Problem 6 - Applied Physics

Question

A circuit contains a 225- \(\Omega\) resistance, a \(5.00-\mu \mathrm{F}\) capacitor, and a \(0.550-\mathrm{H}\) inductance in series with a \(7.50-\mathrm{V}, 60.0\) -Hz source. Find the impedance and the current.

Step-by-Step Solution

Answer

The impedance is approximately 343.74 Ω and the current is approximately 0.0218 A.

Step 1: Calculate Capacitive Reactance

The formula for capacitive reactance \(X_C\) is \(X_C = \frac{1}{2\pi f C}\), where \(f\) is the frequency and \(C\) is the capacitance. Substituting the values \(f = 60.0\text{ Hz}\) and \(C = 5.00 \times 10^{-6} \text{ F}\), we get:\[X_C = \frac{1}{2 \pi \times 60.0 \times 5.00 \times 10^{-6}} \approx 530.52 \Omega\]

Step 2: Calculate Inductive Reactance

The formula for inductive reactance \(X_L\) is \(X_L = 2\pi f L\), where \(L\) is the inductance. Substituting the values \(f = 60.0\text{ Hz}\) and \(L = 0.550 \text{ H}\), we get:\[X_L = 2 \pi \times 60.0 \times 0.550 \approx 207.35 \Omega\]

Step 3: Calculate Total Impedance

The series circuit's total impedance \(Z\) is given by the formula:\[Z = \sqrt{R^2 + (X_L - X_C)^2}\]where \(R = 225\) \(\Omega\), \(X_L = 207.35\) \(\Omega\), \(X_C = 530.52\) \(\Omega\). Substituting these values, we find:\[Z = \sqrt{225^2 + (207.35 - 530.52)^2} \approx 343.74 \Omega\]

Step 4: Calculate Current in the Circuit

The current \(I\) in the circuit can be calculated using Ohm's Law, \(I = \frac{V}{Z}\), where \(V = 7.50\text{ V}\) and \(Z = 343.74 \Omega\). Thus, the current is:\[I = \frac{7.50}{343.74} \approx 0.0218 \text{ A}\]

Key Concepts

Capacitive Reactance

Capacitive reactance (\(X_C\)) plays a crucial role in AC circuits, especially those with capacitors. It is the opposition that a capacitor offers to the flow of AC current. Unlike resistance, which remains constant, capacitive reactance depends on the frequency of the AC signal and the capacitance of the capacitor.
To find capacitive reactance, we use the formula:\[X_C = \frac{1}{2\pi f C}\]Here:

• \(f\) is the frequency in Hertz (Hz)
• \(C\) is the capacitance in Farads (F)
• \(\pi\) is a mathematical constant approximately equal to 3.14159.

This formula tells us that as the frequency increases, the capacitive reactance decreases. This is because, at higher frequencies, the capacitor can charge and discharge cycles quickly, allowing more current to pass through it.
In our example, the circuit had a capacitance of \(5.00\mu F\) and a frequency of \(60.0\) Hz. This resulted in a capacitive reactance of approximately \(530.52\) \(\Omega\).
Understanding capacitive reactance helps when designing and analyzing circuits, as it affects the phase and amplitude of the AC current.

Inductive Reactance

Inductive reactance (\(X_L\)) is an important concept when dealing with AC circuits that include inductors. It represents the opposition that an inductor presents to the alternating current. Much like capacitive reactance, inductive reactance also varies with the frequency of the current.
The formula used to calculate inductive reactance is:\[X_L = 2\pi f L\]Where:

• \(f\) is the frequency in Hz
• \(L\) is the inductance in Henrys (H).

As the frequency increases, so does the inductive reactance. This is because higher frequencies mean the current changes direction more rapidly, which increases the opposition to the flow of current by the inductor.
In the provided example, with an inductance of \(0.550\) H and a frequency of \(60.0\) Hz, the inductive reactance is approximately \(207.35\) \(\Omega\).
Understanding inductive reactance is essential, as it allows us to predict how the inductor will behave in the circuit and how it will affect circuit impedance.

Ohm's Law

Ohm's Law forms the foundation for understanding electrical circuits, especially AC circuits. It relates voltage (\(V\)), current (\(I\)), and resistance, or in this case impedance (\(Z\)) for AC circuits.
The formula is straightforward:\[I = \frac{V}{Z}\]Where:

• \(I\) is the current in Amperes (A)
• \(V\) is the voltage in Volts (V)
• \(Z\) is the impedance in Ohms (\(\Omega\)).

Impedance combines both resistance and reactance (capacitive and inductive), providing a comprehensive picture of opposition to current in an AC circuit.
In our scenario, the circuit has an impedance of approximately \(343.74\) \(\Omega\), resulting in a current of about \(0.0218\) A when the voltage is \(7.50\) V. By using Ohm's Law, we can easily determine how much current flows in the circuit.
Understanding Ohm's Law is fundamental for any linear circuit analysis, as it helps predict how circuit components will interact and influence current and voltage dynamics.