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Chapter 21: Problem 72

A capacitor is rated at \(0.025 \mu \mathrm{F}\). How much rms current flows when the capacitor is connected to a \(110-\mathrm{V}\) rms, \(60.0-\mathrm{Hz}\) line?

Short Answer

Expert verified
Based on the given information, the problem required finding the rms current flowing through a capacitor connected to an AC voltage source. To solve this, we first calculated the capacitive reactance using the formula \(X_C = \frac{1}{2\pi fC}\) and found it to be approximately \(106.22 \Omega\). Then, we used Ohm's law for AC circuits (\(I = \frac{V}{X_C}\)) to find the rms current, which was approximately \(1.036 A\).

Step by step solution

01

Calculate the capacitive reactance

We need to determine the capacitive reactance, denoted as \(X_C\), which can be calculated using the formula: \(X_C = \frac{1}{2\pi fC}\). Here, \(f\) represents the frequency of the AC source and \(C\) represents the capacitance. Plugging in the given values, we have: \(X_C = \frac{1}{2\pi (60.0Hz)(0.025\mu F)}\) Convert the capacitance in farads: \(0.025\mu F = 0.025 \times 10^{-6} F\) Now, calculate \(X_C\): \(X_C = \frac{1}{2\pi (60Hz)(0.025 \times 10^{-6} F)} \approx 106.22 \Omega\)
02

Calculate the rms current using Ohm's law for AC circuits

Ohm's law states that \(I = \frac{V}{X_C}\), where \(I\) is the rms current, \(V\) is the rms voltage, and \(X_C\) is the capacitive reactance calculated in step 1. Plugging in the values, we get: \(I = \frac{110V}{106.22 \Omega} \approx 1.036 A\) Therefore, the rms current flowing through the capacitor is approximately \(1.036 A\).

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Most popular questions from this chapter

In an RLC circuit, these three clements are connected in series: a resistor of \(20.0 \Omega,\) a \(35.0-\mathrm{mH}\) inductor, and a 50.0 - \(\mu\) F capacitor. The ac source of the circuit has an rms voltage of \(100.0 \mathrm{V}\) and an angular frequency of $1.0 \times 10^{3} \mathrm{rad} / \mathrm{s} .$ Find (a) the reactances of the capacitor and inductor, (b) the impedance, (c) the rms current, (d) the current amplitude, (e) the phase angle, and (f) the rims voltages across each of the circuit elements. (g) Does the current lead or lag the voltage? (h) Draw a phasor diagram.
The circuit shown has a source voltage of \(440 \mathrm{V}\) ms, resistance \(R=250 \Omega\), inductance \(L=0.800 \mathrm{H},\) and capacitance $C=2.22 \mu \mathrm{F} .\( (a) Find the angular frequency \)\omega_{0}$ for resonance in this circuit. (b) Draw a phasor diagram for the circuit at resonance. (c) Find these rms voltages measured between various points in the circuit: \(V_{a b}, V_{b c}, V_{c d}, V_{b d},\) and \(V_{c d},\) (d) The resistor is replaced with one of \(R=125 \Omega\). Now what is the angular frequency for resonance? (e) What is the rms current in the circuit operated at resonance with the new resistor?
An x-ray machine uses \(240 \mathrm{kV}\) rms at \(60.0 \mathrm{mA}\) ms when it is operating. If the power source is a \(420-\mathrm{V}\) rms line, (a) what must be the turns ratio of the transformer? (b) What is the rms current in the primary? (c) What is the average power used by the \(x\) -ray tube?
At what frequency is the reactance of a \(20.0-\mathrm{mH}\) inductor equal to \(18.8 \Omega ?\)
A lightbulb is connected to a \(120-\mathrm{V}\) (rms), \(60-\mathrm{Hz}\) source. How many times per second does the current reverse direction?
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