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Chapter 30: Problem 76

In a certain \(\mathrm{RLC}\) circuit, a \(20.0-\Omega\) resistor, a 10.0 - \(\mathrm{mH}\) inductor, and a \(5.00-\mu \mathrm{F}\) capacitor are connected in series with an AC power source for which \(V_{\mathrm{rms}}=10.0 \mathrm{~V}\) and \(f=100 . \mathrm{Hz} .\) Calculate a) the amplitude of the current, b) the phase between the current and the voltage, and c) the maximum voltage across each component.

Short Answer

Expert verified
Answer: The amplitude of the current is \(0.047 A\), the phase difference between the current and the voltage is \(-1.539\) radians, and the maximum voltage across the resistor, inductor, and capacitor are \(0.94 V\), \(0.295 V\), and \(14.96 V\), respectively.

Step by step solution

01

Find the circuit impedance

We need to find the impedance of the circuit, which is given by: $$Z = \sqrt{R^2 + (X_L - X_C)^2}$$ where `\(R\)' is the resistance, `\(X_L\)' is the inductive reactance, and `\(X_C\)' is the capacitive reactance. First, calculate the reactances using the following formulas: $$X_L = 2\pi fL$$ $$X_C = \frac{1}{2\pi fC}$$ where `\(f\)' is the frequency, `\(L\)' is the inductor value, and `\(C\)' is the capacitor value.
02

Calculate the reactances values

Now, we can plug in the given values to calculate the reactances: $$X_L = 2\pi (100 Hz)(0.010 H) = 6.283 Hz$$ $$X_C = \frac{1}{2\pi (100 Hz)(5\times10^{-6} F)} = 318.31 Hz$$
03

Calculate the circuit impedance

With the reactances calculated, find the impedance of the RLC circuit: $$Z = \sqrt{(20.0 \Omega)^2 + (6.283 \Omega - 318.31 \Omega)^2} = 299.22 \Omega$$
04

Find the current amplitude

Use Ohm's Law to calculate the amplitude of the current: $$I_0 = \frac{V_0}{Z} = \frac{\sqrt{2} V_{rms}}{Z} = \frac{\sqrt{2}(10.0 V)}{299.22 \Omega} = 0.047 A$$
05

Calculate the phase difference

The phase difference between the current and the voltage can be found using the relation: $$\tan \phi = \frac{X_L - X_C}{R}$$ $$\phi = \arctan \left(\frac{6.283 \Omega - 318.31 \Omega}{20.0 \Omega}\right) = -1.539 \, \mathrm{rad}$$
06

Calculate the maximum voltage across each component

Finally, determine the maximum voltage across each component using the following formulas: $$V_R = I_0R$$ $$V_L = I_0X_L$$ $$V_C = I_0X_C$$ Now, plug in the values: $$V_R = (0.047 A)(20.0 \Omega) = 0.94 V$$ $$V_L = (0.047 A)(6.283 \Omega) = 0.295 \, V$$ $$V_C = (0.047 A)(318.31 \Omega) = 14.96 V$$ So, the amplitude of the current is \(0.047 A\), the phase difference between the current and the voltage is \(-1.539\) radians, and the maximum voltage across the resistor, inductor, and capacitor are \(0.94 V\), \(0.295 V\), and \(14.96 V\), respectively.

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Most popular questions from this chapter

A particular RC low-pass filter has a breakpoint frequency of \(200 .\) Hz. At what frequency will the output voltage divided by the input voltage be \(0.100 ?\)

In a series RLC circuit, \(V=(12.0 \mathrm{~V})(\sin \omega t), R=10.0 \Omega\) \(L=2.00 \mathrm{H},\) and \(C=10.0 \mu \mathrm{F}\). At resonance, determine the voltage amplitude across the inductor. Is the result reasonable, considering that the voltage supplied to the entire circuit has an amplitude of \(12.0 \mathrm{~V} ?\)

A variable capacitor used in an RLC circuit produces a resonant frequency of \(5.0 \mathrm{MHz}\) when its capacitance is set to \(15 \mathrm{pF}\). What will the resonant frequency be when the capacitance is increased to \(380 \mathrm{pF} ?\)

In an RL circuit with alternating current, the current lags behind the voltage. What does this mean, and how can it be explained qualitatively, based on the phenomenon of electromagnetic induction?

A capacitor with capacitance \(C=5.00 \cdot 10^{-6} \mathrm{~F}\) is connected to an AC power source having a peak value of \(10.0 \mathrm{~V}\) and \(f=100 . \mathrm{Hz} .\) Find the reactance of the capacitor and the maximum current in the circuit.
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