FHSST Physics/Electricity/Simple Series Circuits

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The Free High School Science Texts: A Textbook for High School Students Studying Physics
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Electricity
Flow of Charge - Circuits - Voltage and Current - Resistance - Voltage and Current in a Practical Circuit - How Voltage, Current, and Resistance Relate

- Ohm's Law Analogy - Power in Electric Circuits - Calculating Electric Power - Resistors - Nonlinear Conduction - Circuit Wiring - Polarity of Voltage Drops - Series and Parallel - Simple Series Circuits - Simple Parallel Circuits - Power Calculations - Using Ohm's Law - Conductor Size - Fuses - Important Equations and Quantities

Simple series circuits[edit]

Let's start with a series circuit consisting of three resistors and a single battery:

Fhsst_electricity62.png

The first principle to understand about series circuits is that the amount of current is the same through any component in the circuit. This is because there is only one path for electrons to flow in a series circuit, and because free electrons flow through conductors like marbles in a tube, the rate of flow (marble speed) at any point in the circuit (tube) at any specific point in time must be equal.

From the way that the 9 volt battery is arranged, we can tell that the electrons in this circuit will flow in a counter-clockwise direction, from point 4 to 3 to 2 to 1 and back to 4. However, we have one source of voltage and three resistances. How do we use Ohm's Law here?

An important caveat to Ohm's Law is that all quantities (voltage, current, resistance, and power) must relate to each other in terms of the same two points in a circuit. For instance, with a single-battery, single-resistor circuit, we could easily calculate any quantity because they all applied to the same two points in the circuit:

Fhsst_electricity63.png

Fhsst_electricity64.png

Since points 1 and 2 are connected together with wire of negligible resistance, as are points 3 and 4, we can say that point 1 is electrically common to point 2, and that point 3 is electrically common to point 4. Since we know we have 9 volts of electromotive force between points 1 and 4 (directly across the battery), and since point 2 is common to point 1 and point 3 common to point 4, we must also have 9 volts between points 2 and 3 (directly across the resistor). Therefore, we can apply Ohm's Law (I =\frac{E}{R}) to the current through the resistor, because we know the voltage (E) across the resistor and the resistance (R) of that resistor. All terms (E, I, R) apply to the same two points in the circuit, to that same resistor, so we can use the Ohm's Law formula with no reservation.

However, in circuits containing more than one resistor, we must be careful in how we apply Ohm's Law. In the three-resistor example circuit below, we know that we have 9 volts between points 1 and 4, which is the amount of electromotive force trying to push electrons through the series combination of R_1, R_2, and R_3. However, we cannot take the value of 9 volts and divide it by 3k, 10k or 5k \Omega to try to find a current value, because we don't know how much voltage is across any one of those resistors, individually.

Fhsst_electricity65.png

The figure of 9 volts is a total quantity for the whole circuit, whereas the figures of 3k, 10k, and 5k \Omega are individual quantities for individual resistors. If we were to plug a figure for total voltage into an Ohm's Law equation with a figure for individual resistance, the result would not relate accurately to any quantity in the real circuit.

For R_1, Ohm's Law will relate the amount of voltage across R_1 with the current through R_1, given R_1's resistance, 3k\Omega:

Fhsst_electricity66.png

But, since we don't know the voltage across R_1 (only the total voltage supplied by the battery across the three-resistor series combination) and we don't know the current through R_1, we can't do any calculations with either formula. The same goes for R_2 and R_3: we can apply the Ohm's Law equations if and only if all terms are representative of their respective quantities between the same two points in the circuit.

So what can we do? We know the voltage of the source (9 volts) applied across the series combination of R_1, R_2, and R_3, and we know the resistances of each resistor, but since those quantities aren't in the same context, we can't use Ohm's Law to determine the circuit current. If only we knew what the total resistance was for the circuit: then we could calculate total current with our figure for total voltage (I =\frac{E}{R}).

This brings us to the second principle of series circuits: the total resistance of any series circuit is equal to the sum of the individual resistances. This should make intuitive sense: the more resistors in series that the electrons must flow through, the more difficult it will be for those electrons to flow. In the example problem, we had a 3 k\Omega, 10 k\Omega, and 5 k\Omega resistor in series, giving us a total resistance of 18 k\Omega:

Fhsst_electricity67.png

In essence, we've calculated the equivalent resistance of R_1, R_2, and R_3 combined. Knowing this, we could re-draw the circuit with a single equivalent resistor representing the series combination of R_1, R_2, and R_3:

Fhsst_electricity68.png

Now we have all the necessary information to calculate circuit current, because we have the voltage between points 1 and 4 (9 volts) and the resistance between points 1 and 4 (18 k\Omega):

Fhsst_electricity69.png

Knowing that current is equal through all components of a series circuit (and we just determined the current through the battery), we can go back to our original circuit schematic and note the current through each component:

Fhsst_electricity70.png

Now that we know the amount of current through each resistor, we can use Ohm's Law to determine the voltage drop across each one (applying Ohm's Law in its proper context):

Fhsst_electricity71.png

Notice the voltage drops across each resistor, and how the sum of the voltage drops (1.5 + 5 + 2.5) is equal to the battery (supply) voltage: 9 volts. This is the third principle of series circuits: that the supply voltage is equal to the sum of the individual voltage drops.

However, the method we just used to analyze this simple series circuit can be streamlined for better understanding. By using a table to list all voltages, currents, and resistances in the circuit, it becomes very easy to see which of those quantities can be properly related in any Ohm's Law equation:

Fhsst_electricity72.png

The rule with such a table is to apply Ohm's Law only to the values within each vertical column. For instance, E_R1 only with I_R1 and R_1 E_R2 only with I_R2 and R_2 etc. You begin your analysis by filling in those elements of the table that are given to you from the beginning:

Fhsst_electricity73.png

As you can see from the arrangement of the data, we can't apply the 9 volts of E_T (total voltage) to any of the resistances (R_1, R_2, or R_3) in any Ohm's Law formula because they're in different columns. The 9 volts of battery voltage is not applied directly across R_1, R_2, or R_3. However, we can use our "rules" of series circuits to fill in blank spots on a horizontal row. In this case, we can use the series rule of resistances to determine a total resistance from the sum of individual resistances:

Fhsst_electricity74.png

Now, with a value for total resistance inserted into the rightmost ("Total") column, we can apply Ohm's Law of I=\frac{E}{R} to total voltage and total resistance to arrive at a total current of 500 A:

Fhsst_electricity75.png

Then, knowing that the current is shared equally by all components of a series circuit (another "rule" of series circuits), we can fill in the currents for each resistor from the current figure just calculated:

Fhsst_electricity76.png

Finally, we can use Ohm's Law to determine the voltage drop across each resistor, one column at a time:

Fhsst_electricity77.png