# Solving circuit problems

I'd like to know why in a parallel loop, the voltage, along with the current, isn't divided.

e.g. in this image from a youtube video the voltage in parallel loop is the same across both resistors, for some reason this seems counter-intuitive to me.

Also later on in the video, the resistance was proportional to the voltage. I talked about this in class today, there was some debate on whether the voltage is inversely or directly proportional to the resistance. My argument was that because the resistance is higher, there must be less voltage going through at that point. So as resistance increases, voltage drops, showing an inverse proportion, but then someone in the class brought up the equation V=IR, in the circuits we've looked at, the current is seen to be the same throughout, in series, so it can be simplified to V=(K)R, V(proportion sign)R which would mean as Voltage went up, so would the resistance. Then we were talking about this in parallel, and things got more confusing, he spoke about the current not being the same across both resistors therefore we cant use the above rule, which I understood, but he didn't explain what actually happens. I'm getting confused writing this question.. perhaps I should do a few practice questions.

My teacher was worryingly also unsure. I Hope this question doesn't come across too ambiguously.

• "I'd like to know why in a parallel loop, the voltage, along with the current, isn't divided." KVL If both current and voltage divided, conservation of energy would not hold. Nov 17 '15 at 2:21
• Your question isn't clear. For instance, you say the "voltage in parallel loop is the same across both resistors", but that diagram shows three resistors. You may want to edit to clarify. Nov 17 '15 at 2:33
• en.wikipedia.org/wiki/Series_and_parallel_circuits
– user83548
Nov 17 '15 at 3:55
• "because the resistance is higher, there must be less voltage going through at that point." - Voltage, however is not something that goes through. It is a quantity that is fixed (for your dc circuits) in any given point. So if the voltage on the left of that parallel loop in the leftmost circuit is 9V, and to the right of the 25 Ohm resistor it is 3V, no matter which way you got from the first point to the next, you must have lost 6V along the way, be it across the bigger resistor, or the smaller one. Nov 17 '15 at 7:20

My argument was that because the resistance is higher, there must be less voltage going through at that point.

This is probably the cause of the confusion. In spite of the usual formulation $V=IR$, in an electrical circuit Voltage and Resistance are the "inputs" to the equation and Current is the result or output.

As an analogy, think of Newton's 2nd law $F=ma$. Here mass is a fundamental property of the object being moved, and we are applying a fixed force to it; the result is that it accelerates, and the heavier the object, the slower it accelerates.

In a similar way in $V=IR$, resistance is a "fixed"* property of the resistor, and we apply a given voltage across it, resulting in a current flowing. The higher the resistance, the "slower" the current flows.

(* in fact in real life this is often only an approximation, but for most materials - known as "Ohmic" materials - it is a good one)

Now, the above analogy is useful for remembering which are the causes and which are the effects, but it is not much use for thinking about parallel circuits.

Instead we can think of voltage like a pressure difference in a water pipe, and resistors as being narrow pipes (with a higher resistance being analogous to a narrower pipe). If we have two pipes of different widths open between two points of high and low water pressure, the total flow will be higher than with either one of them by itself.

• Hooray for the water pipe analogy and not being too pedantic to use it :) +1 Nov 17 '15 at 7:10

The problem is really with the parallel diagram on the far left. It shows a 6V drop across the parallel combination of resisters. That is determined backwards from how we read left-to-right. In order to solve for the voltage drop you must:

(1) Solve for the equivalent resister to the pair, the 20 Ohm resister shown in the middle diagram. The equivalent resister doesn't depend on the Voltage. (In reality this isn't really true since a 1 watt resister won't carry 1,000 watts of power...)

(2) Calculate voltage drop across the 20 Ohm and 10 Ohm resisters which are now in series.

(3) Now you can go back and assign the voltage drop across the pair of parallel resisters.

Let's label circuit junction to the left of the 25 ohm resister A, the junction point between the 25 and 10 ohm resisters as B, and C for the point after the 10 Ohm resister.

In essence a without the 100 Ohm resister there would be a 6.42 volt drop across the A-B. With the addition of the 100 Ohm resister then more current flows from A to B. But the voltage at the "end" of the 100 Ohm resister has to be the same as the Voltage at the "end" of 25 Ohm resister. So the "effective" resistance has to be less than either individually.