3.3 Series and parallel resistor circuits

Series and Parallel Connections

Connecting Resistors in Series

A series connection is when resistors are linked end-to-end, creating a single path for current. Because there's only one path, the current through every resistor is identical.

The voltage, however, gets divided up. Each resistor causes a voltage drop, and those drops add up to the total source voltage. This is Kirchhoff's Voltage Law (KVL) in action.

To find the equivalent resistance of resistors in series, just add them:

$R_{eq} = R_1 + R_2 + R_3 + \ldots$

Example: Three resistors of 10Ω, 20Ω, and 30Ω in series give $R_{eq} = 10 + 20 + 30 = 60\,\Omega$. The equivalent resistance is always larger than any individual resistor in the chain.

Connecting Resistors in Parallel

A parallel connection is when resistors are connected across the same two nodes, creating multiple paths for current. Because they share the same nodes, the voltage across every resistor is the same.

Current, on the other hand, splits up. Each branch carries a portion of the total current, and those branch currents add up to the total. This is Kirchhoff's Current Law (KCL).

To find the equivalent resistance of resistors in parallel, use the reciprocal formula:

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots$

Example: Three resistors of 10Ω, 20Ω, and 30Ω in parallel:

$\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{20} + \frac{1}{30} = \frac{6 + 3 + 2}{60} = \frac{11}{60}$

$R_{eq} = \frac{60}{11} \approx 5.45\,\Omega$

Notice the equivalent resistance is always smaller than the smallest individual resistor. Adding parallel paths makes it easier for current to flow.

Equivalent Resistance in Complex Circuits

Most real circuits mix series and parallel connections. The strategy is to simplify from the inside out:

1. Identify a group of resistors that are purely in series or purely in parallel.
2. Replace that group with its equivalent resistance.
3. Redraw the simplified circuit.
4. Repeat until you're left with a single equivalent resistance.

Example: Two 10Ω resistors are in series, and that combination is in parallel with a 20Ω resistor.

1. Combine the series pair: $10 + 10 = 20\,\Omega$
2. Now you have 20Ω in parallel with 20Ω: $R_{eq} = \frac{20 \times 20}{20 + 20} = 10\,\Omega$

Circuit Analysis Techniques

Current and Voltage Division

These are shortcuts that save you from solving full systems of equations in simple circuits.

Voltage division tells you the voltage across one resistor in a series circuit. The voltage splits in proportion to resistance:

$V_x = V_{total} \times \frac{R_x}{R_{total}}$

Example: A 10Ω and a 20Ω resistor are in series across a 15V source. The total resistance is 30Ω.

• Voltage across the 10Ω resistor: $V = 15 \times \frac{10}{30} = 5\,\text{V}$
• Voltage across the 20Ω resistor: $V = 15 \times \frac{20}{30} = 10\,\text{V}$

The larger resistor gets the larger share of the voltage.

Current division tells you the current through one branch of a parallel circuit. A common form for two parallel resistors is:

$I_1 = I_{total} \times \frac{R_2}{R_1 + R_2}$

Notice the "opposite" resistor goes in the numerator. The branch with less resistance carries more current.

Kirchhoff's Laws and Circuit Simplification

Kirchhoff's Current Law (KCL): The total current entering any node equals the total current leaving that node. No charge piles up or disappears.

Kirchhoff's Voltage Law (KVL): The sum of all voltage rises and drops around any closed loop equals zero. What the source puts in, the resistors use up.

When a circuit is too tangled for simple series/parallel reduction, you can use KCL and KVL to write equations and solve for unknowns. For this intro course, though, most problems can be handled with this approach:

1. Look for resistors that are clearly in series or parallel.
2. Combine them into equivalent resistances.
3. Repeat until the circuit is simple enough to apply Ohm's Law ($V = IR$) directly.
4. Then work backward through your simplifications to find individual currents and voltages.

Example: A 10Ω resistor is in series with two 20Ω resistors that are in parallel.

1. Combine the parallel pair: $R_{eq} = \frac{20 \times 20}{20 + 20} = 10\,\Omega$
2. Add the series resistor: $R_{total} = 10 + 10 = 20\,\Omega$
3. If the source is, say, 40V, total current is $I = \frac{40}{20} = 2\,\text{A}$
4. Working backward: the 10Ω series resistor drops $V = 2 \times 10 = 20\,\text{V}$, leaving 20V across the parallel pair. Each 20Ω branch carries $I = \frac{20}{20} = 1\,\text{A}$.