Question

The wiring for a refrigerator contains a starter capacitor. A voltage of amplitude 170 V and frequency 60.0 Hz applied across the capacitor is to produce a current amplitude of 0.850 A through the capacitor. What capacitance \(C\) is required?

Short answer

The required capacitance is approximately 13.33 µF.

Step-by-step solution

Understanding the Problem

We are asked to find the capacitance required for a starter capacitor in a refrigerator. We know the voltage amplitude is 170 V, the frequency is 60.0 Hz, and the current amplitude is 0.850 A. We need to use the relationship between voltage, current, and capacitance for an AC circuit.

Recall the Formula for Current in a Capacitor

The current amplitude through a capacitor in an AC circuit is given by the formula: \(I = V \times 2 \pi f C\), where \(I\) is the current amplitude, \(V\) is the voltage amplitude, \(f\) is the frequency, and \(C\) is the capacitance. We will rearrange this formula to solve for \(C\).

Rearrange the Formula to Find C

Rearrange the formula to solve for capacitance \(C\): \[ C = \frac{I}{V \times 2 \pi f} \]. We will use this formula to find the value of \(C\).

Substitute the Given Values

Now, we substitute the given values into the formula: \(C = \frac{0.850}{170 \times 2 \pi \times 60.0}\).

Calculate the Capacitance

Perform the calculation: \[ C = \frac{0.850}{170 \times 2 \times 3.1416 \times 60} \approx 1.333 \times 10^{-5} \text{ F} \] or \(13.33 \text{ µF}\).

Confirm the Calculation

Finally, verify the result by checking calculation steps and ensuring all units were consistent. The calculations seem correct with correct unit conversions.

Key concepts

Capacitance Calculation

Capacitance in an AC circuit, like a refrigerator, can be determined using a specific formula that links current, voltage, and frequency. This relationship is depicted as \(I = V \times 2\pi f C\), where:

• \(I\) is the current amplitude.
• \(V\) is the voltage amplitude.
• \(f\) is the frequency.
• \(C\) is the capacitance.

To find the needed capacitance, you rearrange the formula: \[ C = \frac{I}{V \times 2 \pi f} \].
Using this formula involves substituting known values for each parameter, performing calculations, and interpreting the units correctly. Capacitance measures how much charge a capacitor can store per unit voltage, which is essential in managing the current flow in AC circuits effectively.

Current Amplitude

Current amplitude in an AC circuit refers to the peak value of alternating current (AC). Unlike DC, where the current is constant, AC varies with time, oscillating between its peak positive and negative values. This peak value is crucial since it determines the necessary voltage and frequency values needed by capacitive elements in the circuit.
In the exercise provided, the current amplitude is given as 0.850 A, which represents this maximum value. In practical terms, this means that during one cycle of the alternating current, the maximum current flowing through the capacitor is 0.850 Amperes. Knowing this helps in designing circuits, ensuring that all components can handle the peak current efficiently.

Voltage Amplitude

Voltage amplitude describes the peak voltage value in an AC circuit. In essence, it is the maximum amount of voltage being applied across components such as capacitors during each AC cycle.
In the problem, the voltage amplitude is 170 V. This peak voltage is significant as it influences the current flowing through the capacitor. Higher voltage amplitudes lead to higher current amplitudes, as per Ohm’s Law for AC circuits. Understanding voltage amplitude also helps in ensuring that circuit components, like capacitors, can withstand and operate effectively under these peak voltage conditions, preventing damage and ensuring optimal performance.

Frequency in AC Circuits

Frequency in AC circuits is a critical factor influencing the behavior of components like capacitors. It is the rate at which the current changes direction per second, measured in hertz (Hz).
In our exercise, the given frequency is 60.0 Hz, a standard value for AC power in many regions. Frequency affects how much reactance a capacitor has, altering the current flowing through it. Specifically, the higher the frequency, the lower the capacitive reactance, allowing more current to pass. This dependency is why frequency must be carefully considered in calculations, as it directly affects the design and functionality of AC circuits.