A \(0.250-\mu \mathrm{F}\) capacitor is connected to a \(220-\mathrm{V}\) rms ac source at \(50.0 \mathrm{Hz}\) (a) Find the reactance of the capacitor. (b) What is the rms current through the capacitor?

Short Answer

Expert verified
Answer: The reactance of the capacitor is approximately 1276.45 Ω, and the rms current through it is approximately 0.172 A.

Step by step solution

01

Calculate the reactance of the capacitor

To find the reactance, we use the formula \(X_C = \frac{1}{2\pi fC}\), where \(f\) is the frequency of 50 Hz and \(C\) is the given capacitance of 0.250 \(\mu F\). Plug the values into the formula: \(X_C = \frac{1}{2\pi (50)(0.250×10^{-6})}\) Now, calculate the reactance: \(X_C = \frac{1}{2\pi (50)(0.250×10^{-6})} ≈ 1276.45 \Omega\)
02

Calculate the RMS current through the capacitor

We can now use Ohm's law for capacitive AC circuits to find the RMS current. The formula is \(I = \frac{V}{X_C}\), where V is the RMS voltage (220 V) and \(X_C\) is the reactance we calculated in step 1 (1276.45 \(\Omega\)). Plug the values into the formula: \(I = \frac{220}{1276.45}\) Now, calculate the RMS current: \(I = \frac{220}{1276.45} ≈ 0.172 A\) The RMS current through the capacitor is approximately 0.172 A.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A capacitor (capacitance \(=C\) ) is connected to an ac power supply with peak voltage \(V\) and angular frequency \(\alpha\) (a) During a quarter cycle when the capacitor goes from being uncharged to fully charged, what is the average current (in terms of \(C . V\), and \(\omega\) )? [Hint: \(\left.i_{\mathrm{av}}=\Delta Q / \Delta t\right]\) (b) What is the rms current? (c) Explain why the average and rms currents are not the same.
An RLC series circuit is driven by a sinusoidal emf at the circuit's resonant frequency. (a) What is the phase difference between the voltages across the capacitor and inductor? [Hint: since they are in series, the same current \(i(t) \text { flows through them. }]\) (b) At resonance, the rms current in the circuit is \(120 \mathrm{mA}\). The resistance in the circuit is \(20 \Omega .\) What is the rms value of the applied emf? (c) If the frequency of the emf is changed without changing its rms value, what happens to the rms current? (W) tutorial: resonance)
The voltage across an inductor and the current through the inductor are related by \(v_{\mathrm{L}}=L \Delta i / \Delta t .\) Suppose that $i(t)=I \sin \omega t .\( (a) Write an expression for \)v_{\mathrm{L}}(t) .$ [Hint: Use one of the relationships of Eq. \((20-7) .]\) (b) From your expression for \(v_{\mathrm{L}}(t),\) show that the reactance of the inductor is \(X_{\mathrm{L}}=\omega L_{\mathrm{r}}\) (c) Sketch graphs of \(i(t)\) and \(v_{\mathrm{L}}(t)\) on the same axes. What is the phase difference? Which one leads?
A coil with an internal resistance of \(120 \Omega\) and inductance of $12.0 \mathrm{H}\( is connected to a \)60.0-\mathrm{Hz}, 110-\mathrm{V}$ ms line. (a) What is the impedance of the coil? (b) Calculate the current in the coil.
A variable inductor with negligible resistance is connected to an ac voltage source. How does the current in the inductor change if the inductance is increased by a factor of 3.0 and the driving frequency is increased by a factor of \(2.0 ?\)
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free