Physics · Ch 1 — Electrostatics
Capacitor in series and parallel
Capacitor in series and parallel
(i) Capacitors in series: consider capacitors C1, C2 and C3 connected end to end in a single chain (series) across a battery of voltage V. Because they form one uninterrupted conducting path, the same charge Q is transferred onto every capacitor in the chain, while each capacitor develops its own voltage Vi = Q/Ci, with the individual voltages summing to the battery's total: V = V1+V2+V3 = Q(1/C1+1/C2+1/C3). Comparing this with the definition of an equivalent single capacitance, V = Q/C_eq, gives 1/C_eq = 1/C1+1/C2+1/C3 -- for a series combination, it is the reciprocals of the individual capacitances that add, and the resulting equivalent capacitance is always smaller than the smallest individual capacitor in the chain (exactly analogous to resistors in parallel in a circuit, not resistors in series). (ii) Capacitors in parallel: consider C1, C2 and C3 instead connected side by side, with both plates of every capacitor tied directly to the same pair of battery terminals. Because every capacitor now sees the same full voltage V, each stores its own charge Qi = CiV, and the total charge supplied by the battery is their sum: Q = Q1+Q2+Q3 = (C1+C2+C3)V. Comparing with Q = C_eq V gives C_eq = C1+C2+C3 -- for a parallel combination, the capacitances themselves simply add, and the equivalent capacitance is always larger than the largest individual capacitor in the group (exactly analogous to resistors in series). More complicated capacitor networks, combining several series and parallel groups (o …
What this figure shows. Three capacitors C1, C2 and C3 are drawn connected end to end in a single chain (series) between the two terminals of a battery of voltage V; because they form one single conducting path with no branch points between them, the same charge Q necessarily flows onto (and is stored by) every one of the three capacitors, while each individual capacitor develops its own share V1, V2, V3 of the total battery voltage, …
What this figure shows. Three capacitors C1, C2 and C3 are instead drawn side by side, each with both of its own plates connected directly to the same pair of battery terminals; because every capacitor's two plates are tied straight to the same two nodes, all three see exactly the same voltage V across them, while each stores its own individual share of charge, Q1 = C1V, Q2 = C2V, Q3 = C3V, with the total charge drawn from the batte …
Worked out. For capacitors in series, the same charge Q sits on every capacitor in the chain (since they share one single conducting path with no other way for charge to enter or leave between them), while the individual voltages add up to the battery's total voltage: V = V1+V2+V3 = Q/C1+Q/C2+Q/C3 = Q(1/C1+1/C2+1/C3). Since the equivalent single capacitor must satisfy V = Q/C_eq, dividing through by Q gives 1/C_eq = 1/C1+1/C2+1/C3 -- the reciprocals of the individual capacitances add, meaning C_eq is always smaller than the smallest individual capacitance in a series combination. For capacitors in parallel, every capacitor instead shares the same voltage V (since all their plates connect directly to the same two nodes), while their individual charges add up to the total charge drawn from the battery: Q = Q1+Q2+Q3 = C1V+C2V+C3V = (C1+C2+C3)V. Since the equivalent capacitor must satisfy Q = C_eq V, dividing through by V gives C_eq = C1+C2+C3 -- the capacitances themselves simply add, meaning C_eq is al …