Question

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

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

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, \(X_L = 2 \pi (60.0)(12.0) = 720 \pi\)

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 \(2255.82 \Omega\).

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 \(0.0488 A\).