The charge and discharge time of an RC circuit is determined by the time constant. With large amounts of capacitance and high resistance the time constant can be several hours, so beware of large capacitors in electronics equipment. In an RC circuit assuming there is no initial charge on the capacitor it takes a time of \(R \times C\) seconds—first time constant (\(\tau\))— to charge a capacitor to 63.2% of its final value.

Note: Charge voltage values are derived from mathemetical constant \(e\). \(63.2\% \approx 1-e^{-1}\) and \(36.8\% \approx e^{-1}\).

A capacitor is typically considered fully charged after 5 time constants (\(5\tau\)).

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This is an electronics definition that you'll have to memorize. For what it's worth, the correct answer is the only one without the word "discharge" in it.

Further reading: https://en.wikipedia.org/wiki/RC_time_constant

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One time constant is the time it takes for a charged capacitor in an RC circuit to discharge to 36.8% of its initial value. In 2 time constants the capacitor will discharge to .368 x .368 = .135 or 13.5%

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Time constant TC or \(\tau\) is given by:

\[TC \:[\text{seconds, s}] = R\:[\text{ohms}, \Omega] \times C\:[\text{farads, F}]\]

This circuit contains two \(220\ \mu\text{F}\) capacitors and two \(1\text{ M}\Omega\) resistors, all in parallel.

The first thing to remember, capacitance in parallel INCREASES and resistance in parallel DECREASES.

For resistors in parallel:

\begin{align} \frac{1}{R_{\text{total}}} &= \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n} \end{align}

So, keeping consistent units in \(\text{M}\Omega\): \begin{align} R_{\text{total}} &= \frac{1}{ \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n}}\\ &= \frac{1}{ \frac{1}{1\text{ M}\Omega} + \frac{1}{1\text{ M}\Omega}}\\ &= \frac{1}{ \frac{2}{1\text{ M}\Omega} }\\ &= \frac{1\text{ M}\Omega}{2}\\ &= 0.5\text{ M}\Omega\\ \end{align}

For capacitors in parallel:

\begin{align} C_{\text{total}} &= C_1 + C_2 + \ldots + C_n\\ &= 220\ \mu\text{F} + 220\ \mu\text{F}\\ &= 440\ \mu\text{F}\\ \end{align}

For the time constant:

\begin{align} \tau = TC &= R_{\text{total}} \times C_{\text{total}}\\ &= 0.5\text{ M}\Omega \times 440\ \mu\text{F}\\ \end{align}

Replace the SI prefixes Mega \(\left(10^6\right)\) and \(\mu\) (micro) \(\left(10^{-6}\right)\):

\begin{align} \tau = TC &= \left(0.5 \times 10^6\ \Omega\right) \times \left(440 \times 10^{-6}\ \text{F}\right)\\ &= 0.5 \times 440\\ &= 220\text{ seconds} \end{align}

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TEST TIP: For questions E5B04, E5B05, and E5B06 the correct answer is the one equal or closest to the microfarads value in the question.

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First, find HOW MANY Time Constants:

#TC = Discharge (or Charge) / Beginning Charge

#TC = 7.36 / 20

#TC = .368 (or 36.8%)

Now look up the value in the following table:

#Time Constant | Charge | Discharge

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1 TC | 63.2% | 36.8%

2 TC | 86.5% | 13.5%

3 TC | 95% | 5%

The answer is "1 Time Constant"...

NOW, figure out the VALUE of the Time Constant for this circuit:

TC = R (ohms) * C (farads)

2x10^6ohms * 0.01x10^-6farads

(SIMPLIFY by canceling x10^6 and x10^-6)

2 * 0.01 = 0.02

The answer is "0.02 seconds".

Now for the final answer:

We have 1 Time Constant and .02 seconds...

1 * .02 seconds = .02 seconds

For a more in depth discussion read

[http://hyperphysics.phy- astr.gsu.edu/hbase/electric/capchg.html](http://hyperphysics.phy- astr.gsu.edu/hbase/electric/capchg.html)

[http://www.physicsforums.com/showthread.php?t=145203](http://www.physicsforum s.com/showthread.php?t=145203)

[http://forum.allaboutcircuits.com/showthread.php?t=6630](http://forum.allabou tcircuits.com/showthread.php?t=6630)

The second link explains what e is sort of.

TEST TIP: For questions E5B04, E5B05, and E5B06 the correct answer is the one equal or closest to the microfarad value in the question.

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One time constant (=R*C) is the time in seconds it takes for a charged capacitor in an RC circuit to discharge to 36.8% of its initial value.

Since 800 x .368 =294 the capacitor has been discharging for 1 time constant which = .000450f x 1,000,000 ohms = 450 seconds.

TEST TIP: For questions E5B04, E5B05, and E5B06 the correct answer is the one equal or closest to the microfarad value in the question.

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Given: \begin{align} X_C &= 500\ \Omega\\ X_L &= 250\ \Omega\\ R &= 1\text{ k}\Omega = 1000\ \Omega \end{align}

We can calculate the phase angle using: \[\text{phase angle }\phi = \arctan{ \Big ( \frac{X_L - X_C}{R} \Big ) }\]

Pay attention to the sign of the angle:

• If the angle is negative then the voltage is lagging.
• If the angle is positive then the voltage is leading.

Plug in our given values: \begin{align} \text{phase angle }\phi &= \arctan{ \Big ( \frac{250\ \Omega - 500\ \Omega}{1000\ \Omega} \Big ) }\\ &= \arctan{ \Big ( \frac{ -250\ \Omega }{1000\ \Omega} \Big ) }\\ &= \arctan{( -0.25 )}\\ &\approx -14^\circ\\ \end{align}

WARNING: If you are using a calculator to calculate the \(\arctan{(\ldots)}\), make sure the calculator mode is set to degrees and not radians. Using the wrong mode will give you the wrong answer!

------ or ------

The total reactance \(X\) in the circuit is \(X_L + X_C = 250\ \Omega +(- 500\ \Omega) = -250\ \Omega\). (capacitive reactance is regarded as negative when we are doing calculations on imaginary numbers).

Since we now know that the circuit reactance is capacitive we can immediately say that the voltage lags the current.

To calculate phase angle we use

\[\tan{ (\text{phase angle }\phi) } = \frac{X}{R}\]

Where: \begin{align} X &= \text{total reactance} = X_L - X_C\\ R &= \text{total series resistance} \end{align}

So \begin{align} \tan{ (\text{phase angle }\phi)} &= \frac{-250\ \Omega}{1000\ \Omega}\\ &= -0.25 \end{align}

To get the phase angle we now use the inverse tangent function of a calculator.

\[\tan^{-1}(-0.25) \approx -14^{\circ}\]

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Test Tip: ALL the answers to these questions are 14 degrees. For the test, remember: if \(X_C > X_L\), voltage lags.

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\[\text{phase angle}\:\phi=\tan ^{-1}\left(\frac{X_{L}-X_{C}}{R}\right)\]

• If the phase angle is negative then the voltage is lagging.

• If the phase angle is positive then the voltage is leading.

\begin{align} \text{phase angle}\:\phi&=\tan ^{-1}\left(\frac{75-100}{100}\right)\\ &=\tan ^{-1}\left(-0.25\right)\\ &=-14.0362435^{\circ}\\ &\approx-14^{\circ} \end{align}

If you are using a calculator make sure it is in degrees and not radians.

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This is the polar coordinates graph we used for the questions where we find impedance point when given frequency and either L or C.

The X axis is resistance and the Y axis is L-C. Draw a line from the origin through the point. The angle is smaller as the point is further from the origin, i.e. larger R. The angle is smaller as the L-C is smaller.

You should be able imagine or draw the graph and estimate the angle. No trigonometry calculation needed.

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Memory tricks:

• ALL the answers to these questions are 14 degrees.
• For the test, remember: if \(X_C > X_L\), voltage lags.
• "If L is Less, it Lags!"
• "eLi the iCe man" is a mnemonic to remember that in inductors the voltage (e) leads the current (i) and in capacitors current (i) leads the voltage (e). In this circuit, the capacitor value is larger than the inductor so we know the voltage is lagging the current.

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A helpful way to remember the relationship of current and voltage in inductors and capacitors is ELI the ICE man:

In a capacitor, symbol C, current (I) leads voltage (E), by 90 degrees. In an inductor, symbol L, voltage (E) leads current (I), by 90 degrees.

Another helpful way to remember this is, capacitor and current both start with a C. And for the inductor, it is opposite, from the capacitor.

To understand what is happening, consider an uncharged capacitor which has no voltage across it. The voltage appears as the charge flows into the capacitor. (Flowing charge is current.) So therefore, current leads voltage. Inductors act opposite as a change in voltage changes the current flow or voltage leads current.

Hint: Current leads Voltage in the question.

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A helpful way to remember the relationship of current and voltage in inductors and capacitors is ELI the ICE man:

In a capacitor, symbol C, current (I) leads voltage (E), by 90 degrees. In an inductor, symbol L, voltage (E) leads current (I), by 90 degrees.

In this case, with an inductor, VOLTAGE (E) leads CURRENT (I).

Also CiViC acronym is helpful. Clv = Capacitor = Current leads Voltage or iVlC=inductor: Voltage leads Current.

Helpful hint: In an inductor, you have to apply voltage for current to flow, so Voltage leads Current.

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\[\text{phase angle}\:\phi=\tan ^{-1}\left(\frac{X_{L}-X_{C}}{R}\right)\]

If the angle is negative, then the voltage is lagging.

If the angle is positive, then the voltage is leading.

\begin{align} \text{phase angle}\:\phi&=\tan ^{-1}\left(\frac{50-25}{100}\right)\\ &=\tan ^{-1}\left(0.25\right)\\ &=14.0362435^{\circ}\\ &\approx14^{\circ} \end{align}

If you are using a calculator, make sure it is in degrees and not radians.

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Test Tip: ALL the answers to these questions are 14 degrees. For the test, remember: if \(X_C > X_L\), voltage lags.

If you are a track and field person, it is race 14. \(X_C\) is s Cross Country guy and \(X_L\) is the big voltage shot put guy. If \(X_C\) is going faster (higher number), Voltage is lagging. If \(X_C\) is going faster, Voltage is leading.

To remember which X goes first in the formula, think "L" for "leading" or \(X_L\) "excels".

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Phase Angle = tan^-1((XL-XC)/R)

If the angle is negative then the voltage is lagging.

If the angle is positive then the voltage is leading.

tan^-1((50-75)/100) = tan^-1(-25/100) = tan^-1(-0.25) = -14

If you are using a calculator make sure it is in Degrees and not Radians.

Test Tip: ALL the answers to these questions are 14 degrees. For the test, remember: if XC > XL, voltage lags.

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Phase Angle = tan^-1((XL-XC)/R)

If the angle is negative then the voltage is lagging.

If the angle is positive then the voltage is leading.

tan^-1((500-250)/1000) = tan^-1(250/1000) = tan^-1(0.25) = 14.04

If you are using a calculator make sure it is in Degrees and not Radians.

                               OR

The total reactance X in the circuit is XL+XC = 500 - 250 = 250 Ohms. (capacitive reactance is regarded as negative when we are doing calculations on imaginary numbers)

Since we now know that the circuit reactance is inductive we can immediately say that the voltage leads the current.

To calculate phase angle we use

Tan (Phase angle) = X / R

Where: X = total reactance = Xl - Xc R = total series resistance.

So Tan(Phase angle) = 250/1000 = 0.25

To get the Phase angle we now use the INV Tan function of a calculator.

INV Tan (0.25) = 14.04 degrees.

Test Tip: ALL the answers to these questions are 14 degrees. For the test, remember: if XC > XL, voltage lags.

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