# Finding the voltage of a circuit composed of a series and parallel?

I was watching a youtube tutorial to understand calculations using ohm's law and I thought I had it down when I ran into this problem: I was asked to find the voltage at points A and B. I was also asked to find the current that runs through both resistors in the parallel section of the circuit:

So I did the following to find those answers:

I first wanted to find the voltage drop in the series section of the problem so I used ohm's law:

$$V=IR=(6A)(5Ω) = 30V$$

So my first answer was correct, point A had a voltage of 30V. Now I thought I could use this voltage to solve for the parallel part of the problem, but before that I thought:

Ok, so 30V going into the parallel circuit, great! so that means that since there is no voltage drop in parallel point B must be 30V as well!

Then I continued with my analysis. I wanted to find the current at the 3Ω resistor, so I used:

$$I = V/R = 30V/3Ω = 10A$$

Using that same analysis I got the current going through the 6Ω as well, which was 5A. It made no sense to me, because I would be saying that there is a current of 15Ω, greater than the initial current. Why is my analysis incorrect? Why do I get incorrect current values for the parallel circuit using Ohms's law?

This is the video's solution to the problem: As you correctly did, you start with the node that has $60\:\textrm{V}$ on it and then notice that they specify the exact current exiting that node ($6\:\textrm{A}$) and by which route (the $5\:\Omega$ resistor.) Since all of this current must "go through" that resistor, it follows that there must be $6\:\textrm{A}\cdot 5\:\Omega = 30\:\textrm{V}$ across the resistor. The only problem at this point is to ask yourself if this means that point A is $V_A=60\:\textrm{V}+30\:\textrm{V}=90\:\textrm{V}$ or if it is at $V_A=60\:\textrm{V}-30\:\textrm{V}=30\:\textrm{V}$. By convention, current is considered "postive" so the current flows from the more positive end to the more negative end. The arrow shows this and it means that $V_A=60\:\textrm{V}-30\:\textrm{V}=30\:\textrm{V}$.

Good job there!

Okay. So now you are faced with a parallel arrranged pair of resistors. What happens here? Well, the current divides. Some goes in one direction and some goes in the other direction. The sum, of course, is still $6\:\textrm{A}$. You go about solving this one by recognizing that the voltage across these two resistors has to be identical, since they are connected at both ends. Node A can have some voltage $V_A$ and node B can have some voltage $V_B$. But node B cannot have two different voltages. Neither can node A. A node can only have one voltage. So the voltage across the $3\:\Omega$ resistor must be exactly the same voltage as is also across the $6\:\Omega$ resistor. However it is that the current divides itself (and for now, the only thing you know is that the sum must be $6\:\textrm{A}$), it must be the case that the voltage developed across the upper resistor must be the same as the voltage developed across the lower resistor. This voltage will be $V_B - V_A$. That's a simple fact.

So let's write this out:

\begin{align*} V_B - V_A &= I_{3\:\Omega}\cdot 3\:\Omega\\\\ V_B - V_A &= I_{6\:\Omega}\cdot 6\:\Omega\\\\ &\therefore ~~~~~I_{3\:\Omega}\cdot 3\:\Omega = I_{6\:\Omega}\cdot 6\:\Omega \end{align*}

But of course, you also know:

$$I_{3\:\Omega}+I_{6\:\Omega} = 6\:\textrm{A}$$

You can solve these two equations to find that:

\begin{align*} I_{3\:\Omega} &= 4\:\textrm{A}\\\\ I_{6\:\Omega} &= 2\:\textrm{A} \end{align*}

And from that you know that the voltage drop must be $12\:\textrm{V}$ so that $V_B=V_A-12\:\textrm{V}=18\:\textrm{V}$.

Now, like everything, there are shortcuts developed when the same thing is done over and over. No one likes to go solve two simultaneous equations every time they meet a parallel resistor pair like this. So they come up with helpful rules to follow. (This is the same thing like learning your multiplication tables and long-hand multiplication -- it saves you a LOT of addition when you face a multiplication problem.)

One rule to learn is that you can replace a parallel pair of resistors with a single equivalent resistor by converting them to conductances (how much they encourage current rather than discourage it, I suppose), adding the conductances in the parallel branch, and then reconverting back to resistance, again. It's easy to convert each way: $G_R=\frac{1}{R}$ and $R=\frac{1}{G_R}$. So here you'd do the following:

\begin{align*} R_{equiv} &= \frac{1}{G_{equiv}}\\\\ &= \frac{1}{G_{3\:\Omega}+G_{6\:\Omega}}\\\\ &= \frac{1}{\frac{1}{3\:\Omega}+\frac{1}{6\:\Omega}}\\\\ &= 2\:\Omega \end{align*}

Now, you can compute the voltage drop easily, just as you did in your first step in fact, as $6\:\textrm{A}\cdot 2\:\Omega=12\:\textrm{V}$. And knowing that, you can easily go back and figure out the currents in each of the two resistors, since you know the voltage drops across them, now.

You can even simplify the above process into the case for just two parallel resistors as:

$$R_{equiv} = \frac{R_1\cdot R_2}{R_1+R_2}$$

The previous process of finding conductances is more general and will handle three, four, and more resistors in parallel. But two resistors occur often enough that the above formula is worth committing to memory.

• Jonk, do you have a cheat sheet somewhere for those nicely formatted character... – Trevor_G Apr 7 '17 at 2:34
• @Trevor No, I'm just fluent in Latex. – jonk Apr 7 '17 at 3:24
• @Trevor Not sure if you know what LaTex is. But it started with Donald Knuth trying to write the next volume of his incredible compendium in the early 1970's. The tools that he wanted weren't there so he (and help) created Tex, MetaFont, and diverted about 30 years of his life and didn't write that next volume until recently (I bought a copy few years back.) In the meantime, LaTex came about. And this site supports it. You should learn some of it. – jonk Apr 7 '17 at 5:24

In your second equation you are assuming B is 0V.. it is not. You need to calculate what that is.

There are two ways to do that, either combine the 3R and 6R resistors to determine the parallel equivalent and then use the first formula you used.

That works out to $2\:\Omega$. So the voltage drop across the parallel section is $\:2*6 = 12V$

So Voltage at B = 60 - 30 - 12 = 18V

The other way to do it is to figure out the ratio of current that passes through the 3 or 6 ohm path

eg. Current ratio for $R1 = \huge\frac{\frac{1}{R1}}{\frac{1}{R1} + \frac{1}{R2}}$

3R path = 2/3, 6R is 2/3

Hence, Current through 3R = 4A, Current through 6R = 2A

So for 3R path, 2/3 of 6A = 4A, so again Voltage across 3R is 4 x 3 = 12V.

• I understand that the total resistance of the parallel part was 2Ω, due to 1/R = 1/3 + 1/6. So I understand how voltage at B is 18V. However, I don't understand why I = VR did not work to find the current passing through the 3Ω and 6Ω resistors. What do you mean the ratio is 2/3? where do you get that 2/3? – Pablo Apr 6 '17 at 22:05
• @Pablo current ratio of a "leg" is (conductance of leg)/(sum of conductance of all legs). V=IR did not work in your question because you assumed V was 30V. It isn't it's 18V across the parallel part. – Trevor_G Apr 7 '17 at 1:02
• Hehe! I think I see some LaTex! And the use of \huge, too! – jonk Apr 7 '17 at 20:07

The circuit values given are inconsistent.

The equivalent resistance of the 3 Ohm and 6 Ohm in parallel is 2 Ohms, so the total resistance of the circuit is 7 Ohms.

60 volts over 7 Ohms = 8.57 Amps, not 6 Amps.

With 8.57 Amps, you will have 8.57 X 5 = 42.8 volts across the 5 Ohm resistor, leaving 17.14 volts for the parallel section.

• Wow, I didn't think that the video would be be wrong. Thanks, so then would my method work if the values were consistent? – Pablo Apr 6 '17 at 21:52
• I solved a different problem. The video apparently assumed there was an unknown resistance to the right of point B, while I assumed point B was zero volts, as there was no other component shown. – Peter Bennett Apr 6 '17 at 21:54
• @PeterBennett Yeah. I worried about that and then noticed that they specified the current exiting the 60 V node. – jonk Apr 6 '17 at 23:23