Problem 23
Question
A maximum voltage of \(34.0 \mathrm{~V}\) is developed by an ac generator that delivers a maximum current of \(0.170 \mathrm{~A}\) to a circuit. (a) What is the effective voltage of the generator? (b) Find the effective current delivered to the circuit by the generator. (c) Find the resistance of the circuit.
Step-by-Step Solution
Verified Answer
(a) 24.05 V, (b) 0.120 A, (c) 200.42 ohms
1Step 1: Understanding the Problem
We are given the maximum voltage (peak voltage) of the AC generator as \( V_{max} = 34.0 \text{ V} \) and the maximum current (peak current) as \( I_{max} = 0.170 \text{ A} \). We need to find (a) the effective voltage (RMS voltage), (b) the effective current (RMS current), and (c) the resistance of the circuit.
2Step 1: Calculate RMS Voltage
The effective voltage (RMS voltage) for an AC circuit can be found using the relation \( V_{rms} = \frac{V_{max}}{\sqrt{2}} \). Substituting the given values, we get:\[V_{rms} = \frac{34.0}{\sqrt{2}} = \frac{34.0}{1.414} \approx 24.05 \text{ V}\]
3Step 2: Calculate RMS Current
The effective current (RMS current) is calculated in a similar manner using: \( I_{rms} = \frac{I_{max}}{\sqrt{2}} \). Plugging the numbers in, we have:\[I_{rms} = \frac{0.170}{\sqrt{2}} = \frac{0.170}{1.414} \approx 0.120 \text{ A}\]
4Step 3: Find the Resistance of the Circuit
Using Ohm's law for resistances in AC circuits, the resistance \( R \) can be found using the RMS values:\[R = \frac{V_{rms}}{I_{rms}}\]Substituting the calculated effective voltage and current, we get:\[R = \frac{24.05}{0.120} \approx 200.42 \text{ ohms}\]
Key Concepts
RMS VoltageRMS CurrentOhm's Law
RMS Voltage
RMS Voltage, or root mean square voltage, is a critical concept in AC circuits. It provides a way to express the effective value of an AC voltage, allowing for meaningful comparisons with DC voltages. In essence, RMS Voltage represents the equivalent DC voltage that would deliver the same amount of power to a load as the AC voltage. This is vital for ensuring circuits function efficiently and safely.
The formula for finding RMS voltage, given the peak (or maximum) voltage, is:
Understanding RMS voltage is essential for tasks involving power calculations and matching AC devices with proper power sources.
The formula for finding RMS voltage, given the peak (or maximum) voltage, is:
- \[ V_{rms} = \frac{V_{max}}{\sqrt{2}} \]
Understanding RMS voltage is essential for tasks involving power calculations and matching AC devices with proper power sources.
RMS Current
Like RMS voltage, RMS current, or root mean square current, is used in AC circuits to provide a standard, measurable, and comparable value. This concept helps us determine the real effective current flowing in an AC system. In many ways, RMS current parallels RMS voltage. It shows the amount of current that delivers the same power to a load as would a DC current of the same RMS value.
To calculate RMS current from peak current, we use:
RMS current is crucial in ensuring electronic devices, and circuits operate within their safe current ratings to avoid overheating and potential damage.
To calculate RMS current from peak current, we use:
- \[ I_{rms} = \frac{I_{max}}{\sqrt{2}} \]
RMS current is crucial in ensuring electronic devices, and circuits operate within their safe current ratings to avoid overheating and potential damage.
Ohm's Law
Ohm's Law is a fundamental principle for understanding AC circuits, relating 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.
For AC circuits, we typically use the RMS values to apply Ohm's Law:
Ohm's Law is valuable for designing circuits and diagnosing issues, making it one of the cornerstones of electrical engineering.
For AC circuits, we typically use the RMS values to apply Ohm's Law:
- \[ R = \frac{V_{rms}}{I_{rms}} \]
Ohm's Law is valuable for designing circuits and diagnosing issues, making it one of the cornerstones of electrical engineering.
Other exercises in this chapter
Problem 22
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A step-up transformer has \(30 \overline{0} 0\) turns in the secondary coil and \(20 \overline{0}\) turns in the primary coil. The primary is supplied with alte
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