As mentioned in the previous section of Lesson 4, two or more electrical devices in a circuit can be connected by series connections or by parallel connections. When all the devices are connected using series connections, the circuit is referred to as a **series circuit**. In a series circuit, each device is connected in a manner such that there is only one pathway by which charge can traverse the external circuit. Each charge passing through the loop of the external circuit will pass through each resistor in consecutive fashion.

A short comparison and contrast between series and parallel circuits was made in the previous section of Lesson 4. In that section, it was emphasized that the act of adding more resistors to a series circuit results in the rather expected result of having more overall resistance. Since there is only one pathway through the circuit, every charge encounters the resistance of every device; so adding more devices results in more overall resistance. This increased resistance serves to reduce the rate at which charge flows (also known as the current).

**Equivalent Resistance and Current**

Charge flows together through the external circuit at a rate that is everywhere the same. The current is no greater at one location as it is at another location. The actual amount of current varies inversely with the amount of overall resistance. There is a clear relationship between the resistance of the individual resistors and the overall resistance of the collection of resistors. As far as the battery that is pumping the charge is concerned, the presence of two 6-Ω ;resistors in series would be equivalent to having one 12-Ω resistor in the circuit. The presence of three 6-Ω resistors in series would be equivalent to having one 18-Ω resistor in the circuit. And the presence of four 6-Ω resistors in series would be equivalent to having one 24-Ω resistor in the circuit.

This is the concept of equivalent resistance. The **equivalent resistance** of a circuit is the amount of resistance that a single resistor would need in order to equal the overall effect of the collection of resistors that are present in the circuit. For series circuits, the mathematical formula for computing the equivalent resistance (R_{eq}) is

**R _{eq}**

**= R**

_{1}

**+ R**

_{2}

**+ R**

_{3}

**+ …**

where R_{1}, R_{2}, and R_{3} are the resistance values of the individual resistors that are connected in series.

The current in a series circuit is everywhere the same. Charge does NOT pile up and begin to accumulate at any given location such that the current at one location is more than at other locations. Charge does NOT become used up by resistors such that there is less of it at one location compared to another. The charges can be thought of as marching together through the wires of an electric circuit, everywhere marching at the same rate. Current – the rate at which charge flows – is everywhere the same. It is the same at the first resistor as it is at the last resistor as it is in the battery. Mathematically, one might write

**I _{battery}**

**= I**

_{1}

**= I**

_{2}

**= I**

_{3}

**= …**

where I_{1}, I_{2}, and I_{3} are the current values at the individual resistor locations.

These current values are easily calculated if the battery voltage is known and the individual resistance values are known. Using the individual resistor values and the equation above, the equivalent resistance can be calculated. And using Ohm’s law (ΔV = I • R), the current in the battery and thus through every resistor can be determined by finding the ratio of the battery voltage and the equivalent resistance.

**I _{battery}**

**= I**

_{1}

**= I**

_{2}

**= I**

_{3}

**= ΔV**

_{battery}

**/ R**

_{eq}**Electric Potential Difference and Voltage Drops**

As discussed in Lesson 1, the electrochemical cell of a circuit supplies energy to the charge to move it through the cell and to establish an electric potential difference across the two ends of the external circuit. A 1.5-volt cell will establish an electric potential difference across the external circuit of 1.5 volts. This is to say that the electric potential at the positive terminal is 1.5 volts greater than at the negative terminal. As charge moves through the external circuit, it encounters a loss of 1.5 volts of electric potential. This loss in electric potential is referred to as a **voltage drop**. It occurs as the electrical energy of the charge is transformed to other forms of energy (thermal, light, mechanical, etc.) within the resistors or loads. If an electric circuit powered by a 1.5-volt cell is equipped with more than one resistor, then the cumulative loss of electric potential is 1.5 volts. There is a voltage drop for each resistor, but the sum of these voltage drops is 1.5 volts – the same as the voltage rating of the power supply. This concept can be expressed mathematically by the following equation:

**ΔV _{battery}**

_{ }**= ΔV**

_{1}

**+ ΔV**

_{2}

**+ ΔV**

_{3}

**+ …**

To illustrate this mathematical principle in action, consider the two circuits shown below in Diagrams A and B. Suppose that you were to asked to determine the two unknown values of the electric potential difference across the light bulbs in each circuit. To determine their values, you would have to use the equation above. The battery is depicted by its customary schematic symbol and its voltage is listed next to it. Determine the voltage drop for the two light bulbs and then click the Check Answers button to see if you are correct.

Earlier in Lesson 1, the use of an electric potential diagram was discussed. An electric potential diagram is a conceptual tool for representing the electric potential difference between several points on an electric circuit. Consider the circuit diagram below and its corresponding electric potential diagram.

The circuit shown in the diagram above is powered by a 12-volt energy source. There are three resistors in the circuit connected in series, each having its own voltage drop. The negative sign for the electric potential difference simply denotes that there is a loss in electric potential when passing through the resistor. Conventional current is directed through the external circuit from the positive terminal to the negative terminal. Since the schematic symbol for a voltage source uses a long bar to represent the positive terminal, location A in the diagram is at the positive terminal or the high potential terminal. Location A is at 12 volts of electric potential and location H (the negative terminal) is at 0 volts. In passing through the battery, the charge gains 12 volts of electric potential. And in passing through the external circuit, the charge loses 12 volts of electric potential as depicted by the electric potential diagram shown to the right of the schematic diagram. This 12 volts of electric potential is lost in three steps with each step corresponding to the flow through a resistor. In passing through the connecting wires between resistors, there is little loss in electric potential due to the fact that a wire offers relatively little resistance to the flow of charge. Since locations A and B are separated by a wire, they are at virtually the same electric potential of 12 V. When a charge passes through its first resistor, it loses 3 V of electric potential and drops down to 9 V at location C. Since location D is separated from location C by a mere wire, it is at virtually the same 9 V electric potential as C. When a charge passes through its second resistor, it loses 7 V of electric potential and drops down to 2 V at location E. Since location F is separated from location E by a mere wire, it is at virtually the same 2 V electric potential as E. Finally, as a charge passes through its last resistor, it loses 2 V of electric potential and drops down to 0 V at G. At locations G and H, the charge is out of energy and needs an energy boost in order to traverse the external circuit again. The energy boost is provided by the battery as the charge is moved from H to A.

In Lesson 3, Ohm’s law (ΔV = I • R) was introduced as an equation that relates the voltage drop across a resistor to the resistance of the resistor and the current at the resistor. The Ohm’s law equation can be used for any individual resistor in a series circuit. When combining Ohm’s law with some of the principles already discussed on this page, a big idea emerges.

In series circuits, the resistor with the greatest resistance has the greatest voltage drop.

Since the current is everywhere the same within a series circuit, the I value of ΔV = I • R is the same in each of the resistors of a series circuit. So the voltage drop (ΔV) will vary with varying resistance. Wherever the resistance is greatest, the voltage drop will be greatest about that resistor. The Ohm’s law equation can be used to not only predict that resistor in a series circuit will have the greatest voltage drop, it can also be used to calculate the actual voltage drop values.

ΔV_{1}_{ }= I • R_{1} | ΔV_{2}_{ }= I • R_{2} | ΔV_{3}_{ }= I • R_{3} |

**Mathematical Analysis of Series Circuits**

The above principles and formulae can be used to analyze a series circuit and determine the values of the current at and electric potential difference across each of the resistors in a series circuit. Their use will be demonstrated by the mathematical analysis of the circuit shown below. The goal is to use the formulae to determine the equivalent resistance of the circuit (R_{eq}), the current at the battery (I_{tot}), and the voltage drops and current for each of the three resistors.

The analysis begins by using the resistance values for the individual resistors in order to determine the equivalent resistance of the circuit.

R_{eq} = R_{1} + R_{2} + R_{3} = 17 Ω + 12 Ω + 11 Ω = **40 Ω**

Now that the equivalent resistance is known, the current at the battery can be determined using the Ohm’s law equation. In using the Ohm’s law equation (ΔV = I • R) to determine the current in the circuit, it is important to use the battery voltage for ΔV and the equivalent resistance for R. The calculation is shown here:

I_{tot} = ΔV_{battery} / R_{eq} = (60 V) / (40 Ω) = **1.5 amp**

The 1.5 amp value for current is the current at the battery location. For a series circuit with no branching locations, the current is everywhere the same. The current at the battery location is the same as the current at each resistor location. Subsequently, the 1.5 amp is the value of I_{1}, I_{2}, and I_{3}.

I_{battery} = I_{1} = I_{2} = I_{3} = **1.5 amp**

There are three values left to be determined – the voltage drops across each of the individual resistors. Ohm’s law is used once more to determine the voltage drops for each resistor – it is simply the product of the current at each resistor (calculated above as 1.5 amp) and the resistance of each resistor (given in the problem statement). The calculations are shown below.

ΔV_{1 }= I_{1} • R_{1}ΔV_{1 }= (1.5 A) • (17 Ω)ΔV_{1 }= 25.5 V | ΔV_{2 }= I_{2} • R_{2}ΔV_{2 }= (1.5 A) • (12 Ω)ΔV_{2 }= 18 V | ΔV_{3 }= I_{3} • R_{3}ΔV_{3 }= (1.5 A) • (11 Ω)ΔV_{3 }= 16.5 V |

As a check of the accuracy of the mathematics performed, it is wise to see if the calculated values satisfy the principle that the sum of the voltage drops for each individual resistor is equal to the voltage rating of the battery. In other words, is ΔV_{battery }= ΔV_{1 }+ ΔV_{2 }+ ΔV_{3} ?

Is ΔV_{battery }= ΔV_{1 }+ ΔV_{2 }+ ΔV_{3} ?

Is 60 V = 25.5 V + 18 V + 16.5 V ?

Is 60 V = 60 V?

Yes!!

The mathematical analysis of this series circuit involved a blend of concepts and equations. As is often the case in physics, the divorcing of concepts from equations when embarking on the solution to a physics problem is a dangerous act. Here, one must consider the concepts that the current is everywhere the same and that the battery voltage is equivalent to the sum of the voltage drops across each resistor in order to complete the mathematical analysis. In the next part of Lesson 4, parallel circuits will be analyzed using Ohm’s law and parallel circuit concepts. We will see that the approach of blending the concepts with the equations will be equally important to that analysis.

**Check Your Understanding**

1. Use your understanding of equivalent resistance to complete the following statements:

a. Two 3-Ω resistors placed in series would provide a resistance that is equivalent to one _____-Ω resistor.

b. Three 3-Ω resistors placed in series would provide a resistance that is equivalent to one _____-Ω resistor.

c. Three 5-Ω resistors placed in series would provide a resistance that is equivalent to one _____-Ω resistor.

d. Three resistors with resistance values of 2-Ω , 4-Ω , and 6-Ω are placed in series. These would provide a resistance that is equivalent to one _____-Ω resistor.

e. Three resistors with resistance values of 5-Ω , 6-Ω , and 7-Ω are placed in series. These would provide a resistance that is equivalent to one _____-Ω resistor.

f. Three resistors with resistance values of 12-Ω, 3-Ω, and 21-Ω are placed in series. These would provide a resistance that is equivalent to one _____-Ω resistor.

**See Answer**

Use the equation for the equivalent resistance of a series circuit:

**R _{eq}**

**= R**

_{1}

**+ R**

_{2}

**+ R**

_{3}

**+ …**

a. Two 3-Ω resistors placed in series would provide a resistance which is equivalent to one **6-Ω** resistor.

b. Three 3-Ω resistors placed in series would provide a resistance which is equivalent to one **9-Ω** resistor.

c. Three 5-Ω resistors placed in series would provide a resistance which is equivalent to one **15-Ω** resistor.

d. Three resistors with resistance values of 2-Ω , 4-Ω , and 6-Ω are placed in series. These would provide a resistance which is equivalent to one **12-Ω** resistor.

e. Three resistors with resistance values of 5-Ω , 6-Ω , and 7-Ω are placed in series. These would provide a resistance which is equivalent to one **18-Ω** resistor.

f. Three resistors with resistance values of 12-Ω , 3-Ω , and 21-Ω are placed in series. These would provide a resistance which is equivalent to one **36-Ω** resistor.

2. As the number of resistors in a series circuit increases, the overall resistance __________ (increases, decreases, remains the same) and the current in the circuit __________ (increases, decreases, remains the same).

**See Answer**

As the number of resistors in a series circuit increases, the overall resistance **increases** and the current in the circuit **decreases**.

3. Consider the following two diagrams of series circuits. For each diagram, use arrows to indicate the direction of the conventional current. Then, make comparisons of the voltage and the current at the designated points for each diagram.

**See Answer**

The conventional current is directed through the external circuit from the positive terminal to the negative terminal. The magnitude of this current (I) is everywhere the same – thus the equal sign in the current comparisons. As charge progresses in the direction of the conventional current, there is a drop in electric potential every time it passes through a light bulb. By the time the charge reaches the negative terminal, its electric potential has dropped to zero volts. The closer a position is to the positive terminal, the higher its potential; and conversely, the closer a position is to the negative terminal, the lower its potential. If two locations are separated by a mere wire (such as location D and F in Diagram B), then their potential is approximately the same since there is relatively little resistance in a wire.

4. Three identical light bulbs are connected to a D-cell as shown at the right. Which one of the following statements is true?

a. All three bulbs will have the same brightness.

b. The bulb between X and Y will be the brightest.

c. The bulb between Y and Z will be the brightest.

d. The bulb between Z and the battery will be the brightest.

**See Answer**

Answer: **A**

The current in a series circuit is the same at each resistor present in the circuit. Since each light bulb has the same resistance (“identical bulbs”) and the same current, they will have the same power output (P = I^{2}R as discussed in the previous Lesson ). Thus, they will shine with the same brightness.

5. Three identical light bulbs are connected to a battery as shown at the right. Which adjustments could be made to the circuit that would increase the current being measured at X? List all that apply.

a. Increase the resistance of one of the bulbs.

b. Increase the resistance of two of the bulbs.

c. Decrease the resistance of two of the bulbs.

d. Increase the voltage of the battery.

e. Decrease the voltage of the battery.

f. Remove one of the bulbs.

**See Answer**

Answer: **C, D, and F**

The current at location X is the same as the current at the battery location. To increase the value of the current at the battery, it would be necessary to increase the battery voltage (choice D) or to decrease the equivalent resistance. Since the equivalent resistance is the sum of the resistance of the individual resistors, any decrease of resistance or removal of a resistor will lead to a decrease in the equivalent resistance.

6. Three identical light bulbs are connected to a battery as shown at the right. W, X,Y and Z represent locations along the circuit. Which one of the following statements is true?

a. The potential difference between X and Y is greater than that between Y and Z.

b. The potential difference between X and Y is greater than that between Y and W.

c. The potential difference between Y and Z is greater than that between Y and W.

d. The potential difference between X and Z is greater than that between Z and W.

e. The potential difference between X and W is greater than that across the battery.

f. The potential difference between X and Y is greater than that between Z and W.

**See Answer**

Answer: **D**

In a series circuit, the current is the same at each resistor. If the light bulbs are identical, then the resistance is the same for each resistor. The voltage drop (I•R) will be the same for each resistor since the current at and the resistance of each resistor is the same. Thus the electric potential difference across any one of the bulbs will be the same as that across any one of the other bulbs. And the electric potential difference across two or three bulbs will be greater than that across one bulb. The voltage boost in the battery will be equal to the sum of the voltage drops across all three resistors. Only choice D meets these criteria.

7. Compare circuit X and Y below. Each is powered by a 12-volt battery. The voltage drop across the 12-ohm resistor in circuit Y is ____ the voltage drop across the single resistor in X.

a. smaller than

b. larger than

c. the same as

**See Answer**

Answer: **A**

The voltage gained at the battery is equal to the accumulative voltage drop when passing through the external circuit. In circuit X, the voltage drop across the single resistor must be 12 V. In circuit Y, the voltage drop across the 12 ohm resistor must be less than 12 V since there will be an additional voltage drop in the 6 ohm resistor. In fact, one might reason that the two voltage drops in circuit Y will be 8 volts and 4 volts respectively, to add to a total voltage drop of 12 volts.

8. A 12-V battery, a 12-ohm resistor and a light bulb are connected as shown in circuit X below. A 6-ohm resistor is added to the 12-ohm resistor and bulb to create circuit Y as shown. The bulb will appear ____.

a. dimmer in circuit X

b. dimmer in circuit Y

c. the same brightness in both circuits

**See Answer**

Answer: **B**

The brightness of a light bulb is related to the power output of the bulb, which can be computed as the I • ΔV (see previous Lesson). Since Circuit Y contains an additional resistor, its equivalent resistance is greater than that of Circuit X. As such, Circuit X has a greater current than that of Circuit Y. The voltage impressed across each circuit is the same – 12 volts (the battery voltage). This 12 volts of electric potential difference is divided among the various circuit elements. There are two resistors and a light bulb in Circuit Y and only one resistor and a light bulb in Circuit X. And so the light bulb of Circuit X will have a greater ΔV than the light bulb of Circuit Y. And so it is reasonable to conclude that the light bulb of Circuit X has the greatest power – that is, the greatest I • ΔV product. And since the bulb brightness will depend upon this power value, one must conclude that the bulb will appear brighter in Circuit X and dimmer in Circuit Y.

close

9. Three resistors are connected in series. If placed in a circuit with a 12-volt power supply. Determine the equivalent resistance, the total circuit current, and the voltage drop across and current at each resistor.

**See Answer**

The analysis begins by using the resistance values for the individual resistors in order to determine the equivalent resistance of the circuit.

R_{eq} = R_{1} + R_{2} + R_{3} = 11 Ω + 7 Ω + 20 Ω = **38 Ω**

Now that the equivalent resistance is known, the current through the battery can be determined using Ohm’s law equation. In using the Ohm’s law equation (ΔV = I • R) to determine the current in the circuit, it is important to use the battery voltage for ΔV and the equivalent resistance for R. The calculation is shown here:

I_{tot} = ΔV_{battery} / R_{eq} = (12 V) / (38 Ω) = **0.31579 Amp**

The 0.316 Amp value for current is the current at the battery location. For a series circuit with no branching locations, the current is everywhere the same. The current at the battery location is the same as the current at each resistor location. Subsequently, the 0.316 Amp (rounded) is the value of I_{1}, I_{2}, and I_{3}.

I_{battery} = I_{1} = I_{2} = I_{3} = **0.316 Amp (rounded)**

There are three values left to be determined – the voltage drops across each of the individual resistors. Ohm’s law is used once more to determine the voltage drops for each resistor – it is simply the product of the current at each resistor (calculated above as 0.31579 Amp) and the resistance of each resistor (given in the problem statement). The calculations are shown below.

ΔV_{1 }= I_{1} • R_{1}ΔV_{1 }= (0.31579 A) • (11 Ω)ΔV_{1 }= 3.47 V | V_{2 }= I_{2} • R_{2}ΔV_{2 }= (0.31579 A) • (7 Ω)ΔV_{2 }= 2.21 V | ΔV_{3 }= I_{3} • R_{3}ΔV_{3 }= (0.31579 A) • (20 Ω)ΔV_{3 }= 6.32 V |

As a check of the accuracy of the mathematics performed, it is wise to see if the calculated values satisfy the principle that the sum of the voltage drops for each individual resistor is equal to the voltage rating of the battery. In other words, is ΔV_{battery }= ΔV_{1 }+ ΔV_{2 }+ ΔV_{3} ?

Is ΔV_{battery }= ΔV_{1 }+ ΔV_{2 }+ ΔV_{3} ?

Is 12 V = 3.47 V + 2.21 V + 6.32 V ?

Is 12 V = 12 V?

Yes!!