Problem

Two capacitors of $4 \, \mathrm{\mu F}$ and $8 \, \mathrm{\mu F}$ are first connected in (a) series and then in (b) parallel. An external source of $200 \, \mathrm{V}$ is applied. Calculate the total capacitance, the potential drop across each capacitor, and the charge on each capacitor.

Data

• Capacitor $C_1 = 4 \, \mathrm{\mu F} = 4 \times 10^{-6} \, \mathrm{F}$
• Capacitor $C_2 = 8 \, \mathrm{\mu F} = 8 \times 10^{-6} \, \mathrm{F}$
• Voltage: $V = 200 \, \mathrm{V}$

Prerequisite Concepts

• Series Combination:

$$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$$

• Parallel Combination:

$$C_{\text{eq}} = C_1 + C_2$$

• Charge:

$$Q = C_{\text{eq}} \cdot V$$

• Voltage in Series:

$$V_i = \frac{Q}{C_i}$$

• Voltage in Parallel:
  $V_1 = V_2 = V$

Answer

(a) Series Combination

Step 1: Calculate Total Capacitance

$$\frac{1}{C_{\text{eq}}} = \frac{1}{4} + \frac{1}{8} = \frac{2 + 1}{8} = \frac{3}{8}$$ $$C_{\text{eq}} = \frac{8}{3} = 2.67 \, \mathrm{\mu F}$$

Step 2: Calculate Total Charge

$$Q = C_{\text{eq}} \cdot V$$ $$Q = 2.67 \times 200 = 534 \, \mathrm{\mu C}$$

Step 3: Calculate Voltage Across Each Capacitor

$$V_1 = \frac{Q}{C_1} = \frac{534}{4} = 133.5 \, \mathrm{V}$$ $$V_2 = \frac{Q}{C_2} = \frac{534}{8} = 66.75 \, \mathrm{V}$$

(b) Parallel Combination

Step 1: Calculate Total Capacitance

$$C_{\text{eq}} = C_1 + C_2 = 4 + 8 = 12 \, \mathrm{\mu F}$$

Step 2: Calculate Total Charge

$$Q = C_{\text{eq}} \cdot V$$ $$Q = 12 \cdot 200 = 2400 \, \mathrm{\mu C}$$

Step 3: Calculate Charge on Each Capacitor

$$Q_1 = C_1 \cdot V = 4 \cdot 200 = 800 \, \mathrm{\mu C}$$ $$Q_2 = C_2 \cdot V = 8 \cdot 200 = 1600 \, \mathrm{\mu C}$$

Final Answer

• Series Combination:
  Total Capacitance: $2.67 \, \mathrm{\mu F}$
  Voltage across $C_1$: $133.5 \, \mathrm{V}$
  Voltage across $C_2$: $66.75 \, \mathrm{V}$
  Total Charge: $534 \, \mathrm{\mu C}$

• Parallel Combination:
  Total Capacitance: $12 \, \mathrm{\mu F}$
  Charge on $C_1$: $800 \, \mathrm{\mu C}$
  Charge on $C_2$: $1600 \, \mathrm{\mu C}$

---