# Parallel and series mix of resistors

#### Brispark

Joined Nov 17, 2017
4
Hello,
I have decided to learn electronics from scratch again and have started with resistors. I was practising simplifying different circuits of parallel and series resistors when I came across this one. After simplifying so far I got stuck. Is this where you use Kirchhoff's loop rule or is there an easier way or am I missing something? (the reason I am unsure whether to use the loop rule is because all the examples I have seen have two different batteries in the circuit). Thanks for any help.

#### Ylli

Joined Nov 13, 2015
1,063

#### dl324

Joined Mar 30, 2015
15,511
Welcome to AAC!

First, I'd suggest that you use paragraphs to organize your thoughts and make your posts easier to read.

Second, I'd suggest that you use more typical conventions when drawing schematics and redraw your circuit like this:

Note that every component has a designator for ease in referencing components and nodes.

What are you trying to solve for?

#### Brispark

Joined Nov 17, 2017
4

#### Brispark

Joined Nov 17, 2017
4
Welcome to AAC!

First, I'd suggest that you use paragraphs to organize your thoughts and make your posts easier to read.

Second, I'd suggest that you use more typical conventions when drawing schematics and redraw your circuit like this:
View attachment 139539
Note that every component has a designator for ease in referencing components and nodes.

What are you trying to solve for?
Not the most welcoming of welcomes.

No need to be so arsey.

It is my first post and I am just staring out.

#### WBahn

Joined Mar 31, 2012
28,177
Hello,
I have decided to learn electronics from scratch again and have started with resistors. I was practising simplifying different circuits of parallel and series resistors when I came across this one. After simplifying so far I got stuck. Is this where you use Kirchhoff's loop rule or is there an easier way or am I missing something? (the reason I am unsure whether to use the loop rule is because all the examples I have seen have two different batteries in the circuit). Thanks for any help.View attachment 139536
I'm assuming that, at least initially, you are looking for the total current provided by the source (and, hence, the effective resistance as seen by the battery).

If that circuit is drawn the way you ran across it, then it was likely drawn intentionally as a way to confuse you or at least make you fail to spot it as being an example of a circuit topology you may have already learned how to analyze.

As dl324 has already shown, this is a classic configuration, usually referred to as a "Wheatstone bridge" topology.

Since it is not balanced, you can't use simple series/parallel combinations to reduce it to an effective resistance.

So you have several approaches available. Perhaps the most fundamental is direct application of KVL and KCL to the circuit as given. As already mentioned, delta-wye transformations are another. There's also a clever little trick using Thevenin equivalents to get at the answer. But both delta-wye and Thevenin are probably in your future still, so that's leaves us with KVL/KCL.

In answer to your question about loops and voltage sources, Kirchhoff's Voltage Law (the "loop rule" you are referring to) applies whether there is one source, five sources, or no sources (although there are situations, involving changing magnetic fields, in which it doesn't apply at all, but that's not the case here).

You have three loops, so try writing the equations for three loops and let's see where that leads.

But before you begin, you might consider some simple approximations that will place hard bounds on the upper and lower values that the effective resistance can be.

For instance, let's say that the 30 kΩ resistor were removed entirely. You know that this can only increase the total resistance seen by the battery, so this is an upper bound on the effective resistance. You should be able to see that this will be 30 kΩ, almost, if not, by inspection.

Next, let's say that the 30 kΩ were replaced by a short circuit (i.e., 0 Ω). You know that this can only decrease the total resistance seen by the battery, so this is a lower bound on the effective resistance. You should be able to satisfy yourself that the total resistance here can't be any lower than 25 kΩ. This is a number that is easy to get by inspection. A bit more effort shows that a tighter bound is 28.88 kΩ.

So (unless I've made a big goof, which is always possible), the total resistance of the actual circuit has to be between 28.8 kΩ and 30 kΩ. So you can expect the answer to be very close to 29.4 kΩ and, in many instances, this answer already would be more than close enough for whatever the purpose is.

After you get the answer, we can revisit how these bounds were arrived at -- knowing how to do them can be a real grade safer as a student and potentially a literal life saver in the real world.

#### WBahn

Joined Mar 31, 2012
28,177
Not the most welcoming of welcomes.

No need to be so arsey.

It is my first post and I am just staring out.

He is merely trying to help you convey what you are trying to communicate more clearly and effectively.

#### neonstrobe

Joined May 15, 2009
182
I tend to use simple nodal analysis. KVL, KCL in my view makes things more complicated. As dl324 has redrawn your circuit, it is clear you only have two nodes to solve. For nodal analysis, all currents into a node are summed to zero. You have Vcc and gnd, let's say, for the +9 and negative return. If the two voltages are V1 between R1 and R2 and V2 between R3 and R4. Taking the first node the currents INTO the node are (vcc-V1)/R1, (V2-V1)/R5 and (gnd-V1)/R2. You can write similar expressions for V2. Notice that by saying INTO a node means all voltage points are treated the same: FROM the further node TO the node in question, hence (gnd-V1). This may be negative in reality, but only one rule to remember. Both sets of equations can be simplified to give you simultaneous expression in V1 and V2. The neatest solution is to write the coefficients as a matrix and solve that, but you can also write a 2-node problem out without too much math, but it is a little long winded.

#### WBahn

Joined Mar 31, 2012
28,177
Note that nodal analysis IS nothing more than a formalized application of KCL.

Most people sum the voltages LEAVING the node as that results in that node's parameter being positive in each term. This can lesson the likelihood of making common silly math errors such as messing up the signs.