Problem 7 - Applied Physics

Question

A circuit contains a \(175-\Omega\) resistance, a \(4.50-\mu \mathrm{F}\) capacitor, and a \(0.735-\mathrm{H}\) inductance in series with a \(5.00-\mathrm{V}, 10 \overline{0}\) -Hz source. Find the impedance and the current.

Step-by-Step Solution

Answer

Impedance is approximately 3079.7 Ω and the current is about 1.625 mA.

Step 1: Calculate Reactance of the Capacitor

The reactance of the capacitor is given by the formula \( X_C = \frac{1}{2\pi f C} \) where \( f \) is the frequency and \( C \) is the capacitance. Here, \( f = 100 \) Hz and \( C = 4.50 \times 10^{-6} \) F. Thus:\[X_C = \frac{1}{2\pi (100) (4.50 \times 10^{-6})} = \frac{1}{2\pi \times 0.00045} \approx 3536.8 \Omega\]

Step 2: Calculate Reactance of the Inductor

The reactance of the inductor is given by \( X_L = 2\pi f L \), where \( L \) is the inductance. Here, \( L = 0.735 \) H:\[X_L = 2\pi (100) (0.735) = 2\pi \times 73.5 \approx 461.8 \Omega\]

Step 3: Determine the Impedance

Impedance \( Z \) in a series RLC circuit is calculated as: \( Z = \sqrt{R^2 + (X_L - X_C)^2} \), where \( R = 175 \) \( \Omega \).Substitute the values:\[Z = \sqrt{(175)^2 + (461.8 - 3536.8)^2} = \sqrt{175^2 + (-3075)^2}\]Calculating further:\[Z = \sqrt{30625 + 9455625} = \sqrt{9486250} \approx 3079.7 \Omega\]

Step 4: Calculate the Current

The current \( I \) in the circuit is given by Ohm's Law: \( I = \frac{V}{Z} \), where \( V \) is the voltage of 5 V:\[I = \frac{5}{3079.7} \approx 0.001625 \text{ A or } 1.625 \text{ mA}\]

Key Concepts

Impedance Calculation

In an RLC circuit, impedance is essentially the overall resistance that the circuit offers to the flow of alternating current (AC). It is a combination of resistance (R), capacitive reactance (X_C), and inductive reactance (X_L). The resistance is straightforward, as it does not vary with frequency. However, the reactance parts change: X_L increases with frequency, while X_C decreases.

To calculate impedance, we use the formula:\[Z = \sqrt{R^2 + (X_L - X_C)^2}\]Here:

• \(Z\) is the impedance in ohms (Ω).
• \(R\) is the resistance in ohms (Ω).
• \(X_L\) is the inductive reactance in ohms (Ω).
• \(X_C\) is the capacitive reactance in ohms (Ω).

The formula essentially combines these elements to give us the equivalent resistance the AC circuit provides.

The term \(X_L - X_C\) reflects the net reactance effect, indicating whether the circuit behaves more as an inductor or a capacitor at certain frequencies. In this exercise, calculating these values lets us find the total opposition the circuit has to AC current flow.

Reactance in Circuits

Reactance is a critical concept in AC circuits and describes how capacitors and inductors respond to AC current. Unlike resistance, reactance varies with frequency, making it unique to AC analysis.

For a capacitor, the reactance (X_C) decreases as the frequency increases. The formula is:\[X_C = \frac{1}{2\pi f C}\]Where:

• \(f\) is the frequency in hertz (Hz).
• \(C\) is the capacitance in farads (F).

As the frequency rises, the capacitor offers less opposition to the current. This exercise involves calculating:\(X_C \approx 3536.8 \, \Omega\), showing substantial resistance at 100 Hz.

For an inductor, the opposite happens: reactance (X_L) increases with frequency. Its formula is:\[X_L = 2\pi f L\]With:

• \(L\) being the inductance in henries (H).

The greater the frequency, the higher the inductor’s opposition. Here, X_L equates to approximately 461.8 \, \Omega, reinforcing the inductor's behavior at the same frequency.

Overall, these values of reactance greatly influence impedance and determine how current flows through the RLC circuit.

Ohm's Law Application

Ohm's Law is a fundamental principle that connects voltage (V), current (I), and resistance (R) in electrical circuits. It's generally articulated as: \[I = \frac{V}{R}\]For AC circuits, we substitute impedance (Z) for resistance, given that impedance encompasses both resistive and reactive properties.

In this exercise, we calculated impedance (Z \approx 3079.7 \, \Omega) using the given frequency, capacitance, and inductance.Utilizing Ohm’s Law:\[I = \frac{V}{Z}\]Given that voltage (V) is 5 volts, the current (I) is found to be approximately 0.001625 \, \text{A} or 1.625 \, \text{mA}. This result showcases how Ohm's Law helps us compute current, reflecting the real-world response of complex AC circuits.

Ohm's Law, despite its simple expression, becomes powerful in AC analysis when used with impedance to understand how voltages drive currents through components that have frequency-dependent reactance.