TWU Physics Laboratory | RC Time Constants

1 Last revised January 2021

RC Time Constants

Introduction

In most experiments with electric circuits we

observe what happens at an appreciable time after

a switch has been closed or opened. In this

experiment, observations will be made of the

behavior of the circuit immediately after a switch is

opened or closed.

When a DC power source is connected across a

capacitor the capacitor becomes charged. The

charging is not instantaneous, but follows an

exponential behavior characteristic of the

capacitance of the capacitor. When the switch S in

Figure 1 is closed to complete the circuit, the

charging starts. At any instant of time after the start

of the charging process, the voltage which causes current to flow in the circuits is called an active

voltage and it is equal to the source voltage Vs minus the voltage Vc across the capacitor at the

instant. Thus, the active voltage VA = Vs –Vc.

When charging is complete, Vc is equal to Vs and

hence the active voltage is zero. Thus no current

passes through the circuit after the charging process

is over. The circuit is then equivalent to two

batteries of the same output connected opposite to

each other. The behavior of Vc and the current I

passing through the circuit as a function of time is

shown in figure 2.

When a charged capacitor is disconnected from the

battery and then connected across a resistor R, it

becomes discharged. The current flows through R

and discharging follows an exponentially decaying

behavior which is characteristic of R and C. In this

experiment we will study the time behavior of the

charging and discharging of a capacitor in an RC

circuit.

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Theory

Discharging

If a capacitor is connected to a resistor R by closing a switch S, as shown in Figure 3, the positive

and negative charges ±Q stored on the plates are no longer prevented from neutralizing each other.

A current will flow in the resistor until the energy stored in the capacitor is dissipated in the

resistor. Let us assume that at time t = 0 we close the switch. The potential difference across the

capacitor is:

𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟏: 𝑉𝑐 =𝑄

𝐶

where V and Q are functions of t. We want to know how fast the capacitor will discharge. Let us

consider the situation some time t after the switch has been closed. We denote by Q (t) the charge

left on the capacitor plates and by I(t) the current through the resistor R. If we consider just the

region enclosed by a surface S as shown in Figure 4, the law of conservation of charge requires

that

𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟐: −𝑑𝑄

𝑑𝑡= 𝐼(𝑡)

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Since the only current anywhere on the surface S is through the resistor, then Ohm’s Law tells us

the potential difference

𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟑: 𝐼(𝑡)𝑅 = 𝑉𝑐 (𝑡)

Substituting the left-hand side of Equation 2 and the right-hand side of Equation 1 into this yields

𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟒: −𝑑𝑄

𝑑𝑡𝑅 =

𝑄(𝑡)

𝐶

Dividing both sides gives the differential equation

𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟓: −𝑑𝑄

𝑑𝑡=

𝑄(𝑡)

𝑅𝐶

The solution to which is

𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟔: 𝑄(𝑡) = 𝑄0 𝑒−𝑡𝑅𝐶

where Q0 is the charge at t = 0.

This can also be written as

𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟕: 𝐼(𝑡) = 𝐼0𝑒−𝑡𝑅𝐶

where I0 is the current at t = 0.

Or

𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟖: (𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑖𝑛𝑔): 𝑉𝑐 (𝑡) = 𝑉0𝑒−𝑡𝑅𝐶

where V0 is the voltage across the capacitor at t = 0.

It can be seen from this equation that after time  = R × C the voltage V will drop by a factor of

1/e. This amount of time,  = RC, is called the time constant of the circuit.

Charging

For charging, the time behavior of the voltage Vc across the capacitor is governed by the

following equation:

𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟗: (𝑐ℎ𝑎𝑟𝑔𝑖𝑛𝑔): 𝑉𝑐 (𝑡) = 𝑉𝑆 (1 − 𝑒−𝑡𝑅𝐶 )

where VS is the source voltage used to charge the capacitor. It is also the limiting value of Vc as

time increases. Equations 8 and 9 for the charging and discharging processes give most of the

information needed to understand the process. One should study them carefully.

𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟏𝟎: (𝑐ℎ𝑎𝑟𝑔𝑖𝑛𝑔): 𝐼(𝑡) = 𝑉𝑆𝑅

𝑒−𝑡𝑅𝐶

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Pre- Lab

1. A 6 µF capacitor is initially charged to 100 V and then discharged across a 600 ohm resistor.

a. What is the initial charge on the capacitor?

b. What is the time constant of the circuit?

c. How much charge is on the capacitor after 6 ms?

2. A 10 mega ohm resistor is connected in series with a 5 µF capacitor and 12 V battery. The capacitor is initially uncharged. After a time equal to one time constant find:

a. The charge on the capacitor

b. The current

3. Graph V of a capacitor when it is charging and discharging as a function of time for a RC circuit. What should this graph look like? No numbers just a qualitative sketch.

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Apparatus

• Regulated DC power supply

• Two switches

• Digital multimeter with 10 MΩ internal resistance

• Electric wires

• Decade Capacitor Box

• Stopwatch- use phones

20 V

20 V

1 µF

1 µF

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Procedure

Part I: charging time constant

1. We will use the circuit in Figure 5 for studying the charging process of a 1 µF capacitor. The

internal resistance of the digital multimeter is 10 mega ohms. This the time constant () of the

circuit is τ = R × C = 10 MΩ × 1µF = 10 s.:

Hence, we want to observe the voltage V across the capacitor at the end of one time constant

(10 seconds), two time constants (20 s), etc. Connect the circuits as shown in Figure 5.

2. Close switch S2. Then close switch S1. Increase the voltage Vs of the power supply until the

multimeter reads 20 V. Now you have set the voltage at 20 V exactly.

3. Open the switch S2 and at the same instant, start your digital clock. When the clock reads 10

seconds, read the voltage from the multimeter. Record the voltage every 10 seconds for 60

seconds. Remember that in this circuit the multimeter reads the active voltage VA.

4. Open S1 and close S2. You have just discharged the capacitor.

5. Close S2 and then S1. Make sure the meter reads 20 V.

6. Repeat steps 2 through 5 for a total of 3 trials. Calculate the average.

t

(s)

Active voltage 𝑉𝐴(V) Experimental Ave, Vc (V)

Computed Vc

(V)

Percent error

Trial 1 Trial 2 Trial 3 Avg 20 V – 𝑉𝑎𝑣 𝑉0(1 − 𝑒−𝑡𝑅𝐶 )

10

20

30

40

50

60

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Part II: Discharging Time constant

1. Connect the circuit as shown in Figure 6. The theoretical  again is 10 s. You want to first

charge the capacitor and then discharge it through the multimeter (R = 10 mega ohms)

2. Close S2 and then S1. Set Vs to 20 V and wait until the multimeter reads 20 V.

3. Open S1 and at the same instant start the digital clock. The capacitor has started discharging

through the 10 mega ohm resistor. Take readings every 10 s.

4. Repeats steps 2 and 3 for a total of 3 trials. Remember that this time the meter directly reads

the voltage Vc across the capacitor.

t

(s)

Vc (V) Computed Vc Percent error

Trial 1 Trial 2 Trial 3 Avg Vc 𝑉𝑐 (𝑡) = 𝑉0𝑒

−𝑡𝑅𝐶

10

20

30

40

50

60

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Questions

1. A way of discharging a fully charged capacitor is to short the two plates with an electrical wire. Explain why this is useful way.

2. To graphically represent the value of , based on your data, consider equation 8:

𝑉𝑐 (𝑡) = 𝑉0𝑒−𝑡

𝑅𝐶

Because RC = , this is equivalent to

𝑉𝑐 (𝑡) = 𝑉0𝑒−𝑡

𝜏 .

Taking the natural logarithm of both sides yields

ln(𝑉𝑐 ) = (−1

𝜏) 𝑡 + ln (𝑉0)

(Recall that ln(𝑒 𝑥 ) = 𝑥 and ln(𝑎𝑏) = ln(𝑎) + ln (𝑏)).

Observe that this has the form of a linear equation 𝑦 = 𝑚𝑥 + 𝑏 for which y = ln(Vc), x = t,

the slope m = (−1

𝜏), and the intercept b = ln(V0).

Graph the natural log of your measured values Vc from part II versus t and draw a line of best fit

for the plotted points. Determine the slope m of the line you drew and use it to make an estimate

of the time constant by calculating τest = -1/m. Compare this estimated value to the theoretical

value of τ = 10 s.