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Chapter 21: Problem 23

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

Short Answer

Expert verified
The current in the inductor decreases by a factor of 6 (1/6) after increasing the inductance by a factor of 3 and the driving frequency by a factor of 2.

Step by step solution

01

Write down the formula for inductive reactance

The formula for inductive reactance (X_L) in an AC circuit is: $$ X_L = 2\pi f L $$ Where \(f\) is the driving frequency, \(L\) is the inductance, and \(2\pi\) is a constant from the derivation.
02

Write down the formula for current using Ohm's law

Using Ohm's law relating voltage (V), current (I), and reactance (X_L) in an inductive AC circuit, we get: $$ I = \frac{V}{X_L} $$
03

Calculate the current change after increasing inductance

Let's assume the initial inductance is \(L\) and the final inductance is \(3L\). The initial reactance (\(X_{L1}\)) is: $$ X_{L1} = 2\pi f L $$ The final reactance (\(X_{L2}\)) after increasing the inductance is: $$ X_{L2} = 2\pi f (3L) = 6\pi f L $$ The initial current (\(I_1\)) is: $$ I_1 = \frac{V}{X_{L1}} $$ The final current (\(I_2\)), after increasing the inductance, is: $$ I_2 = \frac{V}{X_{L2}} $$
04

Calculate the current change after increasing driving frequency

Now we need to factor in the change in driving frequency. Let the initial frequency be \(f\) and the final frequency be \(2f\). The updated reactance (\(X_{L3}\)) after increasing the frequency is: $$ X_{L3} = 2\pi (2f) (3L) = 12\pi f L $$ The final current (\(I_3\)), after increasing both the inductance and driving frequency, is: $$ I_3 = \frac{V}{X_{L3}} $$
05

Compare the final current to the initial current

To find the change in current, we will compare the final current (\(I_3\)) with the initial current (\(I_1\)): $$ \frac{I_3}{I_1} = \frac{\frac{V}{12\pi f L}}{\frac{V}{2\pi f L}} = \frac{2\pi f L}{12\pi f L} = \frac{1}{6} $$ Thus, the current in the inductor decreases by a factor of 6 (\((1/6)\)) after increasing the inductance by a factor of \(3.0\) and the driving frequency by a factor of \(2.0\).

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Most popular questions from this chapter

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 series combination of a resistor and a capacitor are connected to a \(110-\mathrm{V}\) rms, \(60.0-\mathrm{Hz}\) ac source. If the capacitance is \(0.80 \mu \mathrm{F}\) and the rms current in the circuit is $28.4 \mathrm{mA},$ what is the resistance?
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?
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Show that over one complete cycle, the average value of a sine function squared is \(\frac{1}{2} .\) [Hint: Use the following trigonometric identities: \(\sin ^{2} a+\cos ^{2} a=1 ; \cos 2 a=\) \(\left.\cos ^{2} a-\sin ^{2} a .\right]\)
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