Electronic – Is voltage divider applicable on circuits where there is no ground

Voltage Divider Basics and Analysis

I noticed on a later comment that you were really trying to understand how this could (potentially) be done using voltage divider theory. To start, let me remind you what a voltage divider really is:

Take this resistive voltage divider from Wikipedia:

The common equation to find the output is: V_out = R2 / (R1 + R2) * V_in

But what this really boils down to is this: V_out = R2 * I_R2

Of course, in this circuit, the current through R1 and R2 is equal, so finding the value is easily done by dividing the total voltage by the total current: I = V_in / (R1 + R2)

Because R2 is located between V_out and ground, the voltage across R2 is equal to V_out. However, what would happen if a third resistor were placed in parallel to R2? Consider this modified version of the above image:

The current relationship has now changed. In this circuit, I_R1 = I_R2 + I_R3. This makes the equation for the output a bit more complex: V_out = (R2 || R3) / (R1 + (R2 || R3)) * V_in

Because R2 and R3 are in parallel: (R2 || R3) = (R2 * R3) / (R2 + R3)

But the voltage across R2 is still equal to V_out. Adding another series resistor changes this:

V_out = ((R2 || R3) + R4) / (R1 + (R2 || R3) + R4) * Vin

Now, the voltage across R2 is really equal to V_out - V_R4. Adding more resistors will only further skew the relationship between V_out and V_R2. Adding a resistor in series to R3 doesn't change things too much:

V_out = ((R2 || (R3 + R4)) + R5) / (R1 + (R2 || (R3 + R4)) + R5) * Vin

where R2 is in parallel with the series combination of R3 and R4... This is still a big resistive voltage divider with random parallel and series resistors. The voltage across R2 is equal to V_out - V_R5.

The real complication comes with the introduction of the bottom half of the Wheatstone Bridge formed by resistors 2 through 6. Your particular circuit is very simple because all of the resistors are of equal value (the bridge is balanced), but it is a bit harder to formulate equations for a generic circuit which might have a voltage across R4:

If you really want to keep using basic resistive relationships, the next best step is to simplify the circuit. One method is to do a Delta-Wye Transform, as you mentioned in the question. However, it makes a lot more sense to do it on the the bottom half or right side of the bridge where R2 is not involved. For example, converting the bottom half of the bridge yields:

The new resistor values are found as:

• Ra = R4 * R5 / (R4 + R5 + R6)
• Rb = R4 * R6 / (R4 + R5 + R6)
• Rc = R5 * R6 / (R4 + R5 + R6)

But this still doesn't solve the problem because you need to know the voltage across R2 which is now equal to V_out - V_Na. This can still be solved using voltage dividers if you use a couple of different points as outputs:

The final solution (the voltage across R2) can finally be found using a combination of various voltage dividers and the following equations:

• Vx = [((R2 + Ra) || (R3 + Rb)) +Rc +R7] / [R1 + ((R2 + Ra) || (R3 + Rb)) +Rc +R7] * V_in
• Vy = (Rc + R7) / [R1 + ((R2 + Ra) || (R3 + Rb)) + Rc + R7] * V_in
• I_R2 = (Vx - Vy) / (R2 + Ra)
• V_R2 = R2 * I_R2

Since you said you already found the answer using Mesh Analysis, I will go ahead and put the values you should get from the above equations:

• Vx = 10V
• V7 = 6 & 2/3 V = 6.666...V
• I_R2 = 0.025A
• V_R2 = 2.5V

So it is possible to solve the circuit using nothing but voltage dividers, but we needed help from a resistor transformation. All of the basic circuit analysis techniques are based on Ohm's law: V = I*R, some of them just make more sense to use at times than others do.

Using Nodal Analysis

Considering all of your resistors are the same value, there are a few tricks to easily reduce the entire circuit down, but since that is pretty unrealistic in real life, I think it would be better to teach you a real analysis tool. There are many ways to analyse a circuit, but one of the easiest and best for many types of problems is Nodal Analysis. It is very easy to learn.

1) According to Kirchhoff's current law (KCL), the current going into any one node (point in the circuit) must be equal to the current coming out of any one node. So divide the circuit up into individual currents flowing from one node to another. The direction of the current flow is entirely up to you, it will just affect the polarity of the resulting current:

2) Form relationships between the currents:

• I1 = I2 + I3
• I2 = I4 + I5
• I6 = I3 + I4
• I7 = I5 + I6
• I7 = I1

3) Express the currents as voltages/resistances

• I1 = (15V - V3) / R1
• I2 = (V3 - V4) / R2
• I3 = (V3 - V6) / R3
• I4 = (V4 - V6) / R4
• I5 = (V4 - V5) / R5
• I6 = (V6 - V5) / R6
• I7 = (V5 - 0V) / R7

4) Using Algebra, combine and reduce the equations to find the values of the unknown voltages. To answer your specific question, the voltage across R2 would be V3 - V6 which is equal to R2 * I2.

^{simulate this circuit – Schematic created using CircuitLab}

Here's an example, see if this makes sense.

Consider the circuit on the left. Assume the meter VM1 is perfect- it has infinite input resistance. The voltage VM1 measures is \$ 10V \cdot \$ \$R2 \over (R1 + R2) \$ = 5V

Consider the circuit on the right. The meter VM2 has an input resistance of 1M represented by an ideal meter in parallel with 1M ohms. The equivalent resistance of R4 in parallel with R5 is R4||R5 = \$ R4\cdot R5 \over (R4 + R5) \$ = 90.909K ohms. The voltage VM2 reads is then \$ 10V \cdot \$ \$R4 || R5 \over (R3 + R4 || R5) \$ = 4.76V

The loading down of the 1M resistance reduced the reading by about 5%.

If you've been doing Thévenin equivalents, this should come as no surprise, since the unloaded divider could be replaced by a 5V source with 50K in series, and 1M is 20 times the source resistance.