Ohm’s Law

V

I

Slope = conductance

=1

resistance

Units:

[Voltage V] = Volt (V)

[Current I] = Ampere (A)

[Resistance R] = Ohm (Ω)

[Power P] = Watt (W)

Ohm’s Law applies only to metal as a conductor.

V = IR

P = VI = I2R

Example

+ -

Note current direction

100Ω

10V

I

10= 100 I ⇒ I = 0.1A

Power dissipated by the resistor = I2R = (0.1)2(100) = 1W

Power delivered by the battery = VI = (10)(0.1) = 1W

Color code of resistors

Resistor:

Variable Resistor:

Example

Value of this resistor = 2600000 Ω ± 5%

Potentiometer (voltage divider)VH

R1

R2

V = VH ⋅R2

R1+R2

(Only if there is no current in

the green line).

Example10V

99Ω

V = 10 ⋅ = 0.1V1

99+1

1Ω

I = 10/(99+1) = 0.1A

ResistanceThe resistance of a resistor is a constant independent of V. It is also independent of the signal frequency.

V

IFor ALL frequency

Not all devices or components behave like this!

Example

I

V

Resistance at this point = dV

dI

Diode:

Resistance = dV/dI, depends on the applied voltage.

Capacitor:

Impedance of a capacitor does not depend on the applied voltage, but it depends on the frequency:

Xc = iωC1

Kirchhoff’s Law

ΣI = 0

1. Kirchhoff’s current law: (conservation of charge)

2. Kirchhoff’s voltage law: (conservation of energy)

ΣV = 0

Example 1 (Important)

1

2

1

RV I' (1)

,I' gEliminatin-(2)--- )RI'-(IV-(1)--- RI'V

=⇒

==

R1V R2

II’ I-I’

21

2121

21

21

21

21

21

2

21

RRRR//RR

I RR

RR V

IR VR

RR

IR VRR1

)RRV-(IV

:(2) into Substitute

+=∴

+=⇒

=⎟⎟⎠

⎞⎜⎜⎝

⎛ +⇒

=⎟⎟⎠

⎞⎜⎜⎝

⎛+⇒

=

Example 2 (problem 33 of text)

(3)--- 0 I4I32I- 0)III(2000)II(10001000I

(2)--- 0 II2I- 0)I1000(I1000I1000I

(1)--- 0.012 I23I3I 12 000I23000I3000I

)III(2000)I1000(I 12

321

321323

321

2132

321

321

32121

=++⇒=−−−++

=−+⇒=−−−

=−−⇒=−−⇒

−−+−=

V

1000Ω

12V

1000Ω

1000Ω

1000Ω

2000Ω

I1I2I1- I2

I3 I1- I2- I3I2+ I3

I1

10.154mA 1II

5.583mA I6 I

mA923.0 I 0.012 13I 0.012 I2)3(6I)3(11I

(1) into theseeSubstiutut11I II12I II2I I6 I

I3 I21 I2I

23 II2

I2I23

2 I4I3

I (3)

II2I (2)

31

32

3

3333

3321321

32

323232

3232

1

321

==

==

=⇒=⇒=−−∴

=−=⇒−==⇒

=⇒+=−∴

+=+

=⇒

−=⇒

Input impedence and output impedance

Vin Vout

Iin Iout

Input impedance = ∂Vin∂Iin

Output impedance = - ∂Vout∂Iout

Example

10Ω

9V

8V

Iout

Vout

δVout = 8V-9V = -1V

δIout = 8/10 = 0.8A

Output impedance Xout = -δVout / δIout = -(-1)/0.8 = 1.25Ω

The output impedance of a battery is also known as its internal resistance.

An ideal voltage source should have zero output impedance ⇒ δVout =0

VoltmeterAn ideal voltmeter should have infinite input impedance ⇒ δIin =0

An ideal voltmeter A not so ideal voltmeter

V

Xin

V

How you use a voltmeter:

Voltage difference across these two points=?

V

I IYou want to draw as little current as possible.

Easy!

ExampleA voltmeter with an input impedance of 100kΩ is used to measure the voltage across the Ω resistor of a potentiometer. What will be its reading?

10V

90kΩ

10kΩ V

100kΩ 1V 1090

1010V

:terpotentiome theofoutput al theoretic with the thisCompare

V 0.917 9.09190

9.09110

resistors two theseacross Voltage voltmeter theof Reading

9.091k 1001010010 //100kk10

theory =+

×=

=+

×=

=∴

Ω=+×

=ΩΩ

AmmeterAn ideal ammeter should have zero input impedance ⇒ δVin =0

An ideal ammeter

A

How you use an ammeter:

Current along this line =?

I

Voltage drop across the ammeter should be made as small as possible!

Xin

A

A not so ideal ammeter

A

I

Difficult!

If you do not want to break the circuit, use a current clamp!

AmmeterIf you do not want to break the circuit, use a current clamp!

An ideal ammeter

A

How you use an ammeter:

Current along this line =?

I

Voltage drop across the ammeter should be made as small as possible!

Xin

A

A not so ideal ammeter

A

I

Difficult!

Example

0.0901A 100091.9

resistor 100 rough theCurrent th ammeter theof Reading

V091.910100

10010

resistor100 theacross Voltage

==

Ω=∴=

+×=

Ω

An ammeter is constructed by connecting a 100Ω in parallel to an ideal voltmeter and used to measure the current in a simple circuit. What will be its reading?

10Ω

10VI=?

10Ω

10VI=? V100Ω

Ammeter

1A 1010 I

:insertedammeter thout thecurrent wi of valueal theoretic with the thisCompare

theory ==

Current and voltage sourcesAn ideal voltage source should have zero output impedance ⇒ δVout =0

An ideal voltage source

V

V

Xout

A not so ideal voltage source

An ideal current source should have infinite output impedance ⇒ δIout =0

An ideal current source

Xout

A not so ideal current source

e+ -

Textbook symbol:

My symbol:

II

How to measure V and Xout?

How to measure I and Xout?

ExampleWe can construct a current source from a voltage source

106Ω 9V

Current source106Ω 9V

10Ω

I=9/(100+106)

=8.99991×10-7A106Ω 9V

1000Ω

I=9/(1000+106)

=8.9910×10-7A

Thevenin (voltage source)TheoremConsider a circuit made of batteries and resistors only. Any two points on the circuit can be considered as the terminals of a not so ideal voltage source.

Complicated circuit made of batteries and resistors only.

AB

=V

Xout

A B

With respect to points A and B, the whole circuit can be represented by two parameters – the electromotive force V and the internal resistance Xout of the not so ideal voltage source.

Note that if you choose another two points, the parameters V and Xoutwill change their values also!

Norton (current source)TheoremConsider a circuit made of batteries and resistors only. Any two points on the circuit can be considered as the terminals of a not so ideal current source.

Complicated circuit made of batteries and resistors only.

AB

=A B

With respect to points A and B, the whole circuit can be represented by two parameters – the current I and the internal resistance Xout of the not so ideal current source.

Note that if you choose another two points, the parameters I and Xoutwill change their values also!

Xout

I

ExampleWe can construct a current source from a voltage source

106Ω 9V

Current source

106Ω 9V

10-6A

1V

Xout = 106Ω

V

Xout

=

Xout

I=VXout