## Parallel Combination of Capacitors When two or more capacitors are connected between the same two points (A and B), they are said to be in a **parallel combination**. ### Key Characteristics - **Voltage:** The voltage ($V$) supplied by the battery is the **same** across each capacitor in the parallel circuit. - **Charge:** The total charge ($Q$) supplied by the battery is the sum of the charges stored on each individual capacitor ($Q_1, Q_2, Q_3, \dots$). $$ Q = Q_1 + Q_2 + Q_3 $$ ### Derivation of Equivalent Capacitance ($C_e$) 1. The total charge is the sum of individual charges: $Q = Q_1 + Q_2 + Q_3$. 2. Using the formula $Q = CV$, we can substitute for each capacitor: $Q = C_1V + C_2V + C_3V$. 3. Since voltage ($V$) is constant, we can factor it out: $Q = V(C_1 + C_2 + C_3)$. 4. The equivalent capacitance ($C_e$) of the entire circuit is defined as $C_e = \frac{Q}{V}$. 5. Rearranging the equation gives $\frac{Q}{V} = C_1 + C_2 + C_3$. 6. Therefore, the formula for the equivalent capacitance of three capacitors in parallel is: $$ C_e = C_1 + C_2 + C_3 $$ For $n$ number of capacitors connected in parallel, the general formula is: $$ C_e = C_1 + C_2 + C_3 + \dots + C_n \quad (19.10) $$ This shows that the equivalent capacitance of parallel capacitors is **larger** than any of the individual capacitances. ### Comparison for N Identical Capacitors | Combination Type | Effective Capacitance ($C_e$) | | :--------------- | :---------------------------- | | **Series** | $C_e = C / N$ | | **Parallel** | $C_e = CN$ | ### Possible Questions and Answers - **Q:** Find the equivalent capacitance of two capacitors of capacitance $4 \mu \mathrm{F}$ and $8 \mu \mathrm{F}$ when connected in parallel. **A:** For a parallel combination, the equivalent capacitance is the sum of the individual capacitances. $C_e = C_1 + C_2$ $C_e = 4 \mu \mathrm{F} + 8 \mu \mathrm{F} = 12 \mu \mathrm{F}$ - **Q:** Find the equivalent capacitance of the same two capacitors ($4 \mu \mathrm{F}$ and $8 \mu \mathrm{F}$) when connected in series. **A:** For a series combination, the formula for two capacitors is $C_e = \frac{C_1 C_2}{C_1 + C_2}$. $C_e = \frac{(4 \mu \mathrm{F})(8 \mu \mathrm{F})}{4 \mu \mathrm{F} + 8 \mu \mathrm{F}} = \frac{32 \times 10^{-12} F^2}{12 \times 10^{-6} F} = 2.66 \times 10^{-6} F = 2.66 \mu \mathrm{F}$ - **Q:** Find the equivalent capacitance of three capacitors of capacitance $4 \mu \mathrm{F}, 6 \mu \mathrm{F}$ and $8 \mu \mathrm{F}$, connected in series. **A:** For a series combination, the reciprocal of the equivalent capacitance is the sum of the reciprocals of individual capacitances. $\frac{1}{C_e} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{4} + \frac{1}{6} + \frac{1}{8}$ To add these, we find a common denominator (24): $\frac{1}{C_e} = \frac{6}{24} + \frac{4}{24} + \frac{3}{24} = \frac{13}{24}$. Therefore, $C_e = \frac{24}{13} \mu \mathrm{F} \approx 1.85 \mu \mathrm{F}$.