#### Quiz

Q1) When three resistors are combined in series the total resistance of the combination is

A. greater than any of the individual resistance values.

B. less than any of the individual resistance values.

C. the average of the individual resistance values.

Answer) A.

Q2) Several resistors are wired in series. What is true about this circuit?

A. The sum of the currents through each of the resistors is equal to the total circuit current.

B. The total circuit current is the same as the current through any one of the resistors.

C. The voltage across any resistor is the same as the voltage of the power supply.

D. The current in any single resistor is determined by its resistance and the voltage of the power supply.

Answer) B.

A. The sum of the currents through each of the resistors is equal to the total circuit current.

B. The total circuit current is the same as the current through any one of the resistors.

C. The voltage across any resistor is the same as the voltage of the power supply.

D. The current in any single resistor is determined by its resistance and the voltage of the power supply.

Answer) B.

#### Resistors in series total resistance Simulation

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**RESISTORS IN SERIES**

In a circuit that consists of a single bulb and a battery, the potential difference across the bulb equals the terminal voltage. The total current in the circuit can be found using the equation $\Delta V = IR$.

What happens when a second bulb is added to such a circuit, as shown in Figure 1. When moving through this circuit, charges that pass through one bulb must also move through the second bulb. Because all charges in the circuit must follow the same conducting path, these bulbs are said to be connected in series.

**Resistors in series carry the same current**

Light-bulb filaments are resistors; thus, Figure 1(b) represents the two bulbs in Figure 1(a) as resistors. Because charge is conserved, charges cannot build up or disappear at a point. For this reason, the amount of charge that enters one bulb in a given time interval equals the amount of charge that exits that bulb in the same amount of time. Because there is only one path for a charge to follow, the amount of charge entering and exiting the first bulb must equal the amount of charge that enters and exits the second bulb in the same time interval.

Because the current is the amount of charge moving past a point per unit of time, the current in the first bulb must equal the current in the second bulb. This is true for any number of resistors arranged in series.

*When many resistors are connected in series, the current in each resistor is the same.*
The total current in a series circuit depends on how many resistors are present and on how much resistance each offers. Thus, to find the total current, first use the individual resistance values to find the total resistance of the circuit, called the equivalent resistance. Then the equivalent resistance can be used to find the current.

*Figure 1. These two light bulbs are connected in series. Because light-bulb filaments are resistors, (a) the two bulbs in this series circuit can be represented by (b) two resistors in the schematic diagram shown on the right.*

**The equivalent resistance in a series circuit is the sum of the circuit’s resistances**

As described in Section 1, the potential difference across the battery, ΔV, must equal the potential difference across the load, $\Delta V_{1}$ + $\Delta V_{2}$ , where $\Delta V_{1}$ is the potential difference across $\ R_{1}$ and $\Delta V_{2}$ is the potential difference across $\ R_{2}$.

$\Delta V = \Delta V_{1} + \Delta V_{2}$

According to $\ V = IR$, the potential difference across each resistor is equal to the current in that resistor multiplied by the resistance.

$\Delta V = I_{1} R_{1} + I_{2} R_{2}$

Because the resistors are in series, the current in each is the same. For this reason, $\ I_{1}$and $\ I_{2}$ can be replaced with a single variable for the current, $\ I$.

$\Delta V = I ( R_{1} + R_{2})$

*Figure 2. (a) The two resistors in the actual circuit have the same effect on the current in the circuit as (b) the equivalent resistor.*

Finding a value for the equivalent resistance of the circuit is now possible.

If you imagine the equivalent resistance replacing the original two resistors, as shown in Figure 2, you can treat the circuit as if it contains only one resistor and use $\Delta V = IR$ to relate the total potential difference, current, and equivalent resistance.

$\Delta V = I(R_{eq})$

Now set the last two equations for V equal to each other, and divide by the current.

$\Delta V = I(R_{eq}) = I(R_{1} + R_{2})$

$\ R_{eq} = R_{1} + R_{2}$

Thus, the equivalent resistance of the series combination is the sum of the individual resistances. An extension of this analysis shows that the equivalent resistance of two or more resistors connected in series can be calculated using the following equation.

RESISTORS IN SERIES$\ R_{eq} = R_{1} + R_{2} + R_{3} . . .$Equivalent resistance equals the total of individual resistances in series. |

Because $\ R_{eq} $ represents the sum of the individual resistances that have been connected in series, the equivalent resistance of a series combination of resistors is always greater than any individual resistance.

To find the total current in a series circuit, first simplify the circuit to a single equivalent resistance using the boxed equation above; then use $\Delta V = IR$ to calculate the current.

$\ I=\frac{\Delta V}{R_{eq}}$

Because the current in each bulb is equal to the total current, you can also use $\Delta V = IR$ to calculate the potential difference across each resistor.

$\Delta V_{1} = IR_{1}$ and $\Delta V_{2} = IR_{2}$

The method described above can be used to find the potential difference across resistors in a series circuit containing any number of resistors.

What happens to a series circuit when a single bulb burns out? Consider what a circuit diagram for a string of lights with one broken filament would look like. As the schematic diagram in Figure 3 shows, the broken filament means that there is a gap in the conducting pathway used to make up the circuit. Because the circuit is no longer closed, there is no current in it and all of the bulbs go dark.

Why, then, would anyone arrange resistors in series? Resistors can be placed in series with a device in order to regulate the current in that device. In the case of decorative lights, adding an additional bulb will decrease the current in each bulb. Thus, the filament of each bulb need not withstand such a high current.

Another advantage to placing resistors in series is that several lesser resistances can be used to add up to a single greater resistance that is unavailable. Finally, in some cases, it is important to have a circuit that will have no current if any one of its component parts fails. This technique is used in a variety of contexts, including some burglar alarm systems.

**Series circuits require all elements to conduct**What happens to a series circuit when a single bulb burns out? Consider what a circuit diagram for a string of lights with one broken filament would look like. As the schematic diagram in Figure 3 shows, the broken filament means that there is a gap in the conducting pathway used to make up the circuit. Because the circuit is no longer closed, there is no current in it and all of the bulbs go dark.

*Figure 3. A burned-out filament in a bulb has the same effect as an open switch. Because this series circuit is no longer complete, there is no current in the circuit.*Why, then, would anyone arrange resistors in series? Resistors can be placed in series with a device in order to regulate the current in that device. In the case of decorative lights, adding an additional bulb will decrease the current in each bulb. Thus, the filament of each bulb need not withstand such a high current.

Another advantage to placing resistors in series is that several lesser resistances can be used to add up to a single greater resistance that is unavailable. Finally, in some cases, it is important to have a circuit that will have no current if any one of its component parts fails. This technique is used in a variety of contexts, including some burglar alarm systems.