Resistors in Series: Analysis and Calculations - Tenth Grade Science
When designing electrical networks, individual circuit components can be interconnected in several geometric configurations to control current and divide voltage. When components are connected end-to-end sequentially along a single continuous conducting path, they are said to be connected in series. Analyzing series circuits reveals how electrical resistance accumulates, how potential differences distribute across loads, and how circuit current behaves under single-path constraints.
Characteristics of a Series Resistor Network
A series circuit provides a single conducting loop through which electric charges must travel sequentially.
THREE RESISTORS CONNECTED IN SERIES
+--------- R1 --------- R2 --------- R3 ---------+
| [====] [====] [====] |
| <-V1-> <-V2-> <-V3-> |
| |
| I (Current) |
+------------(A)---------[/ ]---------( + - )----+
Ammeter Switch Battery (V)
1. Invariance of Current Throughout the Circuit
Because there are no branching junctions in a series loop, electric charge cannot accumulate, leak, or bypass components. According to the principle of conservation of electric charge, the rate of charge entering any point in the circuit must equal the rate of charge leaving that point. Therefore, the electric current ($I$) flowing through each individual resistor in a series combination is identical:
$$I = I_1 = I_2 = I_3 = \dots = I_n$$
An ammeter placed anywhere in the series loop will register the exact same current reading.
2. Division of Total Potential Difference
While the current remains invariant, the total potential difference ($V$) supplied by the power source is distributed across the series components. The work done in moving a unit charge through the complete series chain is equal to the sum of the work done in transporting that charge across each individual resistor:
$$V = V_1 + V_2 + V_3 + \dots + V_n$$
Each component creates an electrical potential drop ($V_k$) proportional to its resistance.
Derivation of the Equivalent Series Resistance
The equivalent resistance ($R_s$) of a series combination represents the value of a single hypothetical resistor that could replace the entire network while drawing the identical current from the voltage source.
EQUIVALENT CIRCUIT MODEL
Series Network Equivalent Circuit
+--[R1]--[R2]--[R3]--+ +-----[Rs]-----+
| | | |
+-------( V )--------+ ==> +-----( V )----+
Current drawn = I Current drawn = I
Step-by-Step Mathematical Proof
Consider three resistors having individual resistances $R_1$, $R_2$, and $R_3$ connected in series across a source of potential difference $V$. Let a steady current $I$ flow through the series circuit. Applying Ohm's law ($V = I R$) individually across each of the three resistors gives:
$$V_1 = I R_1, \quad V_2 = I R_2, \quad V_3 = I R_3$$
The total potential difference across the entire combination is the sum of the individual potential drops:
$$V = V_1 + V_2 + V_3$$
Substituting the individual Ohm's law expressions into the total potential equation yields:
$$V = I R_1 + I R_2 + I R_3 = I (R_1 + R_2 + R_3)$$
If the complete series network is replaced by an equivalent resistance $R_s$, then by Ohm's law applied to the entire system:
$$V = I R_s$$
Equating the two expressions for the total voltage $V$:
$$I R_s = I (R_1 + R_2 + R_3)$$
Dividing both sides of the equation by the non-zero current $I$ gives the fundamental series resistance formula:
$$R_s = R_1 + R_2 + R_3$$
For an arbitrary collection of $n$ resistors connected in series:
$$R_s = \sum_{k=1}^n R_k = R_1 + R_2 + R_3 + \dots + R_n$$
Core Physical Insights
The equivalent resistance of any series combination is always strictly greater than the resistance of the largest individual resistor in that combination. Connecting resistors in series increases the total resistance of a circuit because it effectively increases the overall length of the conducting path traversed by electrons. If $n$ identical resistors each having resistance $R$ are connected in series, the equivalent resistance simplifies to:
$$R_s = n R$$
The Voltage Divider Principle
In a series circuit, the total applied voltage divides across the resistors in direct proportion to their individual resistances.
THE VOLTAGE DIVIDER PRINCIPLE
Total Voltage V
+---------------------------+
| |
v v
+-------+ +-------+
| R1 | | R2 |
+-------+ +-------+
^ ^
| |
V1 = V*R1/(R1+R2) V2 = V*R2/(R1+R2)
Voltage Distribution Formulation
Because the current is uniform throughout the series branch ($I = V / R_s$), the voltage drop across any individual resistor $R_k$ can be calculated directly:
$$V_k = I R_k = \left(\frac{V}{R_s}\right) R_k = V \left(\frac{R_k}{R_1 + R_2 + \dots + R_n}\right)$$
In a two-resistor network ($R_1$ and $R_2$), the individual voltage drops are:
$$V_1 = V \left(\frac{R_1}{R_1 + R_2}\right), \quad V_2 = V \left(\frac{R_2}{R_1 + R_2}\right)$$
This principle is widely utilized in electronic sensor circuits, potentiometer volume dials, and reference bias networks.
Disadvantages of Series Connections in Practical Wiring
Although series circuits are simple to design, they present significant drawbacks that make them unsuitable for domestic household wiring.
FAILURE PROPAGATION IN A SERIES STRING
Intact Circuit:
+-----(Lamp 1)-----(Lamp 2)-----(Lamp 3)-----+ Current flows
Bulb 2 Filament Breaks:
+-----(Lamp 1)----- -X- -----(Lamp 3)-----+ CIRCUIT BROKEN!
ALL LAMPS GO OUT
1. Complete Failure Upon Single Component Fault
Because a series network has only one conducting path, if any single component fails (for example, if a bulb filament breaks or a plug loosens), the circuit opens immediately. Charge flow stops across the entire line, causing all other connected devices to lose power. Identifying the faulty bulb in a long string of traditional series holiday lights can be time-consuming.
2. Inability to Control Devices Independently
In a series arrangement, all connected electrical devices must share a single common switch. It is impossible to turn on a reading lamp without simultaneously powering the room fan, heater, and television connected in that series branch.
3. Reduced Operating Voltage for Subsequent Loads
Household appliances are engineered to operate at a specific standard voltage (such as 120 V or 230 V). In a series circuit, each added device increases the total circuit resistance and claims a share of the total available voltage. If three identical 230 V light bulbs are wired in series to a 230 V supply, each bulb receives only $230/3 \approx 76.7 \text{ V}$. Operating well below their design voltage, the bulbs glow dimly.
4. Mismatched Current Requirements
Different domestic appliances require vastly different operational currents to function properly. For example, a heavy electric space heater may draw 10 amperes, while a miniature LED indicator requires only 0.02 amperes. Placing them in a series circuit forces both devices to carry the same current, which either starves the high-power load or damages the low-power device.
Chapter Exercises and Solutions
Exercise 1: Three resistors having values of $5.0\ \Omega$, $10.0\ \Omega$, and $15.0\ \Omega$ are connected in series across a steady 12.0-volt DC battery. Calculate the equivalent resistance of the network, the electric current supplied by the battery, and the individual potential difference across each resistor.
Solution 1: The equivalent resistance of a series network is the sum of the individual resistances: $R_s = R_1 + R_2 + R_3 = 5.0\ \Omega + 10.0\ \Omega + 15.0\ \Omega = 30.0\ \Omega$. Applying Ohm's law to the total circuit yields the current: $I = V / R_s = 12.0 \text{ V} / 30.0\ \Omega = 0.40 \text{ A}$. Because the circuit is in series, a uniform current of $0.40 \text{ A}$ flows through each resistor. Using Ohm's law across each component gives: $V_1 = I R_1 = 0.40 \text{ A} \times 5.0\ \Omega = 2.0 \text{ V}$, $V_2 = I R_2 = 0.40 \text{ A} \times 10.0\ \Omega = 4.0 \text{ V}$, and $V_3 = I R_3 = 0.40 \text{ A} \times 15.0\ \Omega = 6.0 \text{ V}$. Summing the individual voltage drops confirms that $V_1 + V_2 + V_3 = 2.0 \text{ V} + 4.0 \text{ V} + 6.0 \text{ V} = 12.0 \text{ V}$, matching the source voltage.
Exercise 2: An electric lamp having an operating resistance of $20\ \Omega$ and a separate conductor of resistance $4\ \Omega$ are wired in series to a 6-volt battery. Calculate the overall circuit resistance, the current in the loop, and the potential difference across the lamp terminals.
Solution 2: The total equivalent resistance of the series combination is $R_s = R_{\text{lamp}} + R_{\text{wire}} = 20\ \Omega + 4\ \Omega = 24\ \Omega$. The steady circuit current drawn from the battery is $I = V / R_s = 6 \text{ V} / 24\ \Omega = 0.25 \text{ A}$. The potential difference developed across the electric lamp is $V_{\text{lamp}} = I R_{\text{lamp}} = 0.25 \text{ A} \times 20\ \Omega = 5.0 \text{ V}$. The remaining $1.0 \text{ V}$ drops across the $4\ \Omega$ conductor ($0.25 \text{ A} \times 4\ \Omega = 1.0 \text{ V}$).
Exercise 3: You are provided with a box containing twenty identical $2.5\ \Omega$ miniature resistors. Explain how you could configure these components in a series network to achieve an equivalent resistance of exactly $50\ \Omega$, and determine what current would flow through each resistor if the combination is connected to a 100-volt line.
Solution 3: In a series network composed of $n$ identical resistors each having resistance $R$, the total equivalent resistance is given by $R_s = n R$. To achieve an equivalent resistance of $50\ \Omega$ using $2.5\ \Omega$ units: $n = R_s / R = 50\ \Omega / 2.5\ \Omega = 20$. Connecting all 20 resistors in a single continuous series chain produces an equivalent resistance of $50\ \Omega$. Connecting this chain across a 100-volt source yields a circuit current of $I = V / R_s = 100 \text{ V} / 50\ \Omega = 2.0 \text{ A}$. Because the components are in series, the exact same current of $2.0 \text{ A}$ flows through each of the twenty resistors, producing a potential drop of $V_k = I R = 2.0 \text{ A} \times 2.5\ \Omega = 5.0 \text{ V}$ across each component.