Problem 16 - Physics

Question

A \(2700-\Omega\) resistor and a \(1.1-\mu F\) capacitor are connected in series across a generator \((60.0 \mathrm{~Hz}, 120 \mathrm{~V}) .\) Determine the power delivered to the circuit.

Step-by-Step Solution

Answer

The power delivered to the circuit is 2.01 W.

Step 1: Calculate Impedance

First, calculate the impedance of the circuit. The impedance in a series resistor-capacitor circuit is given by the formula \[ Z = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} \]where \( R = 2700 \underline{\phantom{xxx}}\Omega \), \( \omega = 2\pi f \) is the angular frequency (with \( f = 60.0 \text{ Hz} \)), and \( C = 1.1 \times 10^{-6} \text{ F} \).First, find \( \omega = 2 \times \pi \times 60 = 120\pi \text{ rad/s} \).Then, calculate \( \frac{1}{\omega C} = \frac{1}{120\pi \times 1.1 \times 10^{-6}} \approx 2.416 \times 10^3 \underline{\phantom{xxx}}\Omega \).Thus, \[ Z = \sqrt{2700^2 + (2.416 \times 10^3)^2} \approx 3570.59 \underline{\phantom{xxx}}\Omega \].

Key Concepts

Impedance Calculation

In AC circuits, impedance is a measure that combines both the resistance and reactance (from components like capacitors and inductors) into a single complex quantity. Impedance is important because it determines how much current will flow in the circuit for a given voltage. It's similar to resistance in DC circuits but applies to AC circuits with additional complexity due to phase changes.To calculate impedance, especially in circuits containing capacitors and resistors, you need to consider both the real and imaginary parts of the components. For a series resistor-capacitor circuit, the impedance formula is:\[ Z = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} \]Here's how you calculate it:

• R represents the resistance. In our example, it is 2700 Ω.
• C is the capacitance. For the exercise, C is 1.1 µF or 1.1 x 10^-6 F.
• \(\omega\) is the angular frequency, found by \(\omega = 2\pi f\), where f is the frequency of oscillation, 60 Hz in this case.

First, compute the angular frequency: \(\omega = 2 \times \pi \times 60 \approx 120\pi\). Then, calculate the reactance of the capacitor: \(\frac{1}{\omega C} \approx 2.416 \times 10^3\). Finally, use the formula above to find the impedance, which gives \(Z \approx 3570.59 \underline{\phantom{xxx}}\Omega\).

Series Resistor-Capacitor Circuit

A series resistor-capacitor (RC) circuit is a type of circuit where resistors and capacitors are connected one after another in a single pathway. This configuration affects how voltage and current behave in the circuit, as well as how energy is stored and released. In such a circuit:

• The capacitor stores electrical energy in an electric field, affecting the phase difference between voltage and current. It introduces reactive opposition to the current flow, known as capacitive reactance. This behavior is frequency-dependent, decreasing as the frequency increases.
• The resistor provides a constant resistive opposition to the flow of current, independent of frequency. The resistor affects the energy being dissipated as heat.

Both these elements together determine the overall impedance of the circuit. A series RC circuit is often used in filters and timing applications because of its ability to form different types of filters and delay circuits.

Power Delivered in AC Circuits

Power in AC circuits signifies how much energy is being transferred or used in the circuit, and it can be more complex than in DC circuits because of the phase difference between voltage and current.In AC circuits, power can be calculated using the formula:\[ P = VI \text{cos}\phi \]Here:

• V is the voltage across the circuit, typically the RMS (root mean square) value.
• I is the current through the circuit, also an RMS value.
• \(\phi\) is the phase difference between the voltage and current, which depends on the reactive components.

In the given example, before calculating the power, you must find the current using Ohm's Law for AC:\[ I = \frac{V}{Z} \]With calculated impedance \(Z \approx 3570.59\) Ω and voltage \(V = 120\) V, find current and then power using the equations above. Considering no phase shift for this approximation (cosφ ≈ 1), you can simplify to find power, giving insight into energy used or transmitted in your AC circuit.