Problem 15 - College Physics

Question

\(\bullet\) The power of a certain \(\mathrm{CD}\) player operating at 120 \(\mathrm{V} \mathrm{rms}\) is 20.0 \(\mathrm{W} .\) Assuming that the \(\mathrm{CD}\) player behaves like a pure resistance, find (a) the maximum instantaneous power, (b) the rms current, and (c) the resistance of this player.

Step-by-Step Solution

Answer

RMS current: 0.167 A, Resistance: 719.16 Ω, Max power: 40.07 W.

Step 1: Calculate RMS Current

To calculate the root mean square (RMS) current, use the formula for power: \[ P = V_{\text{rms}} \times I_{\text{rms}} \]Rearrange the equation to find \(I_{\text{rms}}\): \[ I_{\text{rms}} = \frac{P}{V_{\text{rms}}} = \frac{20.0 \text{ W}}{120 \text{ V}} = 0.167 \text{ A} \]

Step 2: Calculate Resistance

Using Ohm's Law, where \(V_{\text{rms}} = I_{\text{rms}} \times R\), solve for \(R\): \[ R = \frac{V_{\text{rms}}}{I_{\text{rms}}} = \frac{120 \text{ V}}{0.167 \text{ A}} \approx 719.16 \text{ Ω} \]

Step 3: Calculate Maximum Instantaneous Power

The maximum instantaneous power \(P_{\max}\) in an AC circuit with a purely resistive load is given by \[ P_{\max} = V_{\text{peak}} \times I_{\text{peak}} \]First, calculate the peak voltage: \[ V_{\text{peak}} = \sqrt{2} \times V_{\text{rms}} = \sqrt{2} \times 120 \text{ V} \approx 169.7 \text{ V} \]Similarly, find the peak current: \[ I_{\text{peak}} = \sqrt{2} \times I_{\text{rms}} \approx \sqrt{2} \times 0.167 \text{ A} \approx 0.236 \text{ A} \]Now, calculate the maximum instantaneous power: \[ P_{\max} = 169.7 \text{ V} \times 0.236 \text{ A} \approx 40.07 \text{ W} \]

Key Concepts

RMS Current

The root mean square (RMS) current is a way to express the effective value of an alternating current (AC). It represents the equivalent amount of direct current (DC) that would deliver the same power to a load as the AC does. When calculating the RMS current for an appliance like a CD player, you use the formula: \[ I_{\text{rms}} = \frac{P}{V_{\text{rms}}} \] where \( P \) is the power in watts and \( V_{\text{rms}} \) is the RMS voltage. For the CD player described, operating at 120 V RMS with a power of 20 W, the RMS current is calculated as follows: \[ I_{\text{rms}} = \frac{20 \text{ W}}{120 \text{ V}} = 0.167 \text{ A} \] Understanding RMS current helps in designing circuits and ensuring that components can handle the current without overheating.

Ohm's Law

Ohm's Law is foundational in electrical engineering, relating voltage, current, and resistance. It is stated as: \[ V = I \times R \] where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. In AC circuits with a resistive load, this relationship still applies, and it can be rearranged to solve for resistance if the current and voltage are known: \[ R = \frac{V}{I} \] For the example of the CD player, with an RMS voltage of 120 V and an RMS current of 0.167 A, the resistance \( R \) is: \[ R = \frac{120 \text{ V}}{0.167 \text{ A}} \approx 719.16 \text{ Ω} \] This resistance value shows how much the CD player opposes the current flow.

Peak Voltage

In AC circuits, the peak voltage is the maximum voltage reached by the waveform. It can be found using the RMS voltage with the formula: \[ V_{\text{peak}} = \sqrt{2} \times V_{\text{rms}} \] This is because the RMS value is the effective value that relates to the square root of the mean of the squares of the waveform. For our CD player with an RMS voltage of 120 V, the peak voltage comes out to be approximately: \[ V_{\text{peak}} = \sqrt{2} \times 120 \text{ V} \approx 169.7 \text{ V} \] Understanding peak voltage is crucial for assessing how components will handle the voltage peaks in an AC system.

Instantaneous Power

Instantaneous power in an AC circuit is the power at any given moment, which constantly changes as the waveform cycles. In purely resistive loads, maximum instantaneous power occurs at the peak points of the voltage and current waveforms. This is calculated using: \[ P_{\max} = V_{\text{peak}} \times I_{\text{peak}} \] Given that peak values are \( V_{\text{peak}} \approx 169.7 \text{ V} \) and \( I_{\text{peak}} \approx 0.236 \text{ A} \), the maximum instantaneous power is: \[ P_{\max} = 169.7 \text{ V} \times 0.236 \text{ A} \approx 40.07 \text{ W} \] This measurement helps in understanding the power demands at critical moments within the AC cycle and ensuring that electrical components can handle these demands without damage.