Problem

Three capacitors of $4 \, \mathrm{\mu F}$, $6 \, \mathrm{\mu F}$, and $8 \, \mathrm{\mu F}$ are connected in series to a $250 \, \mathrm{V}$ DC supply. Find (i) the total capacitance, (ii) the charge on each capacitor, and (iii) the potential difference across each capacitor.

Data

• Capacitor $C_1 = 4 \, \mathrm{\mu F} = 4 \times 10^{-6} \, \mathrm{F}$
• Capacitor $C_2 = 6 \, \mathrm{\mu F} = 6 \times 10^{-6} \, \mathrm{F}$
• Capacitor $C_3 = 8 \, \mathrm{\mu F} = 8 \times 10^{-6} \, \mathrm{F}$
• Voltage $V = 250 \, \mathrm{V}$.

Prerequisite Concepts

• Series Capacitance:

$$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$

• Charge on Each Capacitor:
  In a series combination, charge remains the same:

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

• Voltage Across Each Capacitor:

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

Answer

Step 1: Calculate Total Capacitance

$$\frac{1}{C_{\text{eq}}} = \frac{1}{4} + \frac{1}{6} + \frac{1}{8}$$

Take LCM:

$$\frac{1}{C_{\text{eq}}} = \frac{6 + 4 + 3}{24} = \frac{13}{24}$$ $$C_{\text{eq}} = \frac{24}{13} = 1.846 \, \mathrm{\mu F}$$

Step 2: Calculate Total Charge

$$Q = C_{\text{eq}} \cdot V$$ $$Q = 1.846 \times 10^{-6} \cdot 250 = 461.54 \, \mathrm{\mu C}$$

Step 3: Calculate Voltage Across Each Capacitor

$$V_1 = \frac{Q}{C_1} = \frac{461.54}{4} = 115.39 \, \mathrm{V}$$ $$V_2 = \frac{Q}{C_2} = \frac{461.54}{6} = 76.92 \, \mathrm{V}$$ $$V_3 = \frac{Q}{C_3} = \frac{461.54}{8} = 57.69 \, \mathrm{V}$$

Final Answer

• Total Capacitance: $1.846 \, \mathrm{\mu F}$
• Charge on Each Capacitor: $461.54 \, \mathrm{\mu C}$
• Potential Differences:
  • $V_1 = 115.39 \, \mathrm{V}$
  • $V_2 = 76.92 \, \mathrm{V}$
  • $V_3 = 57.69 \, \mathrm{V}$

---