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A capacitor holds a charge of 0.15 coulomb when the terminal voltage is 5 V. What is the capacitance value?

Difficulty: Easy

Correct Answer: 0.03 F

Explanation:


Introduction / Context:
Capacitance is defined as the ratio of electric charge stored on the plates of a capacitor to the voltage across the plates. This simple computation appears frequently in electronics labs and exams and is essential for sizing components in power supplies and timing networks.



Given Data / Assumptions:

  • Stored charge Q = 0.15 coulomb.
  • Terminal voltage V = 5 volt.
  • Ideal capacitor behavior is assumed (no leakage or series resistance in the calculation).


Concept / Approach:
The defining equation is C = Q / V. With Q in coulombs and V in volts, the capacitance C is obtained in farads (F). No integration or transient analysis is required for this steady-state snapshot.



Step-by-Step Solution:

Write the formula: C = Q / V.Insert the numbers: C = 0.15 / 5.Compute: C = 0.03 F.


Verification / Alternative check:

Energy check: W = 1/2 * C * V^2 = 0.5 * 0.03 * 25 = 0.375 J. Using W = 1/2 * Q * V gives W = 0.5 * 0.15 * 5 = 0.375 J, confirming consistency.


Why Other Options Are Wrong:

0.75 F or 0.75 μF are an order-of-magnitude mismatch with Q/V.0.03 μF is 10^6 times smaller than the correct value; unit confusion is a common error.3 μF is arbitrary and does not match Q/V.


Common Pitfalls:

Mixing microfarads with farads; always convert μF to F when using SI relations.


Final Answer:

0.03 F
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