Combination of Capacitors

Series combination

Two capacitors $C_1$ and $C_2$ joined end to end across a battery. The outer plates take $+Q$ and $-Q$; the inner plates must then carry $-Q$ and $+Q$, because if the net charge on either capacitor were not zero there would be a field in the connecting conductor and charge would flow until it was. So the charge on every capacitor in a series chain is the same $\pm Q$, while the potential drops add:

$$V=V_1+V_2=\dfrac{Q}{C_1}+\dfrac{Q}{C_2}\ \Rightarrow\ \dfrac{V}{Q}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$$

Regarding the pair as one capacitor with charge $Q$ and potential difference $V$, $C_s=\dfrac{Q}{V}$, so

$$\dfrac{1}{C_s}=\dfrac{1}{C_1}+\dfrac{1}{C_2}\qquad\text{and generally}\qquad \dfrac{1}{C_s}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\cdots+\dfrac{1}{C_n}$$

• For two capacitors, $C_s=\dfrac{C_1C_2}{C_1+C_2}$.
• The equivalent capacitance is less than the smallest capacitance in the chain.
• The largest voltage appears across the smallest capacitor, since $V_i=\dfrac{Q}{C_i}$.

Parallel combination

Both capacitors are connected across the same two points, so both have the same potential difference $V$, while the charges may differ:

$$Q_1=C_1V,\qquad Q_2=C_2V,\qquad Q=Q_1+Q_2=(C_1+C_2)V$$

$$C_p=C_1+C_2\qquad\text{and generally}\qquad C_p=C_1+C_2+\cdots+C_n$$

• The equivalent capacitance is greater than the largest member.
• Charge divides in the ratio of the capacitances: $\dfrac{q_1}{q_2}=\dfrac{C_1}{C_2}$, which is the answer to the MCQ about charge $Q$ given to a parallel pair.

Comparison

Three equal capacitors $C$

• Hence three $4\ \mu\text{F}$ capacitors give $6\ \mu\text{F}$ by putting two in series ($2\ \mu\text{F}$) and the third in parallel with them ($2+4=6\ \mu\text{F}$).
• Three capacitors $2\ \mu\text{F}$, $3\ \mu\text{F}$, $6\ \mu\text{F}$ in series: $\dfrac{1}{C}=\dfrac12+\dfrac13+\dfrac16=1$, so $C=1\ \mu\text{F}$.

Reducing a network

• Redraw the circuit with the nodes labelled, and identify which capacitors share both nodes (parallel) and which share only one node in a chain (series).
• Capacitors between two points joined by a plain wire are short-circuited and carry no charge.
• If the network is a balanced bridge, the bridging capacitor carries no charge and may be removed before reducing.
• NCERT's worked network: three $10\ \mu\text{F}$ capacitors in series give $C'=\dfrac{10}{3}\ \mu\text{F}$; with a fourth $10\ \mu\text{F}$ in parallel, $C=\dfrac{10}{3}+10=13.3\ \mu\text{F}$. Across $500\ \text{V}$, the series branch carries $Q=\dfrac{10}{3}\times10^{-6}\times500=1.7\times10^{-3}\ \text{C}$ and the parallel capacitor carries $Q'=10\times10^{-6}\times500=5.0\times10^{-3}\ \text{C}$.
• Board pattern with five identical $2\ \mu\text{F}$ capacitors: three in parallel across the middle give $6\ \mu\text{F}$, and this in series with the two outer ones gives $\dfrac{1}{C}=\dfrac12+\dfrac16+\dfrac12=\dfrac76$, i.e. $C=\dfrac67\approx0.86\ \mu\text{F}$.

Sharing charge between two capacitors

Connect a capacitor $C_1$ charged to $V_1$ across another at $V_2$, positive plate to positive plate. Charge is conserved and the potentials equalise:

$$V_{\text{common}}=\dfrac{C_1V_1+C_2V_2}{C_1+C_2}$$

• Charged $A$ (capacitance $C$, charge $Q$) connected across uncharged $B$ (capacitance $2C$): $V=\dfrac{Q}{3C}$, the charge remaining on $A$ is $Q_A'=CV=\dfrac{Q}{3}$, so the charge lost by $A$ is $\dfrac{2Q}{3}$.
• Two identical capacitors, one charged to $V$ and one uncharged: the common potential is $\dfrac{V}{2}$ and each ends with half the original charge.

Diagram to describe in words

Series: a battery with two capacitor symbols one after the other along the same wire, the plate charges marked $+Q$, $-Q$, $+Q$, $-Q$ from left to right, and the two voltage drops $V_1$ and $V_2$ bracketed beneath. Parallel: a battery with two capacitor symbols on separate branches between the same pair of nodes, both branches labelled with the same $V$ and with different charges $Q_1$ and $Q_2$.

Standard results