Four identical capacitors, each of 0.15 µF, are connected in parallel. Determine the equivalent (total) capacitance of the parallel network.

Introduction / Context:
Parallel and series combinations alter the effective capacitance of a network. In parallel, plate areas effectively add, increasing total capacitance. This is a fundamental concept in designing filters and decoupling networks.

Given Data / Assumptions:

• Four capacitors, each 0.15 µF.
• All connected in parallel.
• Ideal components assumed.

Concept / Approach:
For capacitors in parallel: C_total = C1 + C2 + C3 + C4. The voltage across all branches is the same, and charges add, making total capacitance the arithmetic sum.

Step-by-Step Solution:
C_total = 0.15 µF + 0.15 µF + 0.15 µF + 0.15 µFC_total = 4 * 0.15 µFC_total = 0.60 µF

Verification / Alternative check:
Group two at a time: two in parallel give 0.30 µF; two such groups in parallel add to 0.30 µF + 0.30 µF = 0.60 µF. Same result confirms correctness.

Why Other Options Are Wrong:

• 0.15 µF: That would be a single capacitor, not four in parallel.
• 0.30 µF: Equals only two in parallel.
• 0.8 µF: Exceeds the sum; not achievable here.

Common Pitfalls:
Confusing series and parallel rules; in series, capacitances combine via reciprocals and result reduces, while in parallel the value increases by addition.

Final Answer:
0.6 µF