Problem 35 - Fundamentals of Physics

Question

A coil of inductance \(88 \mathrm{mH}\) and unknown resistance and a \(0.94 \mu \mathrm{F}\) capacitor are connected in series with an alternating emf of frequency \(930 \mathrm{~Hz}\). If the phase constant between the applied voltage and the current is \(75^{\circ},\) what is the resistance of the coil?

Step-by-Step Solution

Answer

The resistance of the coil is approximately 89.6 Ω.

Step 1: Determine the Inductive Reactance

First, calculate the inductive reactance \(X_L\) using the formula \(X_L = 2 \pi f L\), where \(f = 930\, \mathrm{Hz}\) and \(L = 88\, \mathrm{mH}\). We have: \[X_L = 2 \pi \times 930 \times 88 \times 10^{-3}\]\[X_L \approx 515.1\,\Omega\]

Step 2: Calculate the Capacitive Reactance

Use the formula \(X_C = \frac{1}{2 \pi f C}\) to find the capacitive reactance \(X_C\), where \(C = 0.94 \mu \mathrm{F}\). Substitute to find:\[X_C = \frac{1}{2 \pi \times 930 \times 0.94 \times 10^{-6}}\]\[X_C \approx 180.8\, \Omega\]

Step 3: Set Up the Phase Angle Formula

The phase angle \(\phi\) relates to the reactance and resistance by \(\tan \phi = \frac{X_L - X_C}{R}\). Substitute \(\phi = 75^{\circ}\),\(X_L = 515.1\,\Omega\), and \(X_C = 180.8\,\Omega\) into the equation:\[\tan 75^{\circ} = \frac{515.1 - 180.8}{R}\]

Step 4: Solve for Resistance

Solve the equation from the previous step:\[\tan 75^{\circ} = 3.732, \] so:\[3.732 = \frac{334.3}{R}\]\[R = \frac{334.3}{3.732} \approx 89.6\,\Omega\]

Step 5: Conclusion

The resistance of the coil is found to be approximately \(89.6\, \Omega\).

Key Concepts

Inductive Reactance

In AC circuits, an inductor resists changes in current. This resistance to changes is called inductive reactance. Unlike simple resistance in DC circuits, inductive reactance is frequency-dependent. The formula for calculating inductive reactance is:\[ X_L = 2 \pi f L \]where:

• \( X_L \) is the inductive reactance in ohms (\( \Omega \))
• \( f \) is the frequency (in hertz)
• \( L \) is the inductance (in henrys)

This equation tells us that as frequency increases, the inductive reactance also increases. This means the inductor will oppose the AC signal more effectively at higher frequencies.For instance, in our given exercise, the inductive reactance was calculated using the formula with a frequency of 930 Hz and an inductance of 88 mH, yielding an inductive reactance of approximately 515.1 \( \Omega \). This value acts as a "dynamic resistor" that varies with frequency.

Capacitive Reactance

Capacitive reactance is another essential characteristic of AC circuits. A capacitor does the opposite of an inductor; it stores energy as an electric field and tends to resist changes in voltage. The capacitive reactance is calculated using:\[ X_C = \frac{1}{2 \pi f C} \]where:

• \( X_C \) is the capacitive reactance in ohms (\( \Omega \))
• \( f \) is the frequency (in hertz)
• \( C \) is the capacitance (in farads)

Notice from the equation that as the frequency increases, the capacitive reactance decreases. This indicates that a capacitor allows higher-frequency currents to pass through more easily, behaving like a short circuit at very high frequencies.In the exercise example, with a frequency of 930 Hz and a capacitance of 0.94 µF, the calculated capacitive reactance was approximately 180.8 \( \Omega \). Thus, understanding capacitive reactance helps predict how a capacitor will behave at different frequencies in an AC circuit.

Phase Angle

The phase angle in an AC circuit says a lot about how the voltage across elements and the current through them relate to each other over time. It tells us how out of sync the voltage is with the current. In an inductive circuit, the current lags the voltage, and in a capacitive circuit, the current leads the voltage. The phase angle \( \phi \) in a series RLC circuit is calculated using:\[ \tan \phi = \frac{X_L - X_C}{R} \]where:

• \( \phi \) is the phase angle in degrees
• \( X_L \) is the inductive reactance
• \( X_C \) is the capacitive reactance
• \( R \) is the resistance

This equation demonstrates that the phase angle is a balancing act between the inductive and capacitive effects in a circuit. In our given problem, the phase angle is 75 degrees, indicating a strong inductive presence given that this angle is closer to 90 degrees, where current is almost fully out of phase from voltage. By using the given values, we found the resistance to be approximately 89.6 \( \Omega \), illustrating how these elements interact to define overall circuit behavior.