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Chapter 21: Problem 70

A coil with an internal resistance of \(120 \Omega\) and inductance of $12.0 \mathrm{H}\( is connected to a \)60.0-\mathrm{Hz}, 110-\mathrm{V}$ ms line. (a) What is the impedance of the coil? (b) Calculate the current in the coil.

Short Answer

Expert verified
Answer: The impedance of the coil is approximately 2255.82Ω, and the current in the coil is approximately 0.0488 A.

Step by step solution

01

Find the reactance of the coil

To find the reactance of the coil, we can use the formula \(X_L = 2 \pi f L\), where \(X_L\) is the reactance, \(f\) is the frequency, and \(L\) is the inductance. In this case, \(f = 60.0 Hz\) and \(L = 12.0 H\). So, \(X_L = 2 \pi (60.0)(12.0) = 720 \pi\)
02

Find the impedance of the coil

To find the impedance, we can use the formula \(Z = \sqrt{R^2 + X_L^2}\), where \(R\) is the resistance and \(X_L\) is the reactance. In this case, \(R = 120 \Omega\) and \(X_L = 720 \pi\). So, \(Z = \sqrt{(120)^2 + (720\pi)^2} \approx 2255.82 \Omega\) Thus, the impedance of the coil is approximately \(2255.82 \Omega\).
03

Calculate the maximum current in the coil

To find the maximum current, we can use the formula \(I = \frac{V_m}{Z}\), where \(V_m\) is the maximum voltage and \(Z\) is the impedance. In this case, \(V_m = 110 V\) and \(Z \approx 2255.82 \Omega\). So, \(I \approx \frac{110}{2255.82} \approx 0.0488 A\) Thus, the current in the coil is approximately \(0.0488 A\).

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Most popular questions from this chapter

A lightbulb is connected to a \(120-\mathrm{V}\) (rms), \(60-\mathrm{Hz}\) source. How many times per second does the current reverse direction?
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?
In the crossover network of Problem \(61,\) the inductance \(L\) is $1.20 \mathrm{mH}$. The capacitor is variable; its capacitance can be adjusted to set the crossover point according to the frequency response of the woofer and tweeter. What should the capacitance be set to for a crossover point of $180 \mathrm{Hz} ?[\text {Hint}:$ At the crossover point, the currents are equal in amplitude. \(]\)
The voltage across an inductor and the current through the inductor are related by \(v_{\mathrm{L}}=L \Delta i / \Delta t .\) Suppose that $i(t)=I \sin \omega t .\( (a) Write an expression for \)v_{\mathrm{L}}(t) .$ [Hint: Use one of the relationships of Eq. \((20-7) .]\) (b) From your expression for \(v_{\mathrm{L}}(t),\) show that the reactance of the inductor is \(X_{\mathrm{L}}=\omega L_{\mathrm{r}}\) (c) Sketch graphs of \(i(t)\) and \(v_{\mathrm{L}}(t)\) on the same axes. What is the phase difference? Which one leads?
A 25.0 -mH inductor, with internal resistance of \(25.0 \Omega\) is connected to a \(110-\mathrm{V}\) ms source. If the average power dissipated in the circuit is \(50.0 \mathrm{W},\) what is the frequency? (Model the inductor as an ideal inductor in series with a resistor.)
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