Two capacitors in series across a battery (demo problem)
Capacitance · Series
Problem
Two capacitors of 2 μF and 4 μF are connected in series across a 12 V battery. Find the charge on each capacitor and the voltage across each.
(a) 8 μC, 8 V & 4 V(b) 16 μC, 8 V & 4 V(c) 6 μC, 3 V & 6 V(d) 48 μC, 4 V & 8 V
Given
\(C_1 = 2\,\mu F\), \(C_2 = 4\,\mu F\), \(V = 12\) V, connected in series.
Need to find
Charge on each capacitor (\(Q_1\), \(Q_2\)) and voltage across each (\(V_1\), \(V_2\)).
Formula
\[\dfrac{1}{C_{\text{eq}}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} \quad(\text{series})\]
\[Q = C_{\text{eq}}\,V \qquad V_1 = \dfrac{Q}{C_1},\;V_2 = \dfrac{Q}{C_2}\]
Concept
Think of the two capacitors as two buckets connected end-to-end through a single pipe. Since the same current flowed to fill them, they end up holding the exact same amount of charge — that's the key rule for series. But because the buckets are different sizes, the "water level" (voltage) in each bucket is different: the smaller bucket rises higher for the same amount of water. So the smaller capacitor gets the bigger share of the voltage.
Steps
- Compute equivalent capacitance for the series pair: \(\dfrac{1}{C_{\text{eq}}} = \dfrac{1}{2} + \dfrac{1}{4} = \dfrac{3}{4}\;\mu F^{-1}\) ⇒ \(C_{\text{eq}} = \dfrac{4}{3}\,\mu F\).
- Charge drawn from the battery: \(Q = C_{\text{eq}}\,V = \dfrac{4}{3} \times 12 = 16\,\mu C\). This is the charge on each capacitor (series ⇒ same Q).
- Voltage across \(C_1\): \(V_1 = \dfrac{Q}{C_1} = \dfrac{16}{2} = 8\) V.
- Voltage across \(C_2\): \(V_2 = \dfrac{Q}{C_2} = \dfrac{16}{4} = 4\) V.
- Check: \(V_1 + V_2 = 8 + 4 = 12\) V ✓ (matches battery voltage).
Final answer
Charge on each = 16 μC; voltages = 8 V (across 2 μF) and 4 V (across 4 μF).
Answer: (b)