19.2 Series Circuits

Series Circuits

Series circuits connect components along a single path, so current flows through each element one after another. Because there's only one path, the behavior of every component affects the entire circuit. Getting comfortable with series circuits is essential before moving on to parallel and combination circuits.

The two key rules to remember: current stays the same everywhere in a series circuit, while voltage divides among the components. Everything else follows from these two facts and Ohm's law.

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Analysis of Series Circuits

Ohm's law defines the relationship between voltage ($V$), current ($I$), and resistance ($R$):

$V = IR$

• Voltage (measured in volts, V) is the potential difference that drives charge through a component.
• Current (measured in amperes, A) is the rate of charge flow through the circuit.
• Resistance (measured in ohms, $\Omega$) is how much a component opposes that flow.

Equivalent resistance in a series circuit is simply the sum of all individual resistances:

$R_{eq} = R_1 + R_2 + \cdots + R_n$

This makes intuitive sense: each resistor adds more opposition, so the total resistance only grows as you add components.

Once you know $R_{eq}$, you can find the current through the entire circuit:

$I = \frac{V_{battery}}{R_{eq}}$

Since the current is the same everywhere in a series circuit, you can then find the voltage drop across any individual component:

$V_1 = IR_1, \quad V_2 = IR_2, \quad \text{etc.}$

Power dissipated by each component can be calculated with either form:

$P = IV \quad \text{or} \quad P = I^2R$

These are equivalent (you can verify by substituting $V = IR$ into the first equation). Use whichever form matches the quantities you already know.

Interpretation of Circuit Diagrams

Circuit diagrams use standardized symbols so anyone can read them. Here are the components you'll see most often in series circuit problems:

• Resistors are drawn as zigzag lines, with the resistance value labeled in ohms ($\Omega$). A 100 $\Omega$ resistor appears as a zigzag with "100 $\Omega$" next to it.
• Batteries are drawn as alternating long and short parallel lines. The longer line is the positive terminal, the shorter line is the negative terminal. The voltage is labeled next to the symbol (e.g., "9 V").
• Switches appear as a line with a gap. When the gap is open, the circuit is broken and no current flows. When the switch is closed, the circuit is complete.
• Capacitors are drawn as two equal parallel lines (don't confuse them with batteries, which have unequal lines). Capacitance is labeled in farads (F). For example, a 10 microfarad capacitor would read "10 $\mu$F."

In a series circuit diagram, all components are connected end-to-end forming one single loop. If any component is removed or any switch is opened, the entire circuit stops working.

Current and Voltage in Series Circuits

Current is constant throughout a series circuit. There's only one path, so every coulomb of charge that leaves the battery must pass through every component. The current depends on the battery voltage and the total resistance:

$I = \frac{V_{battery}}{R_{eq}}$

For example, if a 12 V battery is connected to a series circuit with $R_{eq} = 4 \; \Omega$:

$I = \frac{12 \; \text{V}}{4 \; \Omega} = 3 \; \text{A}$

That 3 A flows through every component in the circuit.

Voltage divides among the components proportionally to their resistances. Components with higher resistance get a larger share of the total voltage; components with lower resistance get a smaller share.

Consider a 12 V battery connected to two resistors in series: $R_1 = 6 \; \Omega$ and $R_2 = 3 \; \Omega$.

1. Find the equivalent resistance: $R_{eq} = 6 + 3 = 9 \; \Omega$
2. Find the current: $I = \frac{12 \; \text{V}}{9 \; \Omega} = \frac{4}{3} \; \text{A} \approx 1.33 \; \text{A}$
3. Find each voltage drop:
   • $V_1 = IR_1 = \frac{4}{3} \times 6 = 8 \; \text{V}$
   • $V_2 = IR_2 = \frac{4}{3} \times 3 = 4 \; \text{V}$

Kirchhoff's Voltage Law (KVL) confirms the result: the sum of all voltage drops around a closed loop equals the source voltage.

$V_{battery} = V_1 + V_2 + \cdots + V_n$

In the example above, $8 \; \text{V} + 4 \; \text{V} = 12 \; \text{V}$. This law comes directly from conservation of energy: every joule of energy the battery supplies must be used up by the components in the loop.