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DC Circuits Containing Resistors and Capacitors

When you use a flash camera, it takes a few seconds to charge the capacitor that powers the flash. The light flash discharges the capacitor in a tiny fraction of a second. Why does charging take longer than discharging? This question and a number of other phenomena that involve charging and discharging capacitors are discussed in this module.

RC Circuits

An RC circuit is one containing a resistor R and a capacitor C. The capacitor is an electrical component that stores electric charge.

Figure 21.38 shows a simple RC circuit that employs a DC (direct current) voltage source. The capacitor is initially uncharged. As soon as the switch is closed, current flows to and from the initially uncharged capacitor. As charge increases on the capacitor plates, there is increasing opposition to the flow of charge by the repulsion of like charges on each plate.

In terms of voltage, this is because voltage across the capacitor is given by $V_{c} = Q / C V_{c} = Q / C size 12{V rSub { size 8{c} } =Q/C} {}$ , where $Q Q size 12{Q} {}$ is the amount of charge stored on each plate and $C C size 12{C} {}$ is the capacitance. This voltage opposes the battery, growing from zero to the maximum emf when fully charged. The current thus decreases from its initial value of $I_{0} = \frac{emf}{R} I_{0} = \frac{emf}{R} size 12{I rSub { size 8{0} } = { {"emf"} over {R} } } {}$ to zero as the voltage on the capacitor reaches the same value as the emf. When there is no current, there is no $IR IR size 12{ ital "IR"} {}$ drop, and so the voltage on the capacitor must then equal the emf of the voltage source. This can also be explained with Kirchhoff’s second rule (the loop rule), discussed in Kirchhoff’s Rules, which says that the algebraic sum of changes in potential around any closed loop must be zero.

The initial current is $I_{0} = \frac{emf}{R} I_{0} = \frac{emf}{R} size 12{I rSub { size 8{0} } = { {"emf"} over {R} } } {}$ , because all of the $IR IR size 12{ ital "IR"} {}$ drop is in the resistance. Therefore, the smaller the resistance, the faster a given capacitor will be charged. Note that the internal resistance of the voltage source is included in $R R size 12{R} {}$ , as are the resistances of the capacitor and the connecting wires. In the flash camera scenario above, when the batteries powering the camera begin to wear out, their internal resistance rises, reducing the current and lengthening the time it takes to get ready for the next flash.

Part a shows a circuit with a cell of e m f script E connected in series with a resistor R, a capacitor C, and a switch to close the circuit. The current is shown flowing in a clockwise direction. The capacitor plates are shown to have a charge positive q and negative q respectively. Part b shows a graph of the variation of voltage of the capacitor with time. The voltage is plotted along the vertical axis and the time is along the horizontal axis. The graph shows a smooth upward rising curve which approaches a maximum and flattens out at maximum voltage equal to e m f script E over time.

Figure 21.38 (a) An

RC RC size 12{ ital "RC"} {}

circuit with an initially uncharged capacitor. Current flows in the direction shown (opposite of electron flow) as soon as the switch is closed. Mutual repulsion of like charges in the capacitor progressively slows the flow as the capacitor is charged, stopping the current when the capacitor is fully charged and

Q = C \cdot emf Q = C \cdot emf size 12{Q=C cdot "emf"} {}

. (b) A graph of voltage across the capacitor versus time, with the switch closing at time

t = 0 t = 0 size 12{t=0} {}

. (Note that in the two parts of the figure, the capital script E stands for emf,

q q

stands for the charge stored on the capacitor, and

τ τ

is the

RC RC

time constant.)

Voltage on the capacitor is initially zero and rises rapidly at first, since the initial current is a maximum. Figure 21.38(b) shows a graph of capacitor voltage versus time ( $t t size 12{t} {}$ ) starting when the switch is closed at $t = 0 t = 0 size 12{t=0} {}$ . The voltage approaches emf asymptotically, since the closer it gets to emf the less current flows. The equation for voltage versus time when charging a capacitor $C C size 12{C} {}$ through a resistor $R R size 12{R} {}$ , derived using calculus, is

V = emf (1 - e^{- t / RC}) (charging), V = emf (1 - e^{- t / RC}) (charging), size 12{V="emf" \( 1 - e rSup { size 8{ - t/ ital "RC"} }  \) } {}

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where $V V size 12{V} {}$ is the voltage across the capacitor, emf is equal to the emf of the DC voltage source, and the exponential e = 2.718 … is the base of the natural logarithm. Note that the units of $RC RC size 12{ ital "RC"} {}$ are seconds. We define

τ = RC, τ = RC, size 12{τ= ital "RC"} {}

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where $τ τ size 12{τ} {}$ (the Greek letter tau) is called the time constant for an $RC RC size 12{ ital "RC"} {}$ circuit. As noted before, a small resistance $R R size 12{R} {}$ allows the capacitor to charge faster. This is reasonable, since a larger current flows through a smaller resistance. It is also reasonable that the smaller the capacitor $C C size 12{C} {}$ , the less time needed to charge it. Both factors are contained in $τ = RC τ = RC size 12{τ= ital "RC"} {}$ .

More quantitatively, consider what happens when $t = τ = RC t = τ = RC size 12{t=τ= ital "RC"} {}$ . Then the voltage on the capacitor is

V = emf (1 - e^{- 1}) = emf (1 - 0 . 368) = 0 . 632 \cdot emf . V = emf (1 - e^{- 1}) = emf (1 - 0 . 368) = 0 . 632 \cdot emf . size 12{V="emf" left (1 - e rSup { size 8{ - 1} }  right )="emf" left (1 - 0 "." "368" right )=0 "." "632" cdot "emf"} {}

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This means that in the time $τ = RC τ = RC size 12{τ= ital "RC"} {}$ , the voltage rises to 0.632 of its final value. The voltage will rise 0.632 of the remainder in the next time $τ τ size 12{τ} {}$ . It is a characteristic of the exponential function that the final value is never reached, but 0.632 of the remainder to that value is achieved in every time, $τ τ size 12{τ} {}$ . In just a few multiples of the time constant $τ τ size 12{τ} {}$ , then, the final value is very nearly achieved, as the graph in Figure 21.38(b) illustrates.

Discharging a Capacitor

Discharging a capacitor through a resistor proceeds in a similar fashion, as Figure 21.39 illustrates. Initially, the current is $I_{0} = \frac{V_{0}}{R} I_{0} = \frac{V_{0}}{R} size 12{I rSub { size 8{0} } = { {V rSub { size 8{0} } } over {R} } } {}$ , driven by the initial voltage $V_{0} V_{0} size 12{V rSub { size 8{0} } } {}$ on the capacitor. As the voltage decreases, the current and hence the rate of discharge decreases, implying another exponential formula for $V V size 12{V} {}$ . Using calculus, the voltage $V V size 12{V} {}$ on a capacitor $C C size 12{C} {}$ being discharged through a resistor $R R size 12{R} {}$ is found to be

V = V (discharging). V = V (discharging). size 12{V=`V"" lSub { size 8{0} } `e rSup { size 8{ - t/ ital "RC"} } } {}

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Part a shows a circuit with a capacitor C connected in series with a resistor R and a switch to close the circuit. The current is shown flowing in a counterclockwise direction. The capacitor plates are shown to have a charge positive q and negative q respectively. Part b shows a graph of the variation of voltage across the capacitor with time. The voltage is plotted along the vertical axis and the time is along the horizontal axis. The graph shows a smooth downward falling curve which approaches a minimum and flattens out close to zero over time.

Figure 21.39 (a) Closing the switch discharges the capacitor

C C size 12{C} {}

through the resistor

R R size 12{R} {}

. Mutual repulsion of like charges on each plate drives the current. (b) A graph of voltage across the capacitor versus time, with

V = V_{0} V = V_{0} size 12{V=V rSub { size 8{0} } } {}

t = 0 t = 0

. The voltage decreases exponentially, falling a fixed fraction of the way to zero in each subsequent time constant

τ τ size 12{τ} {}

The graph in Figure 21.39(b) is an example of this exponential decay. Again, the time constant is $τ = RC τ = RC size 12{τ= ital "RC"} {}$ . A small resistance $R R size 12{R} {}$ allows the capacitor to discharge in a small time, since the current is larger. Similarly, a small capacitance requires less time to discharge, since less charge is stored. In the first time interval $τ = RC τ = RC size 12{τ= ital "RC"} {}$ after the switch is closed, the voltage falls to 0.368 of its initial value, since $V = V_{0} \cdot e^{- 1} = 0 . 368 V_{0} V = V_{0} \cdot e^{- 1} = 0 . 368 V_{0} size 12{V=V rSub { size 8{0} } cdot e rSup { size 8{ - 1} } =0 "." "368"V rSub { size 8{0} } } {}$ .

During each successive time $τ τ size 12{τ} {}$ , the voltage falls to 0.368 of its preceding value. In a few multiples of $τ τ size 12{τ} {}$ , the voltage becomes very close to zero, as indicated by the graph in Figure 21.39(b).

Now we can explain why the flash camera in our scenario takes so much longer to charge than discharge; the resistance while charging is significantly greater than while discharging. The internal resistance of the battery accounts for most of the resistance while charging. As the battery ages, the increasing internal resistance makes the charging process even slower. (You may have noticed this.)

The flash discharge is through a low-resistance ionized gas in the flash tube and proceeds very rapidly. Flash photographs, such as in Figure 21.40, can capture a brief instant of a rapid motion because the flash can be less than a microsecond in duration. Such flashes can be made extremely intense.

During World War II, nighttime reconnaissance photographs were made from the air with a single flash illuminating more than a square kilometer of enemy territory. The brevity of the flash eliminated blurring due to the surveillance aircraft’s motion. Today, an important use of intense flash lamps is to pump energy into a laser. The short intense flash can rapidly energize a laser and allow it to reemit the energy in another form.

In the photograph, details of the fast beating wings of the hummingbird taking nectar from a flower have been caught in focus, instead of the blur that our eyes would see in real time.

Figure 21.40 This stop-motion photograph of a rufous hummingbird (Selasphorus rufus) feeding on a flower was obtained with an extremely brief and intense flash of light powered by the discharge of a capacitor through a gas. (credit: Dean E. Biggins, U.S. Fish and Wildlife Service)

Example 21.6

Integrated Concept Problem: Calculating Capacitor Size—Strobe Lights

High-speed flash photography was pioneered by Doc Edgerton in the 1930s, while he was a professor of electrical engineering at MIT. You might have seen examples of his work in the amazing shots of hummingbirds in motion, a drop of milk splattering on a table, or a bullet penetrating an apple (see Figure 21.40). To stop the motion and capture these pictures, one needs a high-intensity, very short pulsed flash, as mentioned earlier in this module.

Suppose one wished to capture the picture of a bullet (moving at $5.0 \times 10^{2} m/s 5.0 \times 10^{2}$ ) that was passing through an apple. The duration of the flash is related to the $RC RC size 12{ ital "RC"} {}$ time constant, $τ τ size 12{τ} {}$ . What size capacitor would one need in the $RC RC size 12{ ital "RC"} {}$ circuit to succeed, if the resistance of the flash tube was $10.0 Ω 10.0 Ω size 12{"10" %OMEGA } {}$ ? Assume the apple is a sphere with a diameter of $8.0 \times 10^{–2} m. 8.0 \times 10^{–2}$

Strategy

We begin by identifying the physical principles involved. This example deals with the strobe light, as discussed above. Figure 21.39 shows the circuit for this probe. The characteristic time $τ τ size 12{τ} {}$ of the strobe is given as $τ = RC τ = RC size 12{τ= ital "RC"} {}$ .

Solution

We wish to find $C C size 12{C} {}$ , but we don’t know $τ τ size 12{τ} {}$ . We want the flash to be on only while the bullet traverses the apple. So we need to use the kinematic equations that describe the relationship between distance $x x size 12{x} {}$ , velocity $v v size 12{v} {}$ , and time $t t size 12{t} {}$ :

x = vt or t = \frac{x}{v} . x = vt size 12{t= {  {x}  over  {v} } } {}

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The bullet’s velocity is given as $5.0 \times 10^{2} m/s 5.0 \times 10^{2}$ , and the distance $x x size 12{x} {}$ is $8.0 \times 10^{–2} m. 8.0 \times 10^{–2}$ The traverse time, then, is

t = \frac{x}{v} = \frac{8.0 \times 10^{–2} m}{5.0 \times 10^{2} m/s} = 1 . 6 \times 10^{- 4} s. t = \frac{x}{v} = \frac{8.0 \times 10^{–2}}{5.0 \times 10^{2}} = 1 . 6 \times 10^{- 4} size 12{t= {  {x}  over  {v} } = {  {0 "." "08"" m"}  over  {"500 m/s"} } =1 "." 6 times "10" rSup { size 8{ - 4} } " s"} {}

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We set this value for the crossing time $t t size 12{t} {}$ equal to $τ τ size 12{τ} {}$ . Therefore,

C = \frac{t}{R} = \frac{1.6 \times 10^{- 4} s}{10.0 Ω} = 16 μF. C = \frac{t}{R} = \frac{1.6 \times 10^{- 4}}{10.0 Ω} = 16 size 12{C= {  {t}  over  {R} } = {  { left (1 "." 6´"10" rSup { size 8{-4} }  right )}  over  {"10"} } ="16" μF} {}

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(Note: Capacitance $C C size 12{C} {}$ is typically measured in farads, $F F$ , defined as Coulombs per volt. From the equation, we see that $C C size 12{C} {}$ can also be stated in units of seconds per ohm.)

Discussion

The flash interval of $160 μs 160 μs size 12{"160" ms} {}$ (the traverse time of the bullet) is relatively easy to obtain today. Strobe lights have opened up new worlds from science to entertainment. The information from the picture of the apple and bullet was used in the Warren Commission Report on the assassination of President John F. Kennedy in 1963 to confirm that only one bullet was fired.

RC Circuits for Timing

$RC RC size 12{ ital "RC"} {}$ circuits are commonly used for timing purposes. A mundane example of this is found in the ubiquitous intermittent wiper systems of modern cars. The time between wipes is varied by adjusting the resistance in an $RC RC size 12{ ital "RC"} {}$ circuit. Another example of an $RC RC size 12{ ital "RC"} {}$ circuit is found in novelty jewelry, Halloween costumes, and various toys that have battery-powered flashing lights. (See Figure 21.41 for a timing circuit.)

A more crucial use of $RC RC size 12{ ital "RC"} {}$ circuits for timing purposes is in the artificial pacemaker, used to control heart rate. The heart rate is normally controlled by electrical signals generated by the sino-atrial (SA) node, which is on the wall of the right atrium chamber. This causes the muscles to contract and pump blood. Sometimes the heart rhythm is abnormal and the heartbeat is too high or too low.

The artificial pacemaker is inserted near the heart to provide electrical signals to the heart when needed with the appropriate time constant. Pacemakers have sensors that detect body motion and breathing to increase the heart rate during exercise to meet the body’s increased needs for blood and oxygen.

Part a shows a charging circuit containing cell of e m f script E connected to a resistor R and capacitor C and a closed switch to complete the circuit. The current is shown to flow clockwise through this arm of the circuit alone. A bulb of high resistance R is connected across the capacitor. Part b shows a discharging circuit containing a cell of e m f script E connected to a resistor R and capacitor C and a closed switch to complete the circuit. A bulb of low resistance R is connected across the capacitor. Current flows clockwise through the arm containing the capacitor and the low resistance bulb. Part c is a graph showing variation of voltage verses time for the bulb in above circuit. The voltage is plotted along the vertical axis and the time is plotted along the horizontal axis. The curve has a smooth rise from the origin, reaches a plateau at threshold value of voltage where it begins to drop and rise as a small sawtooth wave with maxima lying along the threshold line.

Figure 21.41 (a) The lamp in this

RC RC size 12{ ital "RC"} {}

circuit ordinarily has a very high resistance, so that the battery charges the capacitor as if the lamp were not there. When the voltage reaches a threshold value, a current flows through the lamp that dramatically reduces its resistance, and the capacitor discharges through the lamp as if the battery and charging resistor were not there. Once discharged, the process starts again, with the flash period determined by the

RC RC size 12{ ital "RC"} {}

constant

τ τ size 12{τ} {}

. (b) A graph of voltage versus time for this circuit.

Example 21.7

Calculating Time: RC Circuit in a Heart Defibrillator

A heart defibrillator is used to resuscitate an accident victim by discharging a capacitor through the trunk of her body. A simplified version of the circuit is seen in Figure 21.39. (a) What is the time constant if an $8.00-μF 8.00-μF size 12{8 "." "00"-mF} {}$ capacitor is used and the path resistance through her body is $1.00 \times 10^{3} Ω 1.00 \times 10^{3}$ ? (b) If the initial voltage is 10.0 kV, how long does it take to decline to $5.00 \times 10^{2} V 5.00 \times 10^{2}$ ?

Strategy

Since the resistance and capacitance are given, it is straightforward to multiply them to give the time constant asked for in part (a). To find the time for the voltage to decline to $5.00 \times 10^{2} V 5.00 \times 10^{2}$ , we repeatedly multiply the initial voltage by 0.368 until a voltage less than or equal to $5.00 \times 10^{2} V 5.00 \times 10^{2}$ is obtained. Each multiplication corresponds to a time of $τ τ size 12{τ} {}$ seconds.

Solution for (a)

The time constant $τ τ size 12{τ} {}$ is given by the equation $τ = RC τ = RC size 12{τ= ital "RC"} {}$ . Entering the given values for resistance and capacitance (and remembering that units for a farad can be expressed as $s / Ω s / Ω size 12{s/ %OMEGA } {}$ ) gives

τ = RC = (1.00 \times 10^{3} Ω) (8.00 μF) = 8 . 00 ms. τ = RC = (1.00 \times 10^{3} size 12{τ= ital "RC"= \( "1000"  %OMEGA  \)  \( 8 "." "00" μF \) =8 "." "00"" ms"} {}

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Solution for (b)

In the first 8.00 ms, the voltage (10.0 kV) declines to 0.368 of its initial value. That is:

V = 0 . 368 V_{0} = 3.680 \times 10^{3} V at t = 8 . 00 ms. V = 0 . 368 V_{0} = 3.680 \times 10^{3} t = 8 . 00 size 12{t=8 "." "00"" ms"} {}

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(Notice that we carry an extra digit for each intermediate calculation.) After another 8.00 ms, we multiply by 0.368 again, and the voltage is

\begin{array}{l} V' & = & 0.368 V \\ = & (0.368) (3.680 \times 10^{3} V) \\ = & 1.354 \times 10^{3} V at t = 16.0 ms. \end{array} \begin{array}{l} V' & = & 0.368 V \\ = (0.368) (3.680 \times 10^{3}) \\ = 1.354 \times 10^{3} \end{array}

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Similarly, after another 8.00 ms, the voltage is

\begin{array}{l} V ′′ & = & 0.368 V' = (0.368) (1.354 \times 10^{3} V) \\ = & 498 V at t = 24 .0 ms. \end{array} \begin{array}{l} V ′′ & = & 0.368 V' = (0.368) (1.354 \times 10^{3} \\ = 498 V at \end{array}

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Discussion

So after only 24.0 ms, the voltage is down to 498 V, or 4.98% of its original value.Such brief times are useful in heart defibrillation, because the brief but intense current causes a brief but effective contraction of the heart. The actual circuit in a heart defibrillator is slightly more complex than the one in Figure 21.39, to compensate for magnetic and AC effects that will be covered in Magnetism.

Check Your Understanding

When is the potential difference across a capacitor an emf?

Show solution

Solution

Only when the current being drawn from or put into the capacitor is zero. Capacitors, like batteries, have internal resistance, so their output voltage is not an emf unless current is zero. This is difficult to measure in practice so we refer to a capacitor’s voltage rather than its emf. But the source of potential difference in a capacitor is fundamental and it is an emf.

PhET Explorations

Circuit Construction Kit (DC only)

An electronics kit in your computer! Build circuits with resistors, light bulbs, batteries, and switches. Take measurements with the realistic ammeter and voltmeter. View the circuit as a schematic diagram, or switch to a life-like view.

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Figure 21.42