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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 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.
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