How do capacitors behave in series and parallel?

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Multiple Choice

How do capacitors behave in series and parallel?

Explanation:
Capacitance behaves differently depending on how capacitors are connected, and the way charges and voltages distribute explains why the rules are what they are. In parallel, all capacitors experience the same voltage across their plates. The total charge stored is the sum of the individual charges: Q_total = Σ(C_i V). With the same voltage V across each, this becomes Q_total = V Σ(C_i). The total capacitance is C_total = Q_total / V = Σ(C_i). So parallel capacitors simply add their capacitances. In series, the same charge flows through every capacitor, so the charge on each capacitor is the same: Q_i = Q. The voltages add: V_total = Σ V_i, and each V_i = Q / C_i. Therefore V_total = Q Σ(1 / C_i). The total capacitance is C_total = Q / V_total = 1 / Σ(1 / C_i). That’s the inverse-of-sum-of-inverses form. A helpful check: two identical capacitors in parallel give 2C. In series, two identical capacitors give C/2. These align with the formulas above.

Capacitance behaves differently depending on how capacitors are connected, and the way charges and voltages distribute explains why the rules are what they are.

In parallel, all capacitors experience the same voltage across their plates. The total charge stored is the sum of the individual charges: Q_total = Σ(C_i V). With the same voltage V across each, this becomes Q_total = V Σ(C_i). The total capacitance is C_total = Q_total / V = Σ(C_i). So parallel capacitors simply add their capacitances.

In series, the same charge flows through every capacitor, so the charge on each capacitor is the same: Q_i = Q. The voltages add: V_total = Σ V_i, and each V_i = Q / C_i. Therefore V_total = Q Σ(1 / C_i). The total capacitance is C_total = Q / V_total = 1 / Σ(1 / C_i). That’s the inverse-of-sum-of-inverses form.

A helpful check: two identical capacitors in parallel give 2C. In series, two identical capacitors give C/2. These align with the formulas above.

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