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Chapter 28: Problem 19

A \(470-\Omega\) resistor, \(10-\mu \mathrm{F}\) capacitor, and \(750-\mathrm{mH}\) inductor are each connected across \(6.3-\mathrm{V}\) rms, \(60-\mathrm{Hz}\) AC power. Find the rms current in each.

Short Answer

Expert verified
Calculate the RMS current in each device using Ohm’s law. You will find that for the resistor, the RMS current is \(I_{rms} = V_{rms}/R\). For the capacitor and inductor, it's \(I_{rms} = V_{rms}/X_C\) and \(I_{rms} = V_{rms}/X_L\) respectively. To do this, first the reactance for each device is calculated using the provided formulas. The values for \(f\), \(L\), and \(C\) would have to be plugged into the formulas to calculate the reactance. Afterward, each reactance value is plugged into the Ohm's law formula to acquire the desired current.

Step by step solution

01

Determine RMS Current for the Resistor

Apply Ohm's law which states that current equals voltage divided by resistance. Here, RMS voltage is \(6.3 V\) and resistance is \(470 Ω\). Calculate RMS current as: \(I_{rms} = V_{rms}/R = 6.3/470 A\)
02

Find RMS Current for the Capacitor

First calculate the reactance (\(X_C\)) of the capacitor using the formula \(X_C = 1/(2πfC)\) where \(f\) is the frequency and \(C\) is the capacitance. After getting \(X_C\), apply Ohm's law to find the current: \(I_{rms} = V_{rms}/X_C\)
03

Calculate RMS Current for the Inductor

First calculate the reactance (\(X_L\)) of the inductor using the formula \(X_L = 2πfL\) where \(f\) is the frequency and \(L\) is the inductance. After getting \(X_L\), apply Ohm's law to find the current: \(I_{rms} = V_{rms}/X_L\)

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Most popular questions from this chapter

The same AC voltage appears across a capacitor and a resistor, and the same rms current flows in each. Is the power dissipation the same in each?

A series \(R L C\) circuit has resistance \(100 \Omega\) and impedance \(300 \Omega\) (a) What's the power factor? (b) If the rms current is 200 mA, what's the power dissipation?

A series \(R L C\) circuit has \(R=18 \mathrm{k} \Omega, L=20 \mathrm{mH},\) and resonates at \(4.0 \mathrm{kHz}\). (a) What's the capacitance? (b) Find the circuit's impedance at resonance and (c) at \(3.0 \mathrm{kHz}\)

Resonance is a phenomenon which occurs in RLC circuit, during which impedance due to capacitor and inductor become equal and peak current is achieved in the circuit. At resonance the impedance of the circuit is lowest and equal to the resistance in the circuit The value of peak current / is given as follows: \(I=\frac{V}{r}\) Here, \(V\) is voltage in RLC circuit and \(r\) is resistance in the circuit.

Show that the unit of both capacitive and inductive reactance is the ohm.
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