Parallel addition of capacitance: Four capacitors of 10 µF, 20 µF, 22 µF, and 100 µF are connected in parallel. Find the total capacitance of the combination.

Introduction / Context:Capacitors in parallel add directly, a rule exploited to realize non-standard values or to increase total capacitance for energy storage and filtering applications.

Given Data / Assumptions:

• Four capacitors: 10 µF, 20 µF, 22 µF, 100 µF.
• All are connected in parallel.
• Ideal components assumed.

Concept / Approach:For parallel capacitors, the equivalent capacitance is the sum: C_total = sum(Ci). The voltage is common, and the individual stored charges add up, yielding a larger net capacitance.

Step-by-Step Solution:C_total = 10 + 20 + 22 + 100 µFC_total = 152 µF

Verification / Alternative check:Group addition: (10 + 20) = 30 µF; (22 + 100) = 122 µF; then 30 + 122 = 152 µF. Same result confirms correctness.

Why Other Options Are Wrong:

• 2.43 µF and 4.86 µF: Far too small; perhaps confuse with series rules.
• 100 µF: Ignores the additional capacitors in parallel.

Common Pitfalls:Applying series formulas to parallel networks or overlooking unit consistency. In series, capacitance decreases; in parallel, it increases by direct addition.

Final Answer:152 µF