### Let us see how to start!

We shall start with a simple circuit first and then go for more complicated circuits after this. Consider the simple circuit, given below –

As shown above, the effective resistance of the above **Circuit #1 is 15Ω**. A step by step analysis given below can be used to calculate this effective value:

- Resistor R3 and R4 are in series.
- Their series combination is also in series with R2.
- So their effective resistance will be 30Ω.
- Now this effective value 30Ω is in parallel with R1.
- So the total effective resistance across terminals A-B will be 15Ω.

Well, you will say that this is very simple. But this

simple because all the resistors appear either horizontally or vertically connected in the above circuit.is

### What about randomly connected resistors?

Now consider the following circuit. Here R5 is randomly connected in an irregular or say slanted style. **Such connections are always very confusing to the students. But a careful analysis given below will make the things quite clear.**

The effective resistance of **Circuit #2 is 13Ω**. This can be understood with the help of following steps:

- Here R3 and R4 are in series . So their effective value will be, R3+R4 = 10Ω.
- The resistor R5 connected in
with this effective value. But how it is deduced? Well, observe the circuit carefully.*slanted style is in parallel* - The upper and lower terminals of R5 are directly connected to the two terminals of our
*(just calculated)*effective resistance of 10Ω. - So this second effective resistance value will be given as
**R5 || (R3+R4) = R5 || 10Ω = 5Ω**. - Now this second effective value of 5Ω is in series with R2. So the third effective value will be
**R2+5Ω = 25Ω**. - Finally, the resistor R1 is in parallel with this third effective value of 5Ω. So the total effective value of the Circuit #2 across terminal A-B will be (
**R1 || 25Ω) = (30Ω || 25Ω) = 13Ω**.

### What about this combination?

Here we connect R5 only in different combination. Again it is connected in slanted style. But I think, now you will be able to get some idea of this new combination. Well, check it as follows.

- Here R2, R3 and R4 are in series. So their effective resistance will be
**R2+R3+R4 = 30Ω**. - Now this effective resistance is in parallel with R5. How it is deduced? Well, again the same criterion I am applying. The two terminals of R5 are connected to the two terminals of R2+R3+R4 series combination.
- It means that R5 is in parallel with 30Ω. So the second effective value will be R5 || (R2+R3+R4). That gives us:
**10Ω || 30Ω = 7.5Ω**. - And finally, R1 is in parallel with second effective value of 7.5Ω. Hence, the total effective resistance of the circuit across terminals A-B will be: (
**R1 || 7.5Ω) = (30Ω || 7.5Ω) = 225/37.5 = 6Ω**.

### We make it more complicated…

In this circuit, two resistors R5 and R6 are connected in slanted style. But again, if you apply the same, analytical approach, you will come to know the following points:

- First R4 and R6 are in parallel. So,
**(R4 || R6) = (5Ω || 5Ω) = 2.5Ω**. - Now R3 is in series with R2 and our first effective resistance of 2.5Ω. So This series combination gives us the second effective resistance value of 27.5Ω. It is obtained as:
**R2+R3+2.5Ω = 27.5Ω**. - Now there is another slanted resistor, R5. This will be in parallel with our second effective resistance of 27.5Ω. So the third effective value will be:
**(R5 || 27.5Ω) = (10Ω || 27.5Ω) = 275/37.5 = 7.33Ω**. - Finally R1 is in parallel with this third effective value of 7.33Ω. So the total effective resistance of the circuit will be:
**(R1 || 7.33Ω) = (30Ω || 7.33Ω) = 5.89Ω**.

### And finally…

**So my dear students, we shall obtain following conclusions from this discussion:**

- First of all observe the circuit VERY CAREFULLY.
- If the resistors in the circuit do not have numbering, then give it a sequential numbering as per your choice.
- Now it is a general practice, that if the circuit is neatly drawn, like the above ones
*(neatly drawn in rectangular shapes)*, then its OK, otherwise redraw it as per your own style. - What is the importance of this 3rd point above. Actually, when you redraw the circuit, you start getting hints of which resistors are in parallel or in series. Believe me! Just try with any one complicated circuit. And when you redraw it as per your style, your rearrangement, the magic happens, and you get lots of such hints.
- Well, then there is one more thing. Actually I do it like this. For the given circuits above, we had to calculate total effective resistance across A-B
*(given on left)*. So what I do, is that I start my observation and then calculations from OPPOSITE SIDE or say RIGHT SIDE of the circuit. - Go on simplifying by identifying series/parallel combination.
- NOW HERE YOU MUST DO ONE THING. When you come up with certain effective value of a particular combination of resistors, remove all those resistors and redraw the circuit with this effective value. This will give you more CLEAR PICTURE of the circuit and the resistors’ combination.