Problem 44 - Fundamentals of Physics

Question

An ac generator with emf amplitude \(\mathscr{E}_{m}=220 \mathrm{~V}\) and operating at frequency \(400 \mathrm{~Hz}\) causes oscillations in a series \(R L C\) circuit having \(R=220 \Omega, L=150 \mathrm{mH},\) and \(C=24.0 \mu \mathrm{F}\). Find (a) the capacitive reactance \(X_{C},\) (b) the impedance \(Z,\) and \((\mathrm{c})\) the current amplitude \(I\). A second capacitor of the same capacitance is then connected in series with the other components. Determine whether the values of (d) \(X_{C},\) (e) \(Z\), and (f) \(I\) increase, decrease, or remain the same.

Step-by-Step Solution

Answer

X_C and Z increase; I decreases.

Step 1: Calculate the capacitive reactance (X_C)

Capacitive reactance is calculated using the formula:\[ X_C = \frac{1}{2 \pi f C} \]where \( f = 400 \text{ Hz} \) and \( C = 24.0 \times 10^{-6} \text{ F} \).Substitute the values into the formula:\[ X_C = \frac{1}{2 \pi \times 400 \times 24.0 \times 10^{-6}} \approx 16.56 \ \Omega \]

Step 2: Calculate the impedance (Z)

The impedance of a series RLC circuit is given by the formula:\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]First, calculate the inductive reactance \( X_L \):\[ X_L = 2 \pi f L = 2 \pi \times 400 \times 0.150 = 376.99 \ \Omega \]Now use this to find \( Z \):\[ Z = \sqrt{220^2 + (376.99 - 16.56)^2} = \sqrt{220^2 + 360.43^2} \approx 421.30 \ \Omega \]

Step 3: Calculate the current amplitude (I)

The current amplitude in the circuit is calculated using Ohm's Law:\[ I = \frac{\mathscr{E}_m}{Z} \]Given \( \mathscr{E}_m = 220 \text{ V} \), and using the previously calculated \( Z \):\[ I = \frac{220}{421.30} \approx 0.522 \text{ A} \]

Step 4: Determine the effect of adding a second capacitor on X_C

Adding a second capacitor in series increases the total capacitive reactance. The formula is:\[ X_{C_{total}} = \frac{1}{2 \pi f \sum{C}} \]Since \( \sum{C} \) decreases when capacitors are in series:\( X_{C_{new}} > X_C \), so \( X_C \) increases.

Step 5: Determine the effect of adding a second capacitor on Z

With an increased \( X_C \), the value of \( Z \) will also increase because the term \( (X_L - X_C) \) will increase, making \( Z = \sqrt{R^2 + (X_L - X_{C_{new}})^2} \) larger.

Step 6: Determine the effect of adding a second capacitor on I

Since impedance \( Z \) increases, the current amplitude \( I \) will decrease:\[ I = \frac{\mathscr{E}_m}{Z_{new}} \] where \( Z_{new} > Z \).

Key Concepts

Capacitive Reactance

The capacitive reactance, denoted by \(X_C\), is an essential concept in understanding how capacitors behave in AC circuits. It represents the opposition that a capacitor offers to the change of current and is directly related to the frequency of the AC supply and the capacitance itself. Calculated using the formula \(X_C = \frac{1}{2 \pi f C}\), where \(f\) is the frequency and \(C\) is the capacitance, \(X_C\) is inversely proportional to both these quantities.

When capacitance or frequency increases, the capacitive reactance decreases, meaning the capacitor allows more current to pass through. Conversely, a decrease in frequency or capacitance results in higher opposition, or a larger \(X_C\). In practice, this allows for tuning circuits to specific frequencies, making capacitive reactance a critical parameter in designing and analyzing RLC circuits. In exercise step 1, this principle is applied to find \(X_C\) using given values.

Impedance Calculation

Impedance, symbolized as \(Z\), is a key factor in AC circuit analysis, combining the effects of resistance, inductive reactance, and capacitive reactance. Impedance is like resistance but in the AC context, where it includes opposition due to frequency. The impedance in a series RLC circuit is given by the equation \(Z = \sqrt{R^2 + (X_L - X_C)^2}\), where \(R\) is resistance, \(X_L\) is inductive reactance, and \(X_C\) is capacitive reactance.

Inductive reactance \(X_L\) increases with higher frequencies, as calculated by \(X_L = 2 \pi f L\), with \(L\) being inductance. In balancing \(X_L\) and \(X_C\), impedance \(Z\) represents the total opposition to current flow. Therefore, by finding your circuit's \(X_L\) and \(X_C\), one can accurately calculate \(Z\), providing a holistic view of how the circuit causes current to resist motion.

Current Amplitude

Current amplitude, denoted as \(I\), in an AC circuit, refers to the maximum current value achieved during a cycle. In a series RLC circuit, you calculate \(I\) using Ohm’s Law, which is adjusted for impedance: \(I = \frac{\mathscr{E}_m}{Z}\), where \(\mathscr{E}_m\) is the maximum electromotive force and \(Z\) is the total impedance of the circuit.

The current amplitude depends on how much the impedance \(Z\) is able to limit the current. A lower impedance results in a higher current amplitude, while a higher impedance restricts the current. Therefore, any circuit elements such as resistors, inductors, or capacitors that increase \(Z\) will lead to a lower current amplitude in the circuit, as demonstrated in exercise step 3. Adjusting each component can finely control the circuit’s current throughout a range of operating conditions.