Answer to Question #298058 in Electric Circuits for Dreya

Question

What capacitance when connected in series with a 500Ω resistor will limit the current

drawn from a 48-mV 465-kHz source to 20μA?

Answer 1

Solution;

The impedance of the current is;

$Z=\frac VI =\sqrt{X_c^2+R^2}$

Substitute for the known values;

$Z=\frac{48×10^{-3}}{20×10^{-6}}=\sqrt{X_c^2+500^2}$

$2400=\sqrt{X_c^2+500^2}$

$X_c=\sqrt{2400^2-500^2}$

$X_c=2347.34\Omega$

But also;

$X_c=\frac{1}{2πfC}$

$C=\frac{1}{2πfX_c}=\frac{1}{2π×465000×2347.34}$

$C=1.458×10^{-10}=0.1458nF$