Capacitors in AC Circuits: Videos & Practice Problems

In AC circuits, capacitors play a crucial role where the voltage across a capacitor lags the current by 90 degrees. The relationship is defined by the equation for capacitive reactance, X = 1ωC, where ω = 2πf. The maximum current can be calculated using I = VX. Understanding these concepts is essential for analyzing AC circuits effectively.

Capacitors in AC Circuits

Capacitors play a crucial role in alternating current (AC) circuits, particularly in how they interact with current and voltage. In an AC circuit, the current can be expressed as I_max times the cosine of ωt, where ω is the angular frequency. The voltage across a capacitor at any given time is determined by the charge on the capacitor divided by its capacitance, a concept rooted in direct current (DC) circuit principles.

The relationship between charge and current in a capacitor involves calculus, leading to the conclusion that the voltage across the capacitor can be represented as:

$$ V(t) = \frac{I_{max}}{\omega C} \cos(\omega t - \frac{\pi}{2}) $$

This indicates that the voltage across the capacitor lags the current by 90 degrees, meaning that when the current reaches its maximum, the voltage is at zero, and vice versa. This phase difference is critical in understanding how capacitors behave in AC circuits.

Moreover, the maximum voltage across the capacitor can be calculated as V_max = I_max / ωC, which resembles Ohm's Law. Here, the term 1/ωC acts like a resistance, known as capacitive reactance, denoted as X_C:

$$ X_C = \frac{1}{\omega C} $$

Capacitive reactance has units of ohms, similar to resistance, and is essential for analyzing AC circuits.

To illustrate these concepts, consider an example where an AC power source provides a maximum voltage of 120 volts at a frequency of 60 hertz, connected to a 100 microfarad capacitor. First, we calculate the angular frequency:

$$ \omega = 2\pi f = 2\pi \times 60 \approx 377 \, \text{s}^{-1} $$

Next, we find the capacitive reactance:

$$ X_C = \frac{1}{377 \times 100 \times 10^{-6}} \approx 26.5 \, \Omega $$

Finally, the maximum current can be calculated using:

$$ I_{max} = \frac{V_{max}}{X_C} = \frac{120}{26.5} \approx 4.53 \, \text{A} $$

This example highlights the interplay between voltage, current, and capacitive reactance in AC circuits, emphasizing the importance of understanding these relationships for effective circuit analysis.

An AC source operates at a maximum voltage of 120 V and a frequency of 60 Hz. If it is connected to a 175 µF capacitor, what is the maximum charge stored on the capacitor?

• A. 7.92C
• B. 0.021 C
• C. 21 C
• D. 0.0021C

Answer: (B)

Current in a Parallel RC AC Circuit

In this example, we explore the behavior of capacitors in alternating current (AC) circuits, specifically focusing on calculating the root mean square (RMS) current through a capacitor connected in parallel with a resistor. The AC source operates at an angular frequency of \( \omega = 160 \, \text{s}^{-1} \) and has a maximum voltage of \( V_{\text{max}} = 15 \, \text{V} \). The circuit includes a \( 5 \, \Omega \) resistor and a \( 1.5 \, \text{mF} \) capacitor.

To find the RMS current through the capacitor, we first note that the voltage across the capacitor is the same as the maximum voltage of the source due to their parallel connection. The resistor does not influence the current through the capacitor in this scenario, allowing us to focus solely on the capacitor's characteristics.

The maximum current through the capacitor can be calculated using the formula:

\( I_{\text{max}} = \frac{V_{\text{max}}}{X_C} \)

where \( X_C \) is the capacitive reactance, given by:

\( X_C = \frac{1}{\omega C} \)

Here, \( C = 1.5 \, \text{mF} = 0.0015 \, \text{F} \). Substituting the values, we find:

\( X_C = \frac{1}{160 \times 0.0015} = 4.2 \, \Omega \)

Now, substituting \( V_{\text{max}} \) and \( X_C \) into the maximum current formula gives:

\( I_{\text{max}} = \frac{15 \, \text{V}}{4.2 \, \Omega} \approx 3.57 \, \text{A} \)

To find the RMS current, we use the relationship:

\( I_{\text{rms}} = \frac{I_{\text{max}}}{\sqrt{2}} \)

Calculating this yields:

\( I_{\text{rms}} = \frac{3.57 \, \text{A}}{\sqrt{2}} \approx 2.52 \, \text{A} \)

Thus, the RMS current through the capacitor is approximately \( 2.52 \, \text{A} \). This example illustrates the importance of understanding capacitive reactance and the relationship between maximum and RMS values in AC circuits.

A 300 µF capacitor is connected to an AC source operating at an RMS voltage of 120 V. If the maximum current in the circuit is 1.5 A, what is the oscillation frequency of the AC source?

• A. 29.4 Hz
• B. 4.68 Hz
• C. 41.7 Hz
• D. 6.63 Hz

Answer: (B)

Here’s what students ask on this topic:

What is the role of capacitors in AC circuits?

In AC circuits, capacitors play a crucial role by storing and releasing energy. The voltage across a capacitor lags the current by 90 degrees, meaning the current leads the voltage. This phase difference is due to the capacitive reactance, which is given by the equation $X=\frac{1}{\omega C}$, where $\omega$ is the angular frequency and $C$ is the capacitance. This reactance acts like a resistance in the circuit, affecting the current flow and voltage distribution.

How do you calculate the capacitive reactance in an AC circuit?

The capacitive reactance $X$ in an AC circuit is calculated using the formula $X=\frac{1}{\omega C}$, where $\omega$ is the angular frequency and $C$ is the capacitance. The angular frequency $\omega$ is related to the linear frequency $f$ by the equation $\omega =2\pi f$. This reactance is measured in ohms and determines how much the capacitor resists the change in voltage in the AC circuit.

Why does the voltage across a capacitor lag the current in an AC circuit?

The voltage across a capacitor lags the current in an AC circuit by 90 degrees due to the nature of how capacitors store and release energy. When the current changes, the capacitor charges or discharges, causing a delay in the voltage response. This phase difference is a result of the capacitive reactance, which is given by $X=\frac{1}{\omega C}$. The current leads the voltage because the capacitor needs time to build up or release the charge, resulting in the voltage lagging behind the current.

How do you calculate the maximum current in an AC circuit with a capacitor?

The maximum current $I$_max in an AC circuit with a capacitor can be calculated using the formula $I$_max=V_maxX, where $V$_max is the maximum voltage and $X$ is the capacitive reactance. The capacitive reactance is given by $X=\frac{1}{\omega C}$, where $\omega$ is the angular frequency and $C$ is the capacitance. By substituting the values, you can find the maximum current in the circuit.

What is the relationship between capacitive reactance and frequency in an AC circuit?

The capacitive reactance $X$ in an AC circuit is inversely proportional to the frequency $f$. The relationship is given by the formula $X=\frac{1}{2\pi fC}$, where $C$ is the capacitance. As the frequency increases, the capacitive reactance decreases, and vice versa. This means that at higher frequencies, the capacitor offers less opposition to the current flow, while at lower frequencies, it offers more opposition.