# Series Circuits (part 2) - Voltage Divider Equation

## Circuits

Series Circuits (part 2) - Voltage Divider Equation

Video Lectures created by Tim Feiegenbaum at North Seattle Community College.

We are continuing with section 4.2 and in this section, we will be looking at voltage dividers and power equations for series circuits. The voltage drops in a circuit are proportional to the resistance values, so what that is saying is that the voltage drops that we will experience in a given circuit are going to be proportional to the values of resistance in that circuit.

### Voltage Divider Equation

The voltage divider equation is very useful for determining the relationships in a series circuit. Here we have an equation we are solving for V or X and this is what we are solving for and this case we are solving for R1 so we will be solving what is the voltage drop across R1. This is the formula RX in this case R1, divided by the total resistance times the applied voltage. In this case, R1 is 10k and the total resistance will be 60. If we solve this 10 divided by 60 times 120 and in this case we can look at this and see 10 over 60 is going to be one-sixth of 120 is going to be 20 volts, a 20-volt drop… let us just to the next slide I have prepared a circuit for this.

We said this first one was 10 over 60 times 120 and that would be about 20volts and the next one here 20k, again we can do the same thing 20 divided by the total resistance which is 60 times 120 and again this looks like about one-third, and one third times 120 would be about 40 and then we have the final one R3, 30 over the total resistance of 60k times 120 and if we wanted to do… this is one-half as I say 0.5 you could do that, you could say one-half or whatever, but it is going to be half of 120 which is about 60volts. In this equation, we have a 20volt drop here, a 40volt drop here and 60 here and if we were to add these up it would equal 120volts and that equates to our applied voltage.

In the previous section we looked at Kirchhoff's law and what did Kirchhoff's law say? It said the applied voltages will equal the applied voltage so the voltage supply to each of the individual components will equal the voltage source. In this case the voltage across the individual opponents equals the source voltage.

### Power

Power in a resistance circuit can be thought of as heat dissipated by the resistors and we measure power in a quantity called watts. We talked about watts in earlier chapters the more power dissipated the hotter the resistor. Total power in a series circuit is determined by the sum of the individual power dissipations. The total power will equal the sum of all the individual powers within the circuit. We are looking at the same circuit again and we are going to use our little calculator this time. When we calculate power, there are a couple of different formulas I will use in this particular screen. We will look at current times resistance and we will also look at current squared times resistance.

First of all, if we are going to use this formula we need to know what the total current in the circuit is. We have our applied voltage which is 120volts and we will divide that by the total resistance within the circuit to find what is the total current. In this case, we have 120volts supplied and we divide that by the total resistance which is 60K and so we have our value of milliamps. The voltage drops in this circuit we calculated previously: R1 was 20volts, R2 was 40volts and R3 was 60volts. If we wanted to calculate the drops across R1 or the power drop across R1 we could say current times voltage so here we have our 2milliamps times the voltage drop which was 20 and that will equal 40milliwatts, so we could go to this component and document 40milliwatts.

Our next one we can do the same thing, let us put in our value of two exponents minus three which is our 2milliamps and again we would take that times the voltage drop on R2 which we previously calculated as being 40, and there should be no surprises here that it is 80 milliwatts, so we document this component as having an 80milliwatt draw. One the final one let us use the other formula and let us do the current squared times resistance so we are going to go two exponents minus three and we will take that and square it, then we will take that times the total resistance and in this case that is 30k, so we will say times 30 exponent three and we will get 120milliwatts. Here we have our final calculation here and if we were to sum these we would say 40 plus 80 plus 120 and there should be no surprises here, 240milliwatts. One way to test our calculation… in fact, there is a number of ways we could test this, but we know the total current is two milliamps and we could use this to solve for the total current in the circuit. In that case, we would say two milliamps exponent minus three and take that, times 120 and again, no surprises, 240milliwatts. We could use the same equation pie squared time R, we could say two exponents minus three and square that value then takes the times… we multiply it by… in this case we are looking for the total wattage so we would take that times 60K: 60 exponent 3 and again 240 milliwatts. So we have solved power in a couple of different ways.

This lesson we have looked at power in series circuits and we looked at how the power is equal to the sum of all the drops. We also looked at voltage divider calculations, we looked at the individual voltage drops across given components and we found that the voltage drop was proportional to the resistance of the total circuit times the applied voltage. This concludes this lesson.

Video Lectures created by Tim Fiegenbaum at North Seattle Community College.