*Notes to a video lecture on http://www.unizor.com*

__Direct Current - Voltage Drop__

Imagine a waterfall going down the rocks. As it hits each rock, it loses some of its potential energy.

Analogously, consider several radiators in a series connected to a boiler that supplies hot water for them. As the water goes from one radiator to another, it cools down, losing its heat energy.

The same happens with electrons, as they go along the circuit and meet one resistance after another. Going through each of them, they lose their energy.

Consider the following circuit

Assume, we know how to measure the difference in potential (

*voltage*) between any two points on this circuit. Measuring the voltage between the terminals of the battery gives some value

*.*

**U**Measuring the voltage between the positive terminal of the battery (the longer thin line with a plus sign) and the point in-between the resistors gives some other value

*.*

**U**_{1}Measuring the voltage between the negative terminal of the battery (the shorter thick line with a minus sign) and the point in-between the resistors gives yet other value

*.*

**U**_{2}Notice that the amount of electricity going through this circuit per unit of time (

*electric current*or

*amperage*) is the same everywhere since it's a closed loop and equals

*.*

**I**Now let's apply the Ohm's Law to an entire circuit, keeping in mind that the voltage between the terminals of the battery is

*and the total resistance of an entire circuit is*

**U***:*

**R=R**_{1}+R_{2}

**I = U / R = U / (R**_{1}+R_{2})Applying the Ohm's Law to a segment of a circuit that includes only resistor

*and knowing the voltage on its two ends*

**R**_{1}*, we obtain*

**U**_{1}

**I = U**_{1}/ R_{1}Equating two different expressions for the electric current

*, we can find the voltage*

**I***:*

**U**_{1}

**I = U / (R**_{1}+R_{2}) = U_{1}/ R_{1}from which the value of

*is*

**U**_{1}

**U**_{1}= U·R_{1}/ (R_{1}+R_{2})Let's do the same calculation for the second resistor.

**I = U / (R**_{1}+R_{2}) = U_{2}/ R_{2}from which the value of

*is*

**U**_{2}

**U**_{2}= U·R_{2}/ (R_{1}+R_{2})Notice that

*.*

**U**_{1}+ U_{2}= UIndeed,

**U·R**

+ U·R

= U·(R_{1}/ (R_{1}+R_{2}) ++ U·R

_{2}/ (R_{1}+R_{2}) == U·(R

_{1}+R_{2}) / (R_{1}+R_{2}) = UCompletely analogous calculations can be provided with three or more resistors connected in a

*series*.

So, the difference in electric potential or

**voltage drop**between the terminals of a battery

*is split between the voltage drops on each resistor in a*

**U***series*.

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