A series circuit is an electrical configuration in which all components are connected end-to-end in a single, continuous path. This arrangement allows the current to flow through each component sequentially. The total resistance in a series circuit is equal to the sum of the individual resistances of each component. In this type of circuit, the current remains constant throughout all components, while the voltage divides across them in proportion to their resistances. Series circuits are commonly used in various electronic devices, such as holiday light strings and simple battery-operated devices, due to their simplicity and ease of analysis. However, one drawback of series circuits is that if one component fails or is disconnected, the entire circuit is interrupted, causing all components to stop functioning.

## Series Circuits

Characteristics:

- Components are connected end-to-end in a single path.
- The same current flows through all components.
- The total resistance is the sum of the individual resistances.
- The source voltage is distributed across the components based on their resistance values.

Analysis:

- Determine the total resistance: R_total = R1 + R2 + … + Rn
- Calculate the current using Ohm’s Law: I = V_total / R_total
- Determine the voltage drop across each component: V_component = I * R_component
- Calculate power dissipation for each component: P_component = I^2 * R_component

Applications:

- Current limiting: Series resistors are often used to limit the current flowing through a circuit or component.
- Voltage division: Voltage dividers can be created by connecting resistors in series, allowing for the distribution of voltage across multiple components.
- Sensor circuits: Series circuits are used in sensor networks, where multiple sensors are connected in series to detect changes in current flow.

## Example – Calculation of DC Series Circuit

Let’s consider a simple DC series circuit with a voltage source (V) and three resistors (R1, R2, and R3) connected in series. The goal is to calculate the current (I) through the circuit and the voltage across each resistor.

Given values:

- V = 12 V (DC)
- R1 = 4 Ω
- R2 = 6 Ω
- R3 = 2 Ω

Step 1: Determine the total resistance (R_total) of the series circuit:

R_total = R1 + R2 + R3 = 4 Ω + 6 Ω + 2 Ω = 12 Ω

Step 2: Calculate the current (I) through the circuit:

Since it’s a series circuit, the same current flows through all the resistors. We can use Ohm’s Law to find the current:

I = V / R_total = 12 V / 12 Ω = 1 A

Step 3: Calculate the voltage across each resistor:

We can use Ohm’s Law again to find the voltage across each resistor:

V_R1 = I * R1 = 1 A * 4 Ω = 4 V V_R2 = I * R2 = 1 A * 6 Ω = 6 V V_R3 = I * R3 = 1 A * 2 Ω = 2 V

In conclusion, the current (I) through the series circuit is 1 A, and the voltage across R1 (V_R1), R2 (V_R2), and R3 (V_R3) are 4 V, 6 V, and 2 V, respectively. Note that the sum of the individual voltages equals the source voltage:

V = V_R1 + V_R2 + V_R3 = 4 V + 6 V + 2 V = 12 V

## Example – Calculation of AC Series Circuit

Let’s consider a simple series circuit with one resistor (R), one capacitor (C), and one inductor (L) connected in series to an AC voltage source (V). The goal is to calculate the current (I) and voltage across each component. We will use the phasor analysis technique for this calculation.

Given values:

- V = 120 V (rms) at a frequency of 60 Hz
- R = 10 Ω
- L = 200 mH
- C = 100 μF

Step 1: Calculate the angular frequency (ω):

ω = 2 * π * f = 2 * π * 60 ≈ 377 rad/s

Step 2: Calculate the inductive reactance (X_L) and capacitive reactance (X_C):

X_L = ω * L = 377 * 0.2 = 75.4 Ω X_C = 1 / (ω * C) = 1 / (377 * 100 * 10^(-6)) ≈ 26.5 Ω

Step 3: Determine the total impedance (Z) of the series circuit:

Z = R + j(X_L – X_C) = 10 + j(75.4 – 26.5) = 10 + j48.9 Ω

Step 4: Calculate the magnitude of the impedance:

|Z| = √(R² + (X_L – X_C)²) = √(10² + 48.9²) ≈ 50.1 Ω

Step 5: Calculate the current (I) through the circuit:

I = V / |Z| = 120 V / 50.1 Ω ≈ 2.4 A

Step 6: Calculate the voltage across each component:

V_R = I * R = 2.4 A * 10 Ω = 24 V V_L = I * X_L = 2.4 A * 75.4 Ω ≈ 180.96 V V_C = I * X_C = 2.4 A * 26.5 Ω ≈ 63.6 V

In conclusion, the current (I) through the series circuit is approximately 2.4 A, and the voltage across the resistor (V_R), inductor (V_L), and capacitor (V_C) are approximately 24 V, 180.96 V, and 63.6 V, respectively. Note that the sum of the magnitudes of the individual voltages does not equal the source voltage due to the phase differences between the voltages across the reactive components (inductor and capacitor).