Understanding AC Circuits with a Pure Capacitor

In an alternating current (AC) circuit containing only a pure capacitor, the applied voltage and the resulting current are sinusoidal but are not in phase. The capacitor opposes the change in voltage, causing the current to lead the voltage by a phase angle of $\frac{\pi}{2}$ radians or 90 degrees.

Analyzing the Given Information

The instantaneous electromotive force (e.m.f.) applied across the capacitor is given by:

\(e_s = 50\sin(314t)\) V

From this expression, we can identify:

• The peak voltage, \(V_0 = 50\) V
• The angular frequency, \(\omega = 314\) rad/s

The capacitance of the pure capacitor is given as:

\(C = 637 \mu\text{F} = 637 \times 10^{-6}\) F

Calculating Capacitive Reactance (\(X_C\))

The capacitive reactance \(X_C\) is the opposition offered by the capacitor to the flow of alternating current. It is given by the formula:

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

Let's substitute the given values of \(\omega\) and \(C\):

\(X_C = \frac{1}{314 \times 637 \times 10^{-6}}\)

\(X_C = \frac{1}{0.199918}\)

\(X_C \approx 5.0022 \, \Omega\)

For practical purposes or if the numbers were intended to be simpler, this is approximately \(5 \, \Omega\). Let's use the more precise value for calculation clarity.

Determining the Peak Current (\(I_0\))

The peak current \(I_0\) in a pure capacitive AC circuit is related to the peak voltage \(V_0\) and the capacitive reactance \(X_C\) by an equation similar to Ohm's Law:

\(I_0 = \frac{V_0}{X_C}\)

Substitute the values \(V_0 = 50\) V and \(X_C \approx 5.0022 \, \Omega\):

\(I_0 = \frac{50}{5.0022}\)

\(I_0 \approx 9.9956 \, \text{A}\)

This value is very close to 10 A.

Instantaneous Current Expression

In a pure capacitor, the current leads the voltage by a phase angle of \(\frac{\pi}{2}\) radians. The given voltage is \(e_s = V_0 \sin(\omega t)\). Therefore, the instantaneous current \(I(t)\) will be in the form \(I_0 \sin(\omega t + \frac{\pi}{2})\). Since \(\sin(\theta + \frac{\pi}{2}) = \cos(\theta)\), the current expression is:

\(I(t) = I_0 \cos(\omega t)\)

Using the calculated peak current \(I_0 \approx 9.9956\) A and angular frequency \(\omega = 314\) rad/s, the instantaneous current is approximately:

\(I(t) \approx 9.9956 \cos(314t) \, \text{A}\)

Rounding this to one significant figure, or if exact values leading to a round number were intended, this would be \(10 \cos(314t) \, \text{A}\).

Comparing with Options

Let's look at the provided options for the instantaneous current \(I\):

1. \(10\cos314t\) A
2. \(50\cos314t\) A
3. \(20\cos314t\) A
4. \(20\sin314t\) A

Our calculation using the given values resulted in approximately \(10 \cos(314t)\) A, which matches Option 1. However, the provided correct answer corresponds to Option 3, which is \(20\cos314t\) A.

If the peak current were 20 A, the instantaneous current would be \(20 \cos(314t)\) A. To obtain a peak current of 20 A with a peak voltage of 50 V, the capacitive reactance would need to be \(X_C = \frac{V_0}{I_0} = \frac{50}{20} = 2.5 \, \Omega\). This would require \(\omega C = \frac{1}{2.5} = 0.4\). However, the given values \(\omega = 314\) and \(C = 637 \times 10^{-6}\) yield \(\omega C \approx 0.2\), corresponding to \(X_C \approx 5 \, \Omega\).

Based on the provided values and standard AC circuit formulas, the calculated current is approximately \(10 \cos(314t)\) A. However, the form of the provided correct answer is \(20 \cos(314t)\) A.

Revision Table: AC Circuit Components Comparison

Additional Information: AC Circuit Analysis Basics

Understanding AC circuits involves several key concepts:

• Reactance: This is the opposition to current flow by capacitors and inductors due to the storage and release of energy in electric or magnetic fields. Unlike resistance, reactance depends on the frequency of the AC source.
• Impedance (Z): In circuits with resistors, inductors, and capacitors, the total opposition to current flow is called impedance. It is a complex quantity that includes both resistance (real part) and reactance (imaginary part). For a series RLC circuit, \(Z = R + j(X_L - X_C)\). The magnitude is \(|Z| = \sqrt{R^2 + (X_L - X_C)^2}\).
• RMS Values: AC voltage and current are often expressed as Root Mean Square (RMS) values, which represent the equivalent DC value that would dissipate the same average power in a resistor. For sinusoidal waveforms, \(V_{RMS} = \frac{V_0}{\sqrt{2}}\) and \(I_{RMS} = \frac{I_0}{\sqrt{2}}\).
• Phase Angle (\(\phi\)): The phase angle is the difference in phase between the voltage and current in an AC circuit. It is determined by the relative values of resistance and reactance. For an impedance \(Z = R + jX\), the phase angle is \(\phi = \arctan(\frac{X}{R})\), where \(X = X_L - X_C\). In a pure capacitor circuit, \(R=0\) and \(X = -X_C\), leading to a phase angle of \(\arctan(\frac{-X_C}{0}) = -\frac{\pi}{2}\), meaning voltage lags current by \(\pi/2\), or current leads voltage by \(\pi/2\).
• Power Factor: The power factor is \(\cos(\phi)\), where \(\phi\) is the phase angle between voltage and current. It indicates how effectively electrical power is being converted into useful work. For pure capacitive circuits, the phase angle is 90 degrees, so the power factor is \(\cos(90^\circ) = 0\). This means the average power consumed by a pure capacitor is zero.