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Chapter 21: Problem 13

A variable capacitor with negligible resistance is connected to an ac voltage source. How does the current in the circuit change if the capacitance is increased by a factor of 3.0 and the driving frequency is increased by a factor of \(2.0 ?\)

Short Answer

Expert verified
Answer: The current in the circuit increases by a factor of 6.

Step by step solution

01

Write the initial given values and formulas

Initial capacitance, \(C_1\) Final capacitance, \(C_2 = 3C_1\) Initial frequency, \(f_1\) Final frequency, \(f_2 = 2f_1\) Reactance formula for a capacitor is given by: \(X_C = \frac{1}{2\pi f C}\) We will find the reactance before and after the changes and compute the ratio of current after and before the change.
02

Calculate the initial reactance, \(X_{C1}\)

First, we need to calculate the initial reactance (before the changes) using the given capacitance and frequency. \(X_{C1} = \frac{1}{2\pi f_1 C_1}\)
03

Calculate the final reactance, \(X_{C2}\)

Next, we need to calculate the final reactance (after the changes) using the final capacitance and frequency. \(X_{C2} = \frac{1}{2\pi f_2 C_2}\) Substitute: \(f_2 = 2f_1\) and \(C_2 = 3C_1\) \(X_{C2} = \frac{1}{2\pi (2f_1) (3C_1)}\)
04

Find the ratio of reactances

Now, we find the ratio of the final reactance to the initial reactance. \(\frac{X_{C2}}{X_{C1}} = \frac{\frac{1}{2\pi (2f_1)(3C_1)}}{\frac{1}{2\pi f_1 C_1}}\)
05

Simplify the ratio of reactance

Simplify the above ratio: \(\frac{X_{C2}}{X_{C1}} = \frac{1}{(2)(3)} = \frac{1}{6}\)
06

Calculate the change in current

Now we know that the current in the AC circuit is inversely proportional to the reactance. Therefore, the ratio of change in currents is the inverse of the ratio of change in reactances: \(\frac{I_2}{I_1} = \frac{X_{C1}}{X_{C2}} = 6\) So, the current in the circuit increases by a factor of 6 when the capacitance is increased by a factor of 3 and the driving frequency is increased by a factor of 2.

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Most popular questions from this chapter

A 40.0 -mH inductor, with internal resistance of \(30.0 \Omega\) is connected to an ac source $$\varepsilon(t)=(286 \mathrm{V}) \sin [(390 \mathrm{rad} / \mathrm{s}) t]$$ (a) What is the impedance of the inductor in the circuit? (b) What are the peak and rms voltages across the inductor (including the internal resistance)? (c) What is the peak current in the circuit? (d) What is the average power dissipated in the circuit? (e) Write an expression for the current through the inductor as a function of time.
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.
(a) What is the reactance of a 5.00 - \(\mu \mathrm{F}\) capacitor at the frequencies \(f=12.0 \mathrm{Hz}\) and \(1.50 \mathrm{kHz} ?\) (b) What is the impedance of a series combination of the \(5.00-\mu \mathrm{F}\) capacitor and a \(2.00-\mathrm{k} \Omega\) resistor at the same two frequencies? (c) What is the maximum current through the circuit of part (b) when the ac source has a peak voltage of \(2.00 \mathrm{V} ?\) (d) For each of the two frequencies, does the current lead or lag the voltage? By what angle?
For a particular \(R L C\) series circuit, the capacitive reactance is $12.0 \Omega,\( the inductive reactance is \)23.0 \Omega,$ and the maximum voltage across the \(25.0-\Omega\) resistor is \(8.00 \mathrm{V}\) (a) What is the impedance of the circuit? (b) What is the maximum voltage across this circuit?
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?
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