Two capacitors, C₁ = 4 μF and C₂ = 12 μF, are connected in series. What is the equivalent capacitance?
A16 μF — they add directly
B3 μF — from the reciprocal sum formula
C48 μF — from the product of capacitances
D8 μF — their average
For series capacitors: 1/C_eq = 1/C₁ + 1/C₂ = 1/4 + 1/12 = 3/12 + 1/12 = 4/12, so C_eq = 3 μF. Series combination always produces an equivalent capacitance smaller than either individual capacitor. The answer 16 μF (direct addition) is the parallel rule — the most common error is applying parallel addition to a series circuit. Always identify configuration before choosing the formula.
Question 2 Multiple Choice
C₁ = 6 μF and C₂ = 3 μF are connected in series to a 9V battery. What is the charge on each capacitor?
AC₁ has charge 54 μC; C₂ has charge 27 μC — proportional to capacitance
BC₁ has charge 18 μC; C₂ has charge 18 μC — they share the same charge
CC₁ has charge 6 μC; C₂ has charge 3 μC — equal to capacitance value
DC₁ has charge 9 μC; C₂ has charge 9 μC — equal shares of battery voltage
In series, all capacitors carry the same charge: C_eq = 1/(1/6 + 1/3) = 2 μF; Q = C_eq × V = 2 × 9 = 18 μC. Both C₁ and C₂ have Q = 18 μC. Their voltages differ: V₁ = Q/C₁ = 18/6 = 3V and V₂ = Q/C₂ = 18/3 = 6V (summing to 9V ✓). The larger capacitor drops less voltage — an important and often counterintuitive result. Equal charge, not equal voltage, is the defining feature of series capacitors.
Question 3 True / False
Connecting capacitors in parallel increases the equivalent capacitance because you are effectively adding plate area that can store charge at the same voltage.
TTrue
FFalse
Answer: True
Parallel capacitors share the same two nodes and thus the same voltage V. Each accumulates charge independently: Q_i = C_i × V. Total charge = (C₁ + C₂ + ...)V, giving C_eq = C₁ + C₂ + .... Physically, parallel connection is equivalent to one large capacitor whose total plate area is the sum of all individual plate areas — and larger plates store more charge at the same voltage. This is why parallel always increases capacitance.
Question 4 True / False
The combination rules for capacitors follow the same pattern as resistors: capacitors in series add, just as resistors in series add.
TTrue
FFalse
Answer: False
The rules are algebraically mirrored between capacitors and resistors. For resistors: series = add (R_eq = R₁ + R₂), parallel = reciprocal sum (1/R_eq = 1/R₁ + 1/R₂). For capacitors it is exactly reversed: parallel = add (C_eq = C₁ + C₂), series = reciprocal sum (1/C_eq = 1/C₁ + 1/C₂). This swap occurs because capacitors and resistors respond to the same circuit constraints (voltage and charge/current) but in opposite roles. Confusing the two is the most common error in capacitor circuit problems.
Question 5 Short Answer
Why do capacitors in series all carry the same charge, even if they have different capacitances? Explain the physical mechanism.
Think about your answer, then reveal below.
Model answer: The conductor between adjacent series capacitors is electrically isolated — no charge can flow onto or off it from the external circuit. When charge +Q builds up on the outer plate of the first capacitor, it repels an equal +Q off the adjacent inner plate, which flows to the next capacitor. This induction propagates through the chain so every capacitor accumulates exactly Q, regardless of its capacitance. Their different capacitances then determine how the total voltage is divided: V = Q/C, so smaller capacitance receives a larger voltage share.
This charge-conservation argument is the physical basis of the series rule. The isolated middle conductor acts as a charge relay, ensuring equal charge everywhere in the chain. The voltage division is a consequence: series capacitors act like a single capacitor with larger effective gap between outer plates, which reduces total capacitance — consistent with 1/C_eq = Σ(1/C_i).