Question

A \(300.0-\Omega\) resistor and a 2.5 - \(\mu\) F capacitor are connected in series across the terminals of a sinusoidal emf with a frequency of $159 \mathrm{Hz}$. The inductance of the circuit is negligible. What is the impedance of the circuit?

Short answer

Answer: The impedance of the series RC circuit is approximately 517.44 Ohms.

Step-by-step solution

Find the reactance of the capacitor

To find the reactance of the capacitor (\(X_C\)), we can use the given values for frequency (\(f = 159 \mathrm{Hz}\)) and capacitance (\(C = 2.5 \times 10^{-6} \mathrm{F}\)) and the equation for reactance: $$ X_C = \frac{1}{2 \pi f C} $$ Plugging in the given values, we get: $$ X_C = \frac{1}{2\pi(159)(2.5\times 10^{-6})} $$ Calculate the reactance of the capacitor: $$ X_C \approx 422.15\,\Omega $$

Calculate the impedance

Now that we have the reactance of the capacitor, we can find the impedance of the series RC circuit. Using the equation for impedance: $$ Z = \sqrt{R^2 + X_C^2} $$ Plugging in the given resistance (\(R = 300.0\,\Omega\)) and the calculated reactance (\(X_C \approx 422.15\,\Omega\)), we get: $$ Z = \sqrt{(300.0)^2 + (422.15)^2} $$ Calculate the impedance of the circuit: $$ Z \approx 517.44\,\Omega $$ The impedance of the circuit is approximately \(517.44\,\Omega\).