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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

An RLC series circuit is connected to a \(240-\mathrm{V}\) ms power supply at a frequency of \(2.50 \mathrm{kHz}\). The elements in the circuit have the following values: \(R=12.0 \Omega, C=\) \(0.26 \mu \mathrm{F},\) and $L=15.2 \mathrm{mH},$ (a) What is the impedance of the circuit? (b) What is the rms current? (c) What is the phase angle? (d) Does the current lead or lag the voltage? (e) What are the rms voltages across each circuit element?
A \(150-\Omega\) resistor is in series with a \(0.75-\mathrm{H}\) inductor in an ac circuit. The rms voltages across the two are the same. (a) What is the frequency? (b) Would each of the rms voltages be half of the rms voltage of the source? If not, what fraction of the source voltage are they? (In other words, \(V_{R} / \ell_{m}=V_{L} / \mathcal{E}_{m}=?\) ) (c) What is the phase angle between the source voltage and the current? Which leads? (d) What is the impedance of the circuit?
A certain circuit has a \(25-\Omega\) resistor and one other component in series with a \(12-\mathrm{V}\) (rms) sinusoidal ac source. The rms current in the circuit is 0.317 A when the frequency is \(150 \mathrm{Hz}\) and increases by \(25.0 \%\) when the frequency increases to \(250 \mathrm{Hz}\). (a) What is the second component in the circuit? (b) What is the current at $250 \mathrm{Hz} ?$ (c) What is the numerical value of the second component?
A circuit breaker trips when the rms current exceeds \(20.0 \mathrm{A} .\) How many \(100.0-\mathrm{W}\) lightbulbs can run on this circuit without tripping the breaker? (The voltage is \(120 \mathrm{V} \mathrm{rms} .)\)
At what frequency is the reactance of a 6.0 - \(\mu\) F capacitor equal to $1.0 \mathrm{k} \Omega ?$
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