Determing the voltage across a resistor

My professor went over this problem in class today, but I had trouble understanding it even after a few questions were asked. The goal was to find the voltage, $v_i$ across the 8Ω resistor on the right. The math on the right is what was put on the board in class.

What I don't understand is why we would use 2.4Ω in the numerator, when we want to find the voltage across the 8Ω resistor. Isn't this the wrong proportion? If 8Ω was in the numerator, the answer would be 20V, so it definitely makes a difference which one you use.

Why can we use the 2.4Ω resistor in this situation?

• You have to calculate the equivalent resistance to determine the voltage across the equivalent resistor. From that you can calculate the voltage across the series resistors. If you ignore the 2.4Ohm resistor then your equivalent resistance will be off. If you had a voltage source instead of a current source, you could ignore the 2.4Ohm resistor do find Vi. – Wesley Lee Oct 10 '16 at 20:38
• Do a quick sanity check using what you think should be in the numerator. Having 8 ohms would lead to more current down the 9.6 ohm path than the 2.4 ohm path. This logically does not work since we know more current will try and take the path of least resistance. – mmize Oct 10 '16 at 21:21
• Your teacher has simply applied the formula of the current divider. You can do the demonstration; calculate the parallel , and then the voltage drop across equivalent resistor, then calculate the current in the branch and then apply Ohm's law. – Antonio Oct 11 '16 at 12:00

Mathematically, it's pretty straightforward. First you find the voltage across the branches:

$$V_{branch} = 30 \mathrm A \cdot \big(2.4\Omega\ ||\ (1.6\Omega + 8\Omega)\big) = 30 \mathrm A \cdot (2.4\Omega\ ||\ 9.6\Omega)$$

When there are only two resistors in parallel, some algebra gives you a shortcut formula for the parallel resistance:

$$2.4\Omega\ ||\ 9.6\Omega = \frac 1 {\frac 1 {2.4\Omega} + \frac 1 {9.6\Omega}} = \frac {2.4\Omega \cdot 9.6\Omega} {2.4\Omega \cdot 9.6\Omega} \cdot \frac 1 {\frac 1 {2.4\Omega} + \frac 1 {9.6\Omega}} = \frac {2.4\Omega \cdot 9.6\Omega} {9.6\Omega + 2.4\Omega} = 1.92\Omega$$

You might have seen this written as:

$$R_{||} = \frac {R_1 R_2} {R_1 + R_2}$$

Anyway, now you can get the branch voltage:

$$V_{branch} = 30 \mathrm A \cdot 1.92 \Omega = 57.6 \mathrm V$$

Then you divide by the branch resistance to get the branch current:

$$I_{branch} = \frac {V_{branch}} {9.6 \Omega} = \frac {57.6 \mathrm V}{9.6\Omega} = 6 \mathrm A$$

That lets you find the resistor voltage using Ohm's Law. Now when you only have two branches, you can take advantage of that shortcut formula to get another shortcut:

$$I_{branch} = \frac {V_{branch}} {9.6 \Omega} = 30 \mathrm A \cdot \frac {2.4\Omega \cdot 9.6\Omega} {9.6\Omega + 2.4\Omega} \cdot \frac 1 {9.6 \Omega} = 30 \mathrm A \cdot \frac {2.4\Omega} {9.6\Omega + 2.4\Omega}$$

When you have two resistors in parallel, you might see this written as:

$$I_{R1} = \frac {R_2} {R_1 + R_2}$$

This looks like the voltage divider formula. However, this current divider formula only works for two resistors. It's a mathematical artifact, not a general rule. So don't do this:

$$I_{R1} = \frac {R_2 + R_3} {R_1 + R_2 + R_3}\ \ \ \mathrm{WRONG!}$$

That was a lot of algebra, so here's an intuitive way of thinking about it. Imagine the current as being balanced between the two branches. What matters isn't the absolute value of the resistors, it's what fraction of the total resistance each one is. If all of the resistance is on one side, all of the current flows through the other side: simulate this circuit – Schematic created using CircuitLab

If the resistances are equal, the current is split equally between them: simulate this circuit

If three quarters of the resistance is on one side, three quarters of the current flows through the other side: simulate this circuit

I'm not sure how well this works as a general rule, but hopefully it makes you a bit more comfortable with the simple stuff.

Let's call the bottom node "ground" and mark it as 0 V. The top node can be called voltage $V_x$, then. You know that the current into the node must be equal the current flowing back out (current doesn't accumulate in a node.) So the following must be true:

$$(1)~~~~~~~~~~~~30 \textrm{A} = \frac{V_x}{2.4\:\Omega} + \frac{V_x}{1.6\:\Omega+8\:\Omega}$$

In the above, you can see the two branch currents listed on the right side and the incoming 30 A current on the left side. They have to be equal.

The above equation solves into:

$$(2)~~~~~~~~~~~~~~~~~~~~~V_x=\frac{30 \textrm{A}}{\frac{1}{2.4\:\Omega} + \frac{1}{1.6\:\Omega+8\:\Omega}}$$

But you don't care about $V_x$. Just just want the expression that is right-most in the first equation listed above, namely the current through the branch with the two resistors. So you can plug one into the other to get:

$$I=\frac{\left(\frac{30 \textrm{A}}{\frac{1}{2.4\:\Omega} + \frac{1}{1.6\:\Omega+8\:\Omega}}\right)}{1.6\:\Omega+8\:\Omega}=\frac{30 \textrm{A}}{\frac{1.6\:\Omega+8\:\Omega}{2.4\:\Omega} + 1}=\frac{30 \textrm{A}}{\frac{1.6\:\Omega+8\:\Omega+2.4\:\Omega}{2.4\:\Omega}}=30 \textrm{A}\frac{2.4\:\Omega}{1.6\:\Omega+8\:\Omega+2.4\:\Omega}$$

That merely confirms what you've already been told. But it doesn't provide intuition.

In a voltage divider, the voltage across a series element is equal to the total voltage minus all the voltage drops across the other series elements.

Here, we use resistance and compute $I_T=\frac{V_T}{\sum R_i}$ for the divider, so,

\begin{align*} (3)~~~~~~~~~V_T &= \sum^n_1 R_i I_T = I_T\sum^n_1 R_i \\ \\ V_T &= I_T\left[\sum^n_1 R_i\right] - R_x\cdot I_T + V_x \\ \\ V_x &= V_T - I_T\left(\left[\sum^n_1 R_i\right] - R_x\right) \\ \\ V_x &= V_T - \frac{V_T}{\sum^n_1 R_i}\left(\left[\sum^n_1 R_i\right] - R_x\right) \\ \\ V_x &= V_T\cdot\left(1 - \frac{1}{\sum^n_1 R_i}\left(\left[\sum^n_1 R_i\right] - R_x\right)\right) \\ \\ V_x &= V_T\cdot\left(1 - \left(1 - \frac{R_x}{\sum^n_1 R_i}\right)\right) \\ \\ V_x &= V_T\cdot\frac{R_x}{\sum^n_1 R_i} \end{align*}

In a current divider, the current through a parallel element is equal to the total current minus all the currents through other parallel elements.

Here, we use conductance and compute $V_T=\frac{I_T}{\sum G_i}$ for the divider (where $G_i$ is the conductance of each branch), so,

\begin{align*} (4)~~~~~~~~~I_T &= \sum^n_1 G_i V_T = V_T\sum^n_1 G_i \\ \\ I_T &= V_T\left[\sum^n_1 G_i\right] - G_x\cdot V_T + I_x \\ \\ I_x &= I_T - V_T\left(\left[\sum^n_1 G_i\right] - G_x\right) \\ \\ I_x &= I_T - \frac{I_T}{\sum^n_1 G_i}\left(\left[\sum^n_1 G_i\right] - G_x\right) \\ \\ I_x &= I_T\cdot\left(1 - \frac{1}{\sum^n_1 G_i}\left(\left[\sum^n_1 G_i\right] - G_x\right)\right) \\ \\ I_x &= I_T\cdot\left(1 - \left(1 - \frac{G_x}{\sum^n_1 G_i}\right)\right) \\ \\ I_x &= I_T\cdot\frac{G_x}{\sum^n_1 G_i} \end{align*}

The symmetry should be obvious. In your case, you only have two conductances, $G_1=\frac{1}{2.4\:\Omega}$ and $G_2=\frac{1}{1.6\:\Omega+8\:\Omega}$. And in this case, $G_x=G_2$.

If you use those in the bottom equation of (4), you will get the right answer. However, it doesn't look like what the teacher wrote up. But if you plug things in, you can see why:

\begin{align*} (5)~~~~~~~~~I_x &= I_T\cdot\frac{\frac{1}{1.6\:\Omega+8\:\Omega}}{\frac{1}{2.4\:\Omega}+\frac{1}{1.6\:\Omega+8\:\Omega}} \\ \\ &= I_T\cdot\frac{\frac{1}{1.6\:\Omega+8\:\Omega}}{\frac{2.4\:\Omega+1.6\:\Omega+8\:\Omega}{2.4\:\Omega\cdot\left(1.6\:\Omega+8\:\Omega\right)}} \\ \\ &=I_T\cdot\frac{\frac{1}{1}}{\frac{2.4\:\Omega+1.6\:\Omega+8\:\Omega}{2.4\:\Omega}}=I_T\cdot\frac{2.4\:\Omega}{2.4\:\Omega+1.6\:\Omega+8\:\Omega} \end{align*}

However, that is NOT a general approach. It just happens to work in this case. So your intuition perhaps is right, about being confused in class. The general solution looks like this, using resistances:

\begin{align*} (6)~~~~~~~~~I_x &= I_T\cdot\frac{G_x}{\sum^n_1 G_i} \\ \\ &= I_T\cdot\frac{\prod_{i=1,\\ i\ne x}^n R_i}{\sum^n_{i=1} \prod_{j=1,\\ j\ne i}^n R_j} \end{align*}

Keep in mind that in the above, each $R_i$ is not each of all your resistors, helter-skelter. These resistances are the effective branch resistances. So in one of your branches, this value is a sum. Don't forget that, either.

Equation (6) is a lot uglier than you might have imagined from your teacher's solution. It's the correct one, though. So, if you want to commit something to memory, go with the conductances approach. As you can see, using resistances isn't right off the bat intuitive.

If you must use resistances in the current branching case like this, then you have to follow (6) above and DO NOT simply try and extend what your teacher did in some ignorant way. It won't work.

This means I don't agree an answer suggesting, "Your professor has given the formula for a current divider."

Wrong!

Normally, people do as some of the other answers suggest -- work out the series and parallel resistances to a single value (if possible) and then figure out the voltage. Then work through the details for the branch itself to get the current. That's pretty much how most would approach it. (Or use nodal analysis, if the circuit is more complex.)

Your teacher's formula isn't an extensible formula. It's just one that fell out in this case.

Note that source has constant current output, not voltage output. Whatever value resistor network is having, in total there will be 30 A current flowing through them. I did it the following way (probably there're better ways):

1. calculate resistance of the resistor network. You have two lines in parallel: 2.4 Ohm and (1.6 + 8) Ohm. Thus 2.4*9.6/(2.4+9.6)=1.92 Ohm.
2. now when we know total resistance, we can calculate voltage on 2.4 Ohm resistor, it will be 1.92 * 30 = 57.6 V. The same voltage will be on (1.6+8) Ohm series resistor.
3. next, when we know voltage, we can know current flowing through series resistor, it will be 57.6/(1.6+8)=6 A
4. and finally, voltage on 8 Ohm resistor is 6*8=48 V.

To answer your question on why things are there where they are - combine formulas from points 1, 2 and 3 above:

30 * ( 2.4 * (1.6+8) / (2.4+1.6+8) ) / (1.6+8)

and it equals to

30 * 2.4 / (2.4+1.6+8)

Try to think the pair or resistances as one, (after all, there are in series so the current is the same) and then perhaps you see clear how to use the current divisor approach. $$I_{v_i}=30A\frac{2.4Ω}{((8+1.6)+2.4)Ω}=6A$$

then you have the Ohm's Law to simply $$V_i=(6A)(8Ω)=48V$$

You can keep this simple and logical, as is the case in many example problems. Look at the 3 resistors. The 2.4 ohm is in parallel with the 8 and the 1.6. The 8 and 1.6 are in series and add up to 9.6 ohm.

By examination, you have 2.4 ohm in parallel with 9.6 ohms. 9.6 is 4 x 2.4. Isn't that convenient? So 4 times the current will flow through the 2.4 ohm as will flow through the 9.6 ohm combination. This means you can divide the total current into 5 equal portions. 4 portions to the 2.4 ohm and 1 to the 9.6 ohm.

What is the total current? The diagram says 30 amps. 30 divided into 5 equal portions is 6. Isn't that convenient? Therefore 6 amps flow through the 8 and 1.6 ohm resistors.

Voltage is V = IR = 6 amps x 8 ohms = 48 volts.

My advice is look at problems for a while. See if there are symmetries or convenient ratios and other numbers. These problems are designed to be solved, and often without a calculator or complicated equations. The instructor's method is meant to prepare you for more complicated problems. Hope this helps.

You are mixing this up with a voltage divider. Your professor has given the formula for a current divider. I hope you understand the current through the 1.6Ω+8Ω path is higher when the resistance in the other path is higher.

• Perhaps you can more elaborate more for OP and any other future readers of this question as to why the 2.4 ohms is on the numerator and/or any intuition to help understand the material better. Currently, you just stated that the formula is a current divider and that you hope OP understands. – efox29 Oct 10 '16 at 20:51
• Hate to say it, but Wikipedia has a pretty good article about it: en.wikipedia.org/wiki/Current_divider – Janka Oct 10 '16 at 20:56
• @user23885, a good answer on Stack Exchange should be useful by itself. It shouldn't be just an pointer to documentation elsewhere. – The Photon Oct 10 '16 at 20:57
• @Photon: I think the OP's problem is he thought the circuit is a voltage divider. Which it isn't because the source is a current source, not a voltage source. That correction should already be enough for him (and others) to understand the problem. To point out what a current divider is and how it works, I'm grateful there is literature to explain (or Wikipedia, if that one fits, too.) – Janka Oct 10 '16 at 21:01
• I should add I think the reason for not understanding the professor may be the professor being too elaborate in his answer, and the OP not getting the point because of that. – Janka Oct 10 '16 at 21:09