Problem 7 - College Physics

Question

An ac voltage is applied to a \(25.0-\Omega\) resistor so that it dissipates \(500 \mathrm{~W}\) of power. Find the resistor's (a) \(\mathrm{rms}\) and peak currents and (b) rms and peak voltages.

Step-by-Step Solution

Answer

(a) RMS current: 4.47 A, Peak current: 6.32 A; (b) RMS voltage: 111.75 V, Peak voltage: 157.98 V.

Step 1: Determine RMS Current

We start by using the formula for power dissipation in a resistor: \(P = I_{rms}^2 \times R\). Here, \(P = 500 \text{ W}\) and \(R = 25.0 \Omega\). Rearrange the formula to solve for \(I_{rms}\): \[I_{rms} = \sqrt{\frac{P}{R}}\] Substitute the given values: \[I_{rms} = \sqrt{\frac{500}{25.0}} = \sqrt{20} = 4.47\, \text{A}\].

Step 2: Calculate Peak Current

The peak current \(I_0\) is related to the RMS current by the formula: \[I_0 = I_{rms} \times \sqrt{2}\]. Substitute the value of \(I_{rms}\): \[I_0 = 4.47 \times \sqrt{2} = 6.32\, \text{A}\].

Step 3: Calculate RMS Voltage

The RMS voltage \(V_{rms}\) across the resistor can be found using Ohm’s Law: \[V_{rms} = I_{rms} \times R\]. Substitute the value of \(I_{rms}\): \[V_{rms} = 4.47 \times 25.0 = 111.75\, \text{V}\].

Step 4: Calculate Peak Voltage

The peak voltage \(V_0\) is related to the RMS voltage by the formula: \[V_0 = V_{rms} \times \sqrt{2}\]. Substitute the value of \(V_{rms}\): \[V_0 = 111.75 \times \sqrt{2} = 157.98\, \text{V}\].

Key Concepts

RMS Current

In the world of AC circuits, the term RMS, or Root Mean Square, is crucial. RMS current allows us to compare AC current with DC current in terms of power delivery. AC current varies with time, unlike DC current which is constant. However, RMS gives a numerical value to this varying current by averaging its magnitude over time. This value represents the DC current that would produce the same power in a resistor.
For instance, in the original exercise, we calculated the RMS current using the formula: \[I_{rms} = \sqrt{\frac{P}{R}}\]where \(P\) is the power dissipated, and \(R\) is the resistance. Plugging in the numbers, we found \(I_{rms} = 4.47 \text{ A}\). This value helps in understanding the real impact of the current despite its constant fluctuation.
RMS current is immensely useful as it provides a consistent way to measure and calculate outcomes in AC circuits, akin to DC circuits.

Peak Voltage

Peak voltage refers to the maximum value reached by the voltage in an AC circuit. Unlike DC voltage, which remains steady, AC voltage fluctuates between positive and negative peaks. The peak voltage represents the highest point on these fluctuations.
Understanding peak voltage is important because it defines the extremes that the voltage can reach, which is vital for the integrity of electrical components and insulation.
From our calculations in the exercise, we can find the peak voltage \(V_0\) using the formula:\[V_0 = V_{rms} \times \sqrt{2}\].Substituting the RMS voltage we calculated, \(111.75 \text{ V}\), gives us a peak voltage of \(157.98 \text{ V}\).
This concept provides insight into the behavior of the voltage wave and why equipment must be rated to withstand these peak values.

Ohm's Law

Ohm’s Law is a fundamental principle in electrical engineering that relates voltage, current, and resistance. It states that the current through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them.
The formula is succinct: \[V = I \times R\].In the context of AC circuits, although the nature of the current is alternating, Ohm's Law still holds, but it uses the RMS values for current and voltage.
In the exercise, we applied Ohm's Law to calculate RMS voltage across the resistor, given the RMS current. Using the formula \[V_{rms} = I_{rms} \times R\], and substituting \(I_{rms} = 4.47 \text{ A}\) and \(R = 25.0 \Omega\), we found \(V_{rms} = 111.75 \text{ V}\).
This principle serves as a cornerstone for solving many practical problems in circuit analysis and designing systems.