Question

A\(20.0\,{\rm{Hz}},{\rm{ }}16.0\,{\rm{V}}\)source produces a\(2.00\,{\rm{mA}}\)current when connected to a capacitor. What is the capacitance?

Short answer

The capacitance is obtained as, \(0.995\,{\rm{\mu F}}\).

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The frequency is,\(f = 20.0\,{\rm{Hz}}\).
• The voltage is,\(U = 16.0\,{\rm{V}}\).
• The current is, \(I = 2.00\,{\rm{mA}}\).

Evaluating the formula

The inductive reactance is evaluated using the formula:

\({X_C} = \frac{1}{{2\pi fC}}\)

Knowing the reactance, we can then simply apply the Ohm's law and get the intensity as:

\(\begin{aligned} I &= \frac{U}{{{X_C}}}\\ &= 2\pi UCf\end{aligned}\)

Now solving to obtain the capacitance as:

\(C = \frac{I}{{2\pi Uf}}\) ...(1)

Evaluating the capacitance

The numerical values of capacitance and we obtain:

Substitute all the value in the equation (1)

\(\begin{aligned} C &= \frac{{0.002\,{\rm{A}}}}{{2\pi \times 16.0\,{\rm{V}} \times 20.0\,{\rm{Hz}}}}\\ &= 9.95 \times {10^{ - 7}}\,{\rm{F}}\\ &= 0.995\,{\rm{\mu F}}\end{aligned}\)

Therefore, the capacitance is, \(0.995\,{\rm{\mu F}}\).