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6.3 Series and parallel combinations

6.3 Series and parallel combinations

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
🔌Intro to Electrical Engineering
Unit & Topic Study Guides

Capacitors and inductors can be combined in series or parallel, changing how they behave in circuits. Understanding these combinations is key to analyzing and designing electrical systems effectively.

Series connections reduce overall capacitance but increase inductance. Parallel connections do the opposite. Notice the pattern: capacitors combine with the opposite rule that resistors use, while inductors combine with the same rule as resistors. Keeping that analogy in mind makes the formulas much easier to remember.

Series and Parallel Capacitors

Calculating Equivalent Capacitance in Series and Parallel Circuits

Series capacitors are connected end-to-end, so there's only one path for current. The equivalent capacitance is always less than the smallest individual capacitor in the chain. Think of it this way: putting capacitors in series is like increasing the distance between the outer plates, which reduces the ability to store charge.

The formula for series capacitance:

1Ceq=1C1+1C2+...+1Cn\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n}

For the common special case of just two capacitors in series, this simplifies to:

Ceq=C1C2C1+C2C_{eq} = \frac{C_1 \cdot C_2}{C_1 + C_2}

Parallel capacitors share the same two nodes, giving current multiple paths. Here the equivalent capacitance is just the sum of all individual values, because you're effectively increasing the total plate area available to store charge.

Ceq=C1+C2+...+CnC_{eq} = C_1 + C_2 + ... + C_n

Quick example: Two capacitors, C1=4μFC_1 = 4 \, \mu F and C2=12μFC_2 = 12 \, \mu F, in series give Ceq=4×124+12=3μFC_{eq} = \frac{4 \times 12}{4 + 12} = 3 \, \mu F. In parallel, they'd give Ceq=16μFC_{eq} = 16 \, \mu F.

Calculating Equivalent Capacitance in Series and Parallel Circuits, Capacitors in Series and Parallel | Physics

Voltage Division in Series Capacitor Circuits

Voltage divides across series capacitors inversely proportional to their capacitance values. A smaller capacitor gets a larger share of the voltage, and a larger capacitor gets a smaller share. This is the opposite of how voltage divides across series resistors.

You can find the voltage across any individual capacitor using:

VCi=Vtotal×CeqCiV_{C_i} = V_{total} \times \frac{C_{eq}}{C_i}

where VCiV_{C_i} is the voltage across the ii-th capacitor, VtotalV_{total} is the total voltage applied across the series combination, and CeqC_{eq} is the equivalent series capacitance.

The sum of all individual capacitor voltages always equals the total applied voltage, as required by Kirchhoff's Voltage Law.

Calculating Equivalent Capacitance in Series and Parallel Circuits, Capacitors in Series and Parallel | Physics

Series and Parallel Inductors

Calculating Equivalent Inductance in Series and Parallel Circuits

Inductors combine using the same rules as resistors, which makes them a bit more intuitive if you're already comfortable with resistor networks.

Series inductors are connected end-to-end. The equivalent inductance is the sum of all individual values:

Leq=L1+L2+...+LnL_{eq} = L_1 + L_2 + ... + L_n

This makes physical sense: placing inductors in series is like winding a longer coil, which increases total inductance.

Parallel inductors share common nodes. The equivalent inductance is always less than the smallest individual inductor:

1Leq=1L1+1L2+...+1Ln\frac{1}{L_{eq}} = \frac{1}{L_1} + \frac{1}{L_2} + ... + \frac{1}{L_n}

For two inductors in parallel, this simplifies to:

Leq=L1L2L1+L2L_{eq} = \frac{L_1 \cdot L_2}{L_1 + L_2}

Note: These formulas assume no mutual inductance between the inductors (i.e., their magnetic fields don't interact). If inductors are physically close enough to couple magnetically, additional terms are needed. For this course, you can generally assume no mutual coupling unless told otherwise.

Current Division in Parallel Inductor Circuits

Current divides among parallel inductors inversely proportional to their inductance values. A smaller inductor carries more current; a larger inductor carries less. Again, this mirrors how current divides among parallel resistors.

The current through any individual inductor is:

ILi=Itotal×LeqLiI_{L_i} = I_{total} \times \frac{L_{eq}}{L_i}

where ILiI_{L_i} is the current through the ii-th inductor, ItotalI_{total} is the total current entering the parallel combination, and LeqL_{eq} is the equivalent parallel inductance.

The sum of all individual inductor currents equals the total current entering the parallel combination, consistent with Kirchhoff's Current Law.