Help with simple voltage divider question

Question

I have had a complete mind blank on how to solve what should be a relatively simple voltage divider question. Here is an example of the circuit in question.

^{simulate this circuit – Schematic created using CircuitLab}

I need to find the voltage across R3 using the voltage divider formula. So I am trying to somehow combine R1, R2 and R4 into one resistor in series with R3 so I can use the formula, but I am getting confused on how to go about it. Any tips would be appreciated. I am not after the answer, I want to know how to do it so I can also apply it to other problems in the future. Thanks

Answer 1

There are two ways to approach this question.

a) The first is to use the voltage divider rule twice

Put R3 and R4 in series
Combine those in parallel with R2
This now forms the load on R1
So now compute the voltage on the R1-R4 junction
Now compute the voltage on the R4-R3 junction

b) The second is to use the Y-Delta transformation

Make the Y (or Wye, or Star) of R1, R2 and R4 into an equivalent Delta of three resistors. Let's call those resistors Ra, Rb and Rd (1st, 2nd and 4th letter of alphabet, opposite the corresponding Y resistor).

^{simulate this circuit – Schematic created using CircuitLab}

Now Rd is across the source, and doesn't take any part in the sums.

Ra is in parallel with R3, you must combine them to get the proper load on Rb.

Now use the voltage divider formula with Rb feeding the load of Ra||R3.

I've not calculated the values of Ra, b and d, you can do that from the formulae on wikipedia, search for 'wye-delta'.

Which is better?

They give the same answer.

The extra computation needed for Y-delta is probably more than for two applications of the voltage divider rule.

I can remember and check the voltage divider rule, but I need to look up the Y-delta formulae.

With ladder structures, you can always extend the first method.

With lattice or bridge structures, you may need the second method.

Answer 2

The first approach to try would be to follow the simple rules I suspect you already know. (1) If two resistors, \$R_1\$ and \$R_2\$, are alone and in series then replace with a resistor \$R_x = R_1+R_2\$; and, (2) if two resistors are in parallel, then replace with a resistor having a value that is the parallel equivalent \$R_x = \frac{R_1\cdot R_2}{R_1+R_2}\$. Using this, you would:

1. Combine \$R_4\$ with \$R_3\$ to get series equivalent \$R_{43}=130\Omega+100\Omega=230\Omega\$.
2. Combine \$R_{43}\$ with \$R_2\$ to get parallel equivalent \$R_{243}=\frac{230\Omega\cdot 125\Omega}{230\Omega+125\Omega}=\frac{5750}{71}\Omega \approx 81\Omega\$.
3. Treat \$R_{243}\$ and \$R_1\$ as a voltage divider pair, now. For this, you'd get \$V_x = V\cdot\frac{R_{243}}{R_{243}+R_1} = \frac{115}{186}V_0\$, where \$V_0\$ is your voltage source of whatever kind (AC or DC.)
4. Now that you know \$V_x\$ you can plow that back in and see that \$R_4\$ and \$R_3\$ also form a voltage divider for that voltage. In this case, the result is \$V_y = V_x\cdot\frac{130\Omega}{130\Omega+100\Omega} = \frac{115}{186}\cdot\frac{13}{23}V_0 = \frac{65}{186}V_0\$

That's assuming your bottom node is treated as \$0V\$, of course.

A second approach would be to use Thevenin equivalents. In this case, the steps are like:

1. Disconnect \$R_4\$ and \$R_3\$ from the circuit and form a Thevenin from \$R_1\$ and \$R_2\$, computing \$V_{th_1}=V_0\cdot\frac{125\Omega}{125\Omega+50\Omega}=\frac{5}{7}V_0\$ and \$R_{th_1}=\frac{125\Omega\cdot 50\Omega}{125\Omega+50\Omega}=\frac{250}{7}\Omega\$.
2. Apply this new equivalent \$R_{th_1}\$ as being in series with \$R_4\$ and \$R_3\$, to compute a total series resistance of \$R_{tot}= \frac{1860}{7}\Omega\$.
3. Compute the series current as \$I_{tot}=\frac{V_{th_1}}{R_{tot}} = \frac{1}{372}V_0\$.
4. Multiply \$I_{tot}\$ by \$R_3\$ to get the voltage drop across \$R_3\$ as \$V_y=130\Omega\cdot\frac{1}{372}V_0 =\frac{65}{186}V_0\$.

Same answer, either way.

You could also use a variety of other methods, including nodal analysis:

^{simulate this circuit – Schematic created using CircuitLab}

Now, just place your mind at \$V_x\$ for a moment and imagine that there is a voltage there that causes current to spill away. That current would spill outward through all paths available. So you'd have an outward going sum like:

\$\frac{V_x}{R_1} + \frac{V_x}{R_2} + \frac{V_x}{R_4}\$

And current would spill back inward, also through all paths, so:

\$\frac{V_0}{R_1} + \frac{0V}{R_2} + \frac{V_y}{R_4}\$

These must be equal, so:

1. \$\frac{V_x}{R_1} + \frac{V_x}{R_2} + \frac{V_x}{R_4}=\frac{V_0}{R_1} + \frac{0V}{R_2} + \frac{V_y}{R_4}\$
2. \$V_x\cdot\left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_4}\right)=\frac{V_0}{R_1} + \frac{V_y}{R_4}\$
3. \$V_x=\frac{\frac{V_0}{R_1} + \frac{V_y}{R_4}}{\left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_4}\right)}\$

Now looking at \$V_y\$ and applying the same logic we can get:

4. \$\frac{V_y}{R_3} + \frac{V_y}{R_4} = \frac{0V}{R_3} + \frac{V_x}{R_4}\$
5. \$V_y\cdot\left(\frac{1}{R_3} + \frac{1}{R_4}\right) = \frac{V_x}{R_4}\$
6. \$V_y=\frac{\frac{V_x}{R_4}}{\left(\frac{1}{R_3} + \frac{1}{R_4}\right)}\$

You only want \$V_y\$ above, so substitute in \$V_x\$ and you get:

\$V_y = V_0\frac{R_2\cdot R_3}{R_1\cdot R_2 + R_1\cdot R_3 + R_1\cdot R_4 + R_2\cdot R_3 + R_2\cdot R_4}\$

After some algebra stuff. And the answer works out the same.

Answer 3

RS = R3+R4 = 130+100 = 230 ohms

RP = R2||RS = 230||125 = 77.74 ohms

Apply Voltage Divider rule to get voltage across RP,

V(RP) = RP/(RP+R1)*Vin = 77.74 /(77.74+50)*Vin = 0.608 *Vin

Now you have Voltage Across RP = Voltage Across series(R3+R4).

Again Voltage Divider Rule,

V(R3) = R3/(R3+R4)*V(RP) = 130/(130+100)*V(RP) = 0.56 * V(RP) = 0.3436 Vin