Notes - Fisika - Finals

I should take a walk for a bit

Alternating Current

ƒ=1/T

y=A*sin(ω*t)

E=(E_m)*sin(ω*t)=N⋅B⋅A⋅ω⋅sin(ω*t)

ƒ is frequency

T is period of the current

A is amplitude

ω is angular frequency

(V_effective)=(V_max)/√(,2)

(I_effective)=(I_max)/√(,2)

Oscilloscopes measure maximum voltage (V_max), voltmeters measure effective voltage (V_E). Effective voltage is basically the amount of DC voltage needed to match the AC voltage.

How do you read an oscilloscope?

(V_max)=(s_y)⋅A

T=(s_x)⋅λ

The vertical axis uses a scale based on volts per centimeter.

The horizontal axis uses a scale based on miliseconds per centimeter.

R-L-C

(X_L)=ω*L=2*π*ƒ*L

(X_C)=1/(ω*C)=1/(2*π*ƒ*C)

V=√(,(V_R)^2+((V_L)-(V_C))^2)

I⋅Z=√(,(I*R)^2+(I*(X_L)-I*(X_C))^2)

Z=√(,R^2+((X_L)-(X_C))^2)

I=V/Z=(V_L)/(X_L)=(V_R)/R=(V_C)/(X_C)

tan(θ)=((X_L)-(X_C))/R

(X_L)>(X_C) → Circuit is inductive

(X_C)>(X_L) → Circuit is capacitive

(X_L)=(X_C) → Resonance with ƒ

ƒ=1/(2*π√(,L*C))

I(t)=(I_max)*sin(ω*t)

V(t)=(V_max)*sin(ω*t+θ)

P=I^2*R

Z is impedance, basically the total opposing resistance.

R is resistance from a resistor

(X_L) is inductive resistance

(X_C) is capacitive resistance

(V_L) is voltage across the inductor

(V_C) is voltage across the capacitor

P is dissipation, only taking into account R

In a phasor diagram, (V_R) and I have an angle of 0° (right). (V_L) has an angle of 90° (up) and (V_C) has an angle of 270° (down). This means that (V_L) and (V_C) have opposing directions.

Highlighted Problems

Problem 1

∴(V_max)=3⋅80=240volts

T=8⋅2.5=20ms

∴ƒ=1/(20⋅10^(-3))=1000/20=50Hz

(B)

Problem 2

Voltmeters measure effective voltage.

(V_max)=4⋅10=40volts

∴(V_effective)=40/√(,2)=20√(,2)

T=2⋅8=16ms

∴ƒ=1/(16⋅10^(-3))=1000/16=125/2=62.5Hz

Drawing is easy, so I won't showcase it. The vertical scale is basically how many squares a peak or a valley takes up, and the horizontal scale is how many squares 1 wave takes up.

Problem 3

(V_R)=80volts

(V_max)=100volts

80=(I_max)⋅160

(I_max)=0.5A

(V_RL)=100=√(,80^2+(V_L)^2)

(V_L)=√(,100^2-80^2)=60=(X_L)⋅I

(X_L)=60/0.5=120Ω

Z=100=√(,80^2+(Δ(V))^2)

100^2-80^2=(Δ(V))^2

Δ(V)=√(,100^2-80^2)=60

Δ(V)=(V_C)-(V_L)=60

You can swap the order of (V_C) and (V_L) to find a nonzero solution.

(V_C)=60+60=120

120=0.5⋅(X_C)

(X_C)=240Ω

Problem 4

Well, let's just get right to it.

V(t)=100√(,2)*sin(100*t)

(X_L)=2⋅100=200Ω

(X_C)=1/(10⋅100⋅10^(-6))=1000Ω (it's supposed to be micro faradays)

tan(θ)=(200-1000)/600=-4/3

∴θ=-53° (a)

Rangkaian bersifat kapasitif karena (X_C)>(X_L) (b)

Z=√(,600^2+(200-1000)^2)=1000Ω

100√(,2)=1000⋅(I_max)

(I_max)=0.1√(,2)

I(t)=0.1√(,2)*sin(100*t+53°) (c)

(V_R)=600⋅0.1=60volts

(V_L)=200⋅0.1=20volts

(V_C)=1000⋅0.1=100volts

(d)

P=1/10^2⋅600=6W (e)

Black Body Radiation

Use a calculator.

P=e⋅σ⋅A⋅T^4

I=P/(4*π*r^2)

P is radiation, joules per second, aka watts.

e is coefficient of emissivity

σ is the Stefan Boltzmann constant, 5.67⋅10^(-8)

A is the area

T is temperature in Kelvin

I is intensity, with r being distance

Highlighted Problems

Problem 1

A=(10⋅10^(-2))^2=1⋅10^(-2)⋅2

P=5.67⋅10^(-8)⋅8/10⋅1⋅10^(-2)⋅2⋅(727+273)^4=2⋅453.6W

Problem 2

The given variables are R, r, I

I=P/(4*π*r^2)

P=4*I*π*r^2=σ*A*T^4

4*I*π*r^2=σ⋅(4*π*R^2)⋅T^4

I*r^2=σ*R^2*T^4

T^4=(I*r^2)/(σ*R^2)
=(1400⋅(1.5⋅10^11)^2)/(5.67⋅10^(-8)⋅(7⋅10^8)^2)
≈1.13⋅10^15

T≈5800K

Photoelectric Effect

E=h⋅ƒ=(h⋅c)/λ

(KE_max)=e⋅(V_s)

(KE_max)=E-(W_s)

(W_s)=h⋅(ƒ_s)=(h⋅c)/(λ_s)

E is energy from photon, joule or eV

h is 6.62⋅10^(-34)

ƒ is light frequency

(KE_max) is maximum kinetic energy

e is 1.6⋅10^(-19)

(V_s) is stopping potential

(W_s) is work function / energi ambang

(ƒ_s) is work frequency / frekuensi minimum / frekuensi ambang

c is the speed of light

1 eV is 1.6⋅10^(-19) joule

Highlighted Problems

Problem 1

(ƒ_s)=2.30eV

2.30⋅1.6⋅10^(-19)=(6.62⋅10^(-34)⋅3⋅10^8)/(λ_s)

(λ_s)≈5.39⋅10^(-7) (a)

E=(6.62⋅10^(-34)⋅3⋅10^8)/(150⋅10^(-9))=1.324⋅10^(-18)

(KE_max)=1.324⋅10^(-18)-2.30⋅1.6⋅10^(-19)=9.56⋅10^(-19) (joules) (b)

9.56⋅10^(-19)=1.6⋅10^(-19)⋅(V_s)

(V_s)=5.975volt (c)

Problem 2

(W_s)=1.6eV=2.56⋅10^(-19) (a)

2.56⋅10^(-19)=(6.6⋅10^(-34)⋅3⋅10^8)/(λ_s)

(λ_s)=7.73⋅10^(-7)=773nm (b)

(0.4⋅1.6⋅10^(-19))+(2.56⋅10^(-19))=6.6⋅10^(-34)⋅ƒ

ƒ=4.84⋅10^14 (c)

Compton Effect

Photons scatter off electrons.

Δ(λ)=(λ^′)-λ=h/(m⋅c)⋅(1-cos(θ))

E=(h⋅c)/λ

(E_k)=E-(E^′)

At this point the symbols should be obvious, but here they are.

Δ(λ) is change in wavelength

h is 6.6⋅10^(-34)

c is the speed of light

m is mass of an electron, 9.1⋅10^(-31)

λ and (λ^′) are the wavelengths before and after scattering

θ is the angle of scattering

E is an energy of a photon

(E_k) is the amount of kinetic energy given to the electron

Highlighted Problems

Problem 1

λ=3.5⋅10^(-2)⋅10^(-9)=3.5⋅10^(-11)

cos(37°)≈4/5

Δ(λ)=(6.6⋅10^(-34))/(9.1⋅10^(-31)⋅3⋅10^8)⋅(1-4/5)=4.83⋅10^(-13)

(λ^′)=4.83⋅10^(-13)+3.5⋅10^(-11)=3.54⋅10^(-11)=0.0354nm (a)

(E^′)=(3⋅10^8⋅6.6⋅10^(-34))/(3.54⋅10^(-11))=5.59⋅10^(-15) (b)

E=(3⋅10^8⋅6.6⋅10^(-34))/(3.5⋅10^(-11))=5.65⋅10^(-15)

(E_k)=5.65⋅10^(-15)-5.59⋅10^(-15)=0.06⋅10^(-15) (c)

Problem 2

λ=1.000⋅10^(-10)

Δ(λ)=(6.6⋅10^(-34))/(9.1⋅10^(-31)⋅3⋅10^8)⋅1=2.41⋅10^(-12)=0.00241nm (a)

(λ^′)=1⋅10^(-10)+2.41⋅10^(-12)=1.024⋅10^(-10)

(E_k)=3⋅10^8⋅6.6⋅10^(-34)⋅(1/(1.000⋅10^(-10))-1/(1.024⋅10^(-18)))
=4.640⋅10^(-17)

4.64⋅10^(-17)=(m⋅v^2)/2
=1/2⋅9.1⋅10^(-31)⋅v^2

v=10099096=1.0099096⋅10^7