18-19 Circuit and Capacitors

Physics 12 Honors/Regular – Notes – Circuit and Capacitors V2.0 (KWP) Page 1 of 16

For more information, refer to….

1. Giancoli - Physics: Principles with Applications (5th Edition): Chapter 17 – Electric Potential and Electric Energy; Capacitance; Chapter 18 – Electric Currents; Chapter 19 –

DC Circuits 2. Materials regarding AP Physics B & C Exams

… as well as other resources.

Electric Circuits

I : Current is amount of charge Q flowing through per second. It is measured in Amperes (A) or (C/s)

1 V = 1 J/C

V : Voltage or potential difference . It is the energy change across the device in each

coulomb of charge. It is measured in Volts (V) or (J/C).

Ohm’s Law

where R is the resistance of the device. It is measured in Ohms (Ω). …. Devices with a large resistance will have a small current flowing through it.

…. Devices with a greater potential difference V across them will have more current going

through them.

Electric Power

Power is energy produced or energy used per unit time.

𝐼 = 𝑄

𝑡 6𝑉

6𝑉

0𝑉

𝑅 0𝑉

+ −

(∆ 𝑉)

𝐼 ∝ 1

𝑅

𝑅 = 𝑉

𝐼

𝐼 = 𝑉

𝑅

𝑉 ∝ 𝐼 𝑉 = 𝐼𝑅

𝑃 = 𝐸

𝑡

It is measured in Watts (W) or (J/s).

1V = 1J/C

1J = 1VC


Since

But

So

The power of the device can be determined from the product of the potential drop across

the device and the current going through it.

Question

Solution:

(a) the reading on the voltmeter 6V (potential difference of the battery)

(b) the reading on the ammeter

(c) how many electrons flow through the resistor every second

∆ 𝐸𝑃 = 𝑞 ∆𝑉

𝑃 =𝑞 ∆𝑉

𝑡

𝑞

𝑡= 𝐼

𝑃 = 𝐼 ∆𝑉

𝑃 = 𝑉𝐼

𝑉 × 𝐼 = 𝑃

(𝐽

𝐶) × (

𝐶

𝑠) = (

𝐽

𝑠)

or P = 𝐼2𝑅 or P =𝑉2

𝑅

𝐼 = 𝑉

𝑅

𝑉 = 𝐼𝑅

voltage

Device

V

A

Voltmeter

Ammeter

Negligible potential

drop across the

ammeter (R –> 0)

Negligible current

flows through the

voltmeter (R –> ∞)

Resistivity ρ

ρ𝑇 = ρ0(1 + 𝛼(∆𝑇))

𝑅 =

𝜌𝑙

𝐴

Consider the blowing

of a light bulb.

(+) charge “hole” A 6V battery is connected to 220 Ω resistor as shown. Determine: (a) the reading on the voltmeter (b) the reading on the ammeter (c) how many electrons flow through the resistor every second

(d) the power given off by the resistor (e) the power produced by the battery assuming

negligible resistance in the wires and the ammeter. (f) the amount of hear energy produced in 5 minutes.

V

A

− +

𝑒− I

𝐼 = 𝑉

𝑅=

6 𝑉

220 Ω= 0.0273A = 27.3 mA

0.0273𝐴 = 0.0273 𝐶

𝑠 0.0273

𝐶

𝑠×

1 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛

1.60 × 10−19𝐶= 1.7 × 1017𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠/s

*Refer to Notes P.4 for

details of resistivity.


(d) the power given off by the resistor

(e) the power produced by the battery assuming negligible resistance in the wires and the

ammeter.

(f) the amount of heat energy produced in 5 minutes.

In other words, heat produced by the circuit can heat up 1 kg of water by 85.4! Wow!

Question

Calculate:

(a) the current flowing through the light bulb. (b) the current flowing through the hair dryer.

(c) the house circuit is connected to a 13A circuit breaker which will allow a maximum of

13A to flow though the main circuit. How many hair dryers can be connected to the circuit.

Solution: (a)

(b)

(c)

Only one hair dryer can be connected to the circuit at one time.

Read p.573 and p.574 on Electric Hazards.

𝑃 = 𝑉𝐼 = (6V)(0.0273A) = 0.16 W

𝑃𝑜𝑢𝑡 = 𝑃𝑖𝑛 = 0.16 W

𝐸 = 𝑃𝑡 = (0.16 W)(5 mins) (60𝑠

1 min) = 48𝐽

Bonus: Using the equation of specific heat: 𝑄 = 𝑚𝑐∆𝑇

∆𝑇 =𝑚𝑐

𝑄=

(1 kg) (4186J

kg ∙ )

49𝐽= +85.4

120𝑉

120𝑉

120𝑉 ~

120𝑉

circuit breaker

breaker

𝐷. 𝐶. 60𝑊

closed

circuit

Device

connected

in parallel

Also see: Ex 18-10 P541

1000𝑊

𝐼 = 𝑃

𝑉=

60𝑊

120V= 0.5A

𝐼 = 𝑃

𝑉=

1000𝑊

120V= 8.3A

8.3𝑥 ≤ 13 𝑥 ≤ 1


Resistivity, ρ

ρ𝑇 = ρ0(1 + 𝛼(∆𝑇))

Resistivity is a material property. It tells us how easily charge flows through the material.

e.g. Copper

Glass

Resistance, R

Equivalent Resistant, Re

(i) Series resistances

Conservation of energy: Kirchhoff’s Second Law

Sum of emf’s (due to the batteries) is equal to the sum of

potential drops (due to the resistors) around one loop

In this case,

Then with Ohm’s Law: or

p.534 constant at a certain reference

temperature

temperature

coefficient

ρ0 = 1.68 × 10−8Ωm α = 0.0068

ρ0 = 109 − 1012Ωm 𝑎𝑡 20

A

𝑙

What is its resistance, R?

𝑅 =𝜌𝑙

𝐴

𝜌

𝑅1 𝑅2 𝑅3

𝑉1 𝑉2 𝑉3

𝑉 𝐼

𝑅𝑆

𝐼 𝑉

≡

𝑒𝑚𝑓(Electromotive force),𝜀

∑ 𝜀𝑙𝑜𝑜𝑝 = ∑ 𝐼𝑅𝑙𝑜𝑜𝑝 𝑉 = 𝑉1 + 𝑉2 + 𝑉3

𝑅 =

𝑉

𝐼 𝑉 = 𝐼𝑅

𝑉 = 𝐼𝑅1 + 𝐼𝑅2 + 𝐼𝑅3

𝐼𝑅𝑆 = 𝐼𝑅1 + 𝐼𝑅2 + 𝐼𝑅3

𝑅𝑆 = 𝑅1 + 𝑅2 + 𝑅3


(ii) Parallel resistances

Conservation of charge: Kirchhoff’s First Law

At any junction, sum of currents going in is equal to the sum of

currents coming out.

In this case, at junction A,

Each resistor in the parallel circuit has the same potential drop which is equal to the

potential difference of the battery, V.

But and (Ohm’s Law)

Therefore

When two resistors are connected in a parallel circuit, we know that

𝐼

𝑉

𝐴

𝐼3

𝑅1

𝑅2

𝑅3

𝐼1

𝐼2

Potential: +V volts Potential: 0 volts

𝐵 ≡

𝑅𝑃

𝐼 𝑉

Potential Difference

∑ 𝐼𝑖𝑛 = ∑ 𝐼𝑜𝑢𝑡

𝐼 = 𝐼1 + 𝐼2 + 𝐼3

𝑅 =𝑉

𝐼 𝐼 =

𝑉

𝑅

𝑉′

𝑅𝑃=

𝑉′

𝑅1+

𝑉′

𝑅2+

𝑉′

𝑅3

1

𝑅𝑃=

1

𝑅1+

1

𝑅2+

1

𝑅3

𝑅1

𝑅2

1

𝑅𝑃=

1

𝑅1+

1

𝑅2

1

𝑅𝑃=

𝑅2

𝑅1𝑅2+

𝑅1

𝑅1𝑅2

1

𝑅𝑃=

𝑅2 + 𝑅1

𝑅1𝑅2 𝑅𝑃 =

𝑅1𝑅2

𝑅2 + 𝑅1


Question

Calculate

(a) the equivalent resistance, , of the circuit

(b) the current, I, produced by the battery

Solution:

(a) the equivalent resistance Re,

(parallel)

(series)

(b) the current, I, from the battery

(Ohm’s Law)

Question

What is the potential difference across the power supply?

1Ω

2Ω

3Ω

𝐼

6𝑉

≡

𝑅𝑒

6𝑉

𝑅𝑒

1

𝑅𝑃=

1

𝑅1+

1

𝑅2 𝑅𝑃 =

𝑅1𝑅2

𝑅2 + 𝑅1=

1 × 2

1 + 2=

2

3Ω

𝑅𝑒 = 𝑅𝑃 + 𝑅3 =2

3+ 3 = 3.6Ω

𝐼 = 𝑉

𝑅 =

6𝑉

3.6Ω= 1.7 𝐴

𝑅1 = 6Ω

𝑅2 = 10Ω 𝑅3 = 40Ω

𝑅4 = 4Ω

𝐴 𝐼3 = 1.5𝐴 𝑉1 𝐼1

(𝑃1)

𝑉2 𝐼2 (𝑃2)

𝐼4 𝑉4 (𝑃4)

𝑉3 = 𝐼3𝑅3 (𝑃3 = 𝐼3𝑉3)

∑ 𝑃𝑖𝑛 = ∑ 𝑃𝑜𝑢𝑡


Solution:

Method 1 Kirchhoff’s Laws and system of equations.

(Kirchhoff’s 2nd

Law)

(Kirchhoff’s 1st Law)

At A,

Method 2 Kirchhoff’s Laws II

(Ohm’s Law)

Parallel circuits:

Ohm’s Law

Kirchhoff’s 1st Law:

Ohm’s Law

Kirchhoff’s 2nd

Law:


𝐼 = 𝐼2 + 𝐼3 = 𝐼2 + 1.5A (2)

∑ 𝜀 = ∑ 𝐼𝑅𝑙𝑜𝑜𝑝2


𝜀 = 𝐼1𝑅1 + 𝐼3𝑅3 + 𝐼4𝑅4

𝜀 = (𝑅1 + 𝑅4)𝐼 + 𝐼3𝑅3 = (6Ω + 4Ω)𝐼 + (1.5𝐴)(40Ω) = 10𝐼 + 60

(∗ 𝐼1 = 𝐼4 = 𝐼 [𝑠𝑒𝑟𝑖𝑒𝑠])

(1)

𝜀 = 𝐼1𝑅1 + 𝐼2𝑅2 + 𝐼4𝑅4 (∗ 𝐼1 = 𝐼4 = 𝐼 [𝑠𝑒𝑟𝑖𝑒𝑠])

𝜀 = (𝑅1 + 𝑅4)𝐼 + 𝐼2𝑅2 = (10Ω)𝐼 + (𝐼2)(10Ω) = 10𝐼 + 10𝐼2

= 10𝐼 + 10(𝐼 − 1.5𝐴) = 10𝐼 + 10𝐼 − 15 = 20𝐼 − 15

(3)

𝐹𝑟𝑜𝑚 (2): (4)

𝐶𝑜𝑚𝑏𝑖𝑛𝑖𝑛𝑔 (1) 𝑎𝑛𝑑 (4):

10𝐼 + 60 = 20𝐼 − 15 10𝐼 = 75 𝐼 = 7.5𝐴

𝑉 = 𝑉1 + 𝑉2 + 𝑉4 = 𝐼𝑅1 + 𝐼2𝑅2 + 𝐼𝑅4 = 7.5𝐴(6Ω + 4Ω) + (1.5𝐴)(40Ω)

= 75𝑉 + 60 𝑉 = 135𝑉

𝑉3 = 𝐼3𝑅3 = 60𝑉

𝑉2 = 60𝑉

𝐼2 =𝑉2

𝑅2=

60𝑉

10Ω= 6𝐴


𝐼 = 𝐼2 + 𝐼3 = 6𝐴 + 1.5A = 7.5A At A,

𝑉1 = 𝐼𝑅1 = (7.5𝐴)(6Ω) = 45𝑉

𝑉4 = 𝐼𝑅4 = (7.5𝐴)(4Ω) = 30𝑉

= 𝑉1 + 𝑉2(𝑜𝑟 𝑉3) + 𝑉4 = 45𝑉 + 60𝑉 + 30𝑉 = 135𝑉


(part A)

(part B)


Method 3 Re (Equivalent Resistance)

In a parallel circuit:

In a series circuit:

Ohm’s Law:

Method 4 Conservation of energy

and

Question: Solving circuit problems with Kirchhoff’s Laws

Determine I1, I2 and I3 in this circuit.

Solution:

KI: at a junction.

At A,

KII:

As with method 2 (part A) to determine

the current I out of the battery, I = 7.5 A

Useful to find current leaving

the battery when solving circuit

problems.

1

𝑅𝑃=

1

𝑅2+

1

𝑅3 𝑅𝑃 =

𝑅2𝑅3

𝑅2 + 𝑅3=

10Ω × 40Ω

10Ω + 40Ω= 8Ω

𝑅𝑒 = 𝑅1 + 𝑅𝑃 + 𝑅4 = 6Ω + 8Ω + 4Ω = 18Ω

𝜀 = 𝐼𝑅𝑒 = (7.5𝐴)(18Ω) = 135𝑉

As with method 2 (part A) to determine the current circuit,

I = 7.5 A ; I2 = 6A; I3 = 1.5A

𝑃 = 𝐼𝑉 = 𝐼2𝑅 ∑ 𝑃𝑖𝑛 = ∑ 𝑃𝑜𝑢𝑡

𝐼𝜀 = 𝐼2𝑅1 + 𝐼22𝑅2 + 𝐼3

2𝑅3 + 𝐼2𝑅4

(7.5A)𝜀 = (7.5A)2(6Ω) + (6A)2(10Ω) + (1.5A)2(40Ω) + (7.5A)2(4Ω)

𝜀 =337.5 + 360 + 90 + 225

7.5= 135V


𝐼1 = 𝐼2 + 𝐼3 (1)

∑ 𝐸𝑙𝑜𝑜𝑝𝐴 = ∑ 𝐼𝑅𝑙𝑜𝑜𝑝𝐴

9𝑉 = 𝐼1𝑅4 + 𝐼2𝑅2 + 𝐼2𝑅3 + 𝐼1𝑅5

9𝑉 = 4𝐼1 + 2𝐼2 + 3𝐼2 + 5𝐼1

9 = 9𝐼1 + 5𝐼2 (2)


Pay attention to the assumed direction of the loops, which leads to some (-) signs here.

(2) + (3):

Substitute (4) into (1):

Substitute (5) into (2):

Substitute I1 into (5):

From equation (1):

As a result, current I3 is in fact flowing opposite of the assumed direction, with a

magnitude of 0.60A.

Emf, ε, and Internal Resistance, r, of a Battery

−6𝑉 = 𝐼3𝑅1 + (−𝐼2)𝑅2 + (−𝐼2)𝑅2

−6 = 𝐼3 − 2𝐼2 − 3𝐼2 = 𝐼3 − 5𝐼2

∑ 𝜀𝑙𝑜𝑜𝑝𝐵 = ∑ 𝐼𝑅𝑙𝑜𝑜𝑝𝐵

(3)

3 = 9𝐼1 + 𝐼3

𝐼3 = 3 − 9𝐼1 (4)

𝐼1 = 𝐼2 + (3 − 9𝐼1)

10𝐼1 = 𝐼2 + 3

𝐼2 = 10𝐼1 − 3 (5)

9 = 9𝐼1 + 5𝐼2

9 = 9𝐼1 + 5(10𝐼1 − 3)

(2)

9 = 9𝐼1 + 50𝐼1 − 15

𝐼1 =24

59= 0.407𝐴

𝐼2 = 10 (24

59) − 3 = 1.07𝐴

𝐼3 = 𝐼1 − 𝐼2 = 0.407𝐴 − 1.07𝐴 = −0.60𝐴

V

I

R

𝑟 ε

Battery

r is the “internal resistance” of the

battery

Voltmeter measures the terminal voltage, VT of the battery.

(External Resistance)


What is emf, ε?

It is the terminal voltage of the battery when zero current flows through it.

, when (So internal resistance [r] = 0)

*where I = 0*

Question

Determine the internal resistance of a battery which has an emf of 12 V and a terminal

Voltage of 8.8 V when the external resistor draws 60A from the battery. Then find the

external resistance.

A special case

In fact, the following equation is true in specific conditions.

, when a battery is being charged (current flowing to the positive

side of the battery)

𝜀 = 𝑉𝑇 𝐼 = 0

𝑉𝑇 = 𝜀 − 𝐼𝑟 0

VT

I = 60A

R

𝑟

ε = 12V

(External Resistance)

= 8.8V

Solution:

Since K2, ∑ 𝜀 = ∑ 𝐼𝑅𝑙𝑜𝑜𝑝

ε = 𝐼𝑟 + 𝐼𝑅 ε = 𝐼𝑟 + 𝑉𝑇

So, 𝑉𝑇 = 𝜀 − 𝐼𝑟 (which is same as equation shown above)


(*IR (potential difference across the ER) is equal the terminal voltage measured

by the voltmeter)

(8.8𝑉) = 12𝑉 − (60𝐴)𝑟


𝑟 = 0.05Ω


𝑅 =𝑉𝑇

𝐼, (Ohm’s Law)

=8.8𝑉

60𝐴= 0.15Ω

𝑉𝑇 = 𝜀 + 𝐼𝑟

V I

R

𝑟 ε This equation summarizes all possible cases:

𝑉𝑇 = 𝜀 ± 𝐼𝑟

recall that VT is the terminal voltage recorded by

the voltmeter, 𝜀 is the electromotive force of the battery, I is the current flowing through it and r is the its internal resistance.


A concept question

Solution:

Parallel circuit:

Thus, the parallel pair of light R1 || R2 has a resistance smaller than just a single light bulb R1.

As a result, V1 is reduced in the parallel set up (RP < R1).

According to K2,

As stated above, V1 is reduced in the parallel circuit, thus V3 is increased. (𝜀 is constant)

Finally, since (Power determines the brightness, the higher the power the brighter the light bulb.)

P1 is reduced due to the drop of V1 and thus is less bright;

P3 is increased due to the increase of V3 and thus is brighter.

Example

: symbol of a light bulb in circuits 𝑅1

𝑅2

𝑅3

What happens to the brightness

of the three light bulbs once the switch is closed?

Explain.

(Note: all three light bulbs are

identical. In other words,

R1 = R2 = R3)

∑ 𝜀 = ∑ 𝐼𝑅𝑙𝑜𝑜𝑝

𝜀 = 𝑉1 + 𝑉3

𝑅𝑃 =𝑅1𝑅2

𝑅2 + 𝑅1=

(𝑅1)2

2𝑅1=

𝑅1

2< 𝑅1

𝑃 =𝑉2

𝑅

𝑃1 = 𝑃2 =𝑉1

2

𝑅1

𝑃3 =𝑉3

2

𝑅1

V

𝑅1 = 15Ω


𝑅3 = 7Ω


𝑅2 = 6Ω


𝑅5 = 5Ω


𝑅4 = 9Ω


𝜀 = 8𝑉


Find the potential

difference across the

6 Ω resistor.


Moving on, we would then calculate the equivalent resistance Re of the circuit.

….and then the current flowing through the battery.

Capacitors

A capacitor is a pair of parallel plates often sandwiched by a material called a dielectric

which affects the charge storage capacity.

Compare these two circuit diagrams: why is it that electrons are travelling in different

directions in the two circuits? What components of the circuits result in the difference?

(Note: Battery [Charging] vs. Resistance [Charged Used Up])

The amount of charge stored depends on the capacitance, C of the capacitor.

Solution: In this section of the circuit, 𝑅𝑆 = 𝑅1 + 𝑅2 + 𝑅3 = 15Ω + 6Ω + 9Ω = 30Ω

𝑅𝑃 =𝑅𝑆𝑅4

𝑅𝑆 + 𝑅4=

30 × 7

30 + 7=

210

37Ω = 5.68Ω

𝑅𝑒 = 𝑅𝑃 + 𝑅5 = 5.68Ω + 5Ω = 10.68Ω

𝐼 =𝑉

𝑅𝑒=

8 V

10.68Ω= 0.75𝐴

𝑉5Ω = 𝐼𝑅 = (0.75𝐴)(5Ω) = 3.75V

𝑉7Ω = 8𝑉 − 3.75𝑉 = 4.25𝑉 (loop and K2)

𝑉Outer Loop = 𝑉7Ω = 4.25𝑉 (Parallel to 7Ω)

𝐼Outer Loop =𝑉Outer Loop

𝑅=

4.25𝑉

30Ω= 0.14A

𝑉6Ω = 𝐼𝑅 = (0.14𝐴)(6Ω) = 0.85𝑉

𝑜𝑢𝑡𝑒𝑟 𝑙𝑜𝑜𝑝

𝑖𝑛𝑛𝑒𝑟 𝑙𝑜𝑜𝑝

𝑉


+Q - - - -

−Q +

+ +

+

e−

e−

𝑑𝑖𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐

+Q - - - -

−Q + +

+

+

e−

e−

𝑄 = 𝐶𝑉


where Q is amount of charge in C

C is capacitance in Farads (F) V is the potential difference across the capacitor in V.

Capacitors in series

*Conservation of charge Q = Q = Q

Capacitors in Parallel

When two capacitors are connected in a parallel circuit,

(Conservation of charge)

Dielectrics and Capacitor

+Q

+Q

−Q

−Q

𝐶1


𝐶2


𝐶3

So, 𝑉𝑇 = 𝜀 − 𝐼𝑟 (which is same as equation shown above) +Q

−Q

𝑉1 𝑉2


𝑉3


e−

e−

Note that charge distribution across the capacitors.

Each capacitor stores the same amount of charge Q.

𝑉


𝑉 = 𝑉1 + 𝑉2 + 𝑉3

𝑄

𝐶𝑠=

𝑄

𝐶1+

𝑄

𝐶2+

𝑄

𝐶3

1

𝐶𝑠=

1

𝐶1+

1

𝐶2+

1

𝐶3

Capacitance in series:

+Q

−Q

𝑄1

+Q

−Q

𝑄1

𝑄 = 𝑄1 + 𝑄2 + 𝑄3

𝐶𝑃𝑉 = 𝐶1𝑉 + 𝐶2𝑉 + 𝐶3𝑉 (KII)

𝐶𝑃 = 𝐶1 + 𝐶2 + 𝐶3

𝐶 =𝑘𝜖0𝐴

𝑑

charge storage capacity

Dielectric constant depends on dielectric used (For vacuum K=1)

A is the area of each plate, m2 d is the separation distance, m

ϵ0 is the permittivity of free space, = 8.85 × 10 -12 C2/Nm2


Energy stored in a capacitor

(Energy)

or

*Compare with EP =1

2𝑘𝑥2

Charging and discharging a capacitor

(a) Charging

Example

A 1 nF capacitor is set up in series with a 200 Ω resistor and a 12 V battery. How long does

it take for the capacitor to be charged up to 6V?

Solution:

𝐸 = 𝑞∆𝑉 Q

V 0

Q = CV

slope = C

E

𝐸 = 𝑒𝑛𝑒𝑟𝑔𝑦 𝑢𝑛𝑑𝑒𝑟 𝑔𝑟𝑎𝑝ℎ

𝐸 =1

2𝑄𝑉

𝐸 =1

2𝐶𝑉2

𝐴 +Q

−Q

𝐶

𝜀


𝑅

So, 𝑉𝑇 = 𝜀 − 𝐼𝑟 (which is same as equation shown above) e− e−

𝑉 = 𝜀[1 − 𝑒−(𝑡

𝑅𝐶)]

RC is called the time constant, τ, which affects how quickly the capacitor gets charged up.

Large RC … longer charging (and discharging)

𝑉 = 𝜀 [1 − 𝑒−(𝑡

𝑅𝐶)]

6𝑉 = (12𝑉) [1 − 𝑒−(

𝑡(200Ω)(1nF)

)]

6

12= 1 − 𝑒

−(𝑡

2.00×10−9)

ln (0.5) = ln (𝑒−(

𝑡2.00×10−9)

)

𝑡 = −ln (0.5)(2.00 × 10−9)

𝑡 = 1.39 × 10−9𝑠


(b) Discharging

Example

If R = 220 Ω resistor and C = 14 pF, determine

a) The “half-life”

b) Determine when t = τ Solution: (a) “Half-life” ->

(b) When t = τ,

𝐴 +Q

−Q

𝐶

𝑅


𝑉 = 𝜀𝑒−(𝑡

𝑅𝐶)

Math tip: Solving for t?

ln (𝑉) = ln (𝜀𝑒−(𝑡

𝑅𝐶))

ln(𝑉) = ln(𝜀) −𝑡

𝑅𝐶

τ

𝑉

𝜀

𝑉 =𝜀

2

𝑉

𝜀= 𝑒−(

𝑡𝑅𝐶

) ln (1

2) = − (

𝑡

𝑅𝐶) 𝑡 = −ln (

1

2)(𝑅𝐶) → →

𝑡 = − ln (1

2) (220Ω)(14 × 10−12𝐹) = 2.1 × 10−9𝑠

𝑉 = 𝜀𝑒−1

𝑉

𝜀=

1

𝑒


Data Table Charge on proton/electron e (±)𝟏. 𝟔𝟎 × 𝟏𝟎−𝟏𝟗𝑪

Coulomb’s constant k 9.0 × 109 Nm2/C2 Permittivity of free space 𝜖0 = (1/𝑐2𝜇0) 8.85 × 10−12 C2/Nm2

Electron rest mass 𝑚𝑒 9.11 × 10−31 kg Proton rest mass 𝑚𝑝 1.6726 × 10−27 kg

Neutron rest mass 𝑚𝑛 1.6749 × 10−27 kg

Equations Ohm’s Law:

Power:

Resistance and Resistivity:

Kirchhoff’s Law and equivalence resistance

Emf and terminal voltage:

Capacitance:

Charging and discharging:

𝑅 = 𝑉

𝐼

𝑃 = 𝐸

𝑡= 𝑉𝐼 = 𝐼2𝑅 =

𝑉2

𝑅

𝑉 = 𝐼𝑅

𝑅 =𝜌𝑙

𝐴 ρ𝑇 = ρ0(1 + 𝛼(∆𝑇))

∑ 𝐼𝑖𝑛 = ∑ 𝐼𝑜𝑢𝑡 ∑ 𝜀𝑙𝑜𝑜𝑝 = ∑ 𝐼𝑅𝑙𝑜𝑜𝑝

𝑅𝑆 = 𝑅1 + 𝑅2 + 𝑅3

1

𝑅𝑃=

1

𝑅1+

1

𝑅2+

1

𝑅3

𝑉𝑇 = 𝜀 ± 𝐼𝑟

𝑄 = 𝐶𝑉

1

𝐶𝑠=

1

𝐶1+

1

𝐶2+

1

𝐶3 𝐶𝑃 = 𝐶1 + 𝐶2 + 𝐶3

𝐸 =1

2𝑄𝑉 =

1

2𝐶𝑉2

𝑉 = 𝜀𝑒−(𝑡

𝑅𝐶) 𝑉 = 𝜀[1 − 𝑒−(

𝑡𝑅𝐶

)]

𝐶 =𝑘𝜖0𝐴

𝑑

Documents

18-19 Circuit and Capacitors