Physics Electricity Flashcards
A red LED is only lit when an ammeter gives a positive reading and the green LED is only lit when the ammeter gives a negative reading, explain why.
The LEDs only emit light when they are forward biased.
The change in polarity of the voltage changes the biasing.
Determine the peak voltage of the output signal generator.
Vpeak= (no. of boxes * y-gain setting)
Determine the frequency of the output of the signal generator.
f= 1/T
T= No. of boxes for a complete wave
The power dissipated in a 120 Ω resistor is 4.8 W. What is the current in the resistor
P= I²R
I²= P/R
I = √P/R > √4.8/120
I = 0.2 A
State what is meant by the term electromotive force (e.m.f.)
The energy given to each coulomb of charge passing through the battery.
LEDs emit light when electrons fall from the conduction band into the valance band of the p-type semiconductor. Explain using band theory, why the blue LED will not operate with this battery.
The electrons in the n-type conduction band do not gain enough energy to move towards the conduction band of the p-type.
1.) 12V/2 = 6.0V
2.) I = V/R = 6V/10 = 0.6A
3.) Rp = 1/10 + 1/10 = 1/5 Rp = 5Ω
Rs = 5Ω + 10Ω = 15Ω
I = V/R > I = 12V/15Ω > I = 0.8A
r = - gradient
r = - (y2 - y1/ x2 - x1)
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1.) When the capacitor is fully charged, the potential difference across it equals the supply voltage. 12V
2.) E = 1/2CV²
C = 150mF = 150x10-3F (capacitance)
V = 12²V (Voltage across the capacitor)
E = 1/2 x (150x10-3)x(12)² > E = 10.8J
I = Vs/Rt > 12/75 > I = 0.16A
Vs = 12V (Supply Voltage)
Rt = 56 + 19 = 75 (resistance total) the resistors are in series because the current must flow through both resistors in sequence before completing the circuit.
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The resistance of the variable resistor is now increased. State what happens to the reading of the voltmeter. Justify your answer
The reading on the voltmeter increases because the current in the circuit decreases, reducing the voltage lost across the internal resistance of the cells.
Adjust the variable resistor and take readings of V and I
Plot a graph of V against I
Gradient of the graph = - r
Step 1.) find total EMF
EMF = 1.5V x 4 = 6.0V
Step 2.) find total internal resistance
Rseries = 0.5 + 0.5 + 0.5 + 0.5 = 2ohms
Step 3.) the total circuit resistance consists of the internal resistance (2.0) + the motor resistance (20.0) + the variable resistor (?)
V = IR > I = V/R > I = 6V/0.20A = 30ohms
Step 4.) find R
R total = R + 20 + 2
30 = R + 22
R = 30 - 22 = 8 ohms
17.) B
18.) D - Irms = P/Vrms
> 24/12 Irms= 2A
> Irms = Ipeak/root2
> Ipeak = Irms x root2
> Ipeak = 2 x 1.414(root2)
> Ipeak = 2.8A
19.) E
D
To determine the potential difference across the 7.0 kΩ resistor, we will use the potential divider formula.
Step 1: Understanding the Circuit
The circuit consists of two resistors in series:
• 3.0 kΩ resistor
• 7.0 kΩ resistor
• The total supply voltage is 12V.
Since they are in series, they share the same current.
Step 2: Calculate the Total Resistance
The total resistance in the circuit is:
R total = 3.0kohms + 7.0kohms = 10ko
Step 3: Use the Potential Divider Formula
The voltage across a resistor in a series circuit is given by:
Vr = V total x R/R total
For the 7.0 kΩ resistor:
V7.0ko = 12V x 7.0ko/10ko
V7.0ko = 12V x 0.7
V7.0ko = 8.4V
19.) E
20.) B
C = Q/V
18.) A
19.) E
In conductors the conduction band is partially filled allowing free movement of electrons.
Insulators have a large band gap, preventing electrons from easily moving to the conduction band.
In semiconductors increasing the temperature increases the conductivity
V82 = 3 x 82/150
V82 = 3 x 0.5467
V82 = 1.64 V
Vrms82 = 1.64/root2
Vrms82 = 1.64/1.414
Vrms82 = 1.16 V
20.) C
21.) E
22.) A
23.) A
24.) E
25.) D
21.) B
Doubling the frequency would halve the wavelength
Changing the time base to 0.5ms/div means we are zooming in, stretching the wavelength back out to twice its width.
The two effects cancel each other out meaning the waveform’s appearance would remain unchanged.
22.) C
23.) D
24.) B
25.) E
23.) E
Voltage across the resistor starts at the maximum and decreases to 0
Voltage across the capacitor starts at 0 and increases to the supply voltage
Current starts at the maximum and decreases to 0
19.) C
V = 3
R = 1/6 + 1/6 = 3 > 3 + 9 = 12
P = ?
P = V(squared)/R
P = 3(squared)/12
P = 0.75W
20.) B
R = 1/10 + 1/10 = 5 > 5 + 5 + 5 = 15ohms
V1 = (R1/R1 + R2) x Vs
Vxy = (R parallel/R total) x Vs
Vxy = (5/15) x 12v = 4V
21.) B
I = V/R > I = 12/10 = 1.2A
Vlost = I x r(internal resistance only)
Vlost = 1.2 x 4 = 4.8V
22.) D
Q = C x V
24.) C
Photovoltaic effect
P = IV
P = (35x10-3)x2.1
P = 0.074W
Greater number of photons strike the solar cell per second
17.) D
Q = C x V
18.) D
E = 1/2CV(squared)
19.) E
2a.) P = V(squared)/R >
12(squared)/9.6 = 15W
21.) D
1/6 + 1/6 = 3
3 + 8 = 11
22.) C
P = V(squared)/R
V(squared) = PR
V(squared) = 4x100 = 400
V = Root400
V = 20V
23.) A
24.) C
Q = CV
Q = (20x10-6)x6
Q = 1.2x10-4
25.) D
10ai.) 12.8J of energy is gained to 1 coulomb of charge passing through the battery.
ii.)
V = 12.8
R = 6.0x10-3 + 0.05 = 0.056
I = V/R
I = 12.8/0.056
I = 228.6A
iii.) wire of large diameter has a low resistance.
To prevent overheating.
To prevent wires melting.
10bi.) 12.6V
ii.) gradient = -r
gradient = (12.2 - 12.4) / (40 - 20) = -0.01
Internal resistance = 0.01 ohms
A.) V = IR
I = V/R
I = (15 - 11.5) / (0.45 + 0.09)
I = 6.5A
B.) the EMF of the battery increases.
Difference between the two EMF’s decrease
C = Q/V
Q = CV
Q = (64 x 10-6) x ( 2.5 x 10+3)
Q = 0.16 C
11b.) E = 1/2 x QV
E = 1/2 x 0.16 x 2.5x10+3
E = 200J
C.) V = IR
R = V/I
R = 2.5x10+3 / 35
R = 71.4 ohms
ii.) the voltage decreases
iii.) smaller initial current, time to reach 0A is longer
The battery provides a voltage that moves charges around the circuit it does not send out electricity.
The resistor does not use up energy but converts some to heat
The LED does not convert all the remaining energy into light - some is dissipated as heat.
Energy is conserved in the circuit it is transformed but not lost
12ai.) V = IR
V = 1.8 x (4.8 + 0.1)
V = 8.82
Voltmeter reading = 12.8 - 8.82
Voltmeter reading = 4.0V
ii.) reading on the voltmeter decreases .
Total resistance decreases/current increases.
Lost volts increases
Voltage applied causes electrons to move towards the conduction band of the p-type.
Electrons drop from the conduction band to the valence band.
Photons are emitted when the electrons drop.
12a.)
E = h(plancks constant)f
f = E/h
f = (3.03 x 10-19) / (6.63 x 10-34)
f = 4.57 x 10+14 hz
v(velocity) = f λ
λ = v/f
λ = (3.00 x 10+8) / ( 4.57 x 10+14)
λ = 6.56 x 10-7 m
B.) Red
12a.) 1.5 J of energy is supplied by each coulomb of charge passing through the cell.
bi.) Vlost = Ir
Vlost = (64 x 10-3) x (5.4)
Vlost = 0.35 V
ii.) 3 - 0.35 = 2.65 V
iii.) P = IV
P = (64 x 10-3) x (2.65)
P = 0.17 W
E = V + Ir
V = E - Ir
V = 6 - (26 x 10-3) x (2.7 x 4)
V = 5.7 V
V = IR
R = V/I
R = (5.7 - 3.6) / ( 26 x 10-3)
R = 81 ohms
13.a) V = IR
I = V/R
I = 12 / 6800
I = 0.00176
I = 1.76 x 10-3 A
b.) the total resistance is less.
Initial charging current is greater
Gradient = - r
Gradient = ( 310x10-3 - 600x10-3 / 100x10-6 - 20x10-6)
Gradient = -3625
r = 3625 ohms
12ai.) 1.5 V
ii.) Vlost = Ir
r = Vlost / I
r = 0.2 / 0.88
r = 0.23 ohms
iii.) when the switch is closed there is a current in the circuit.
Voltage is dropped across the internal resistance
12bi.) V = IR
I = V/R
I = 9 / 3.6
I = 2.5 A
ii.) P = I(squared)R
P = 2.5(squared) x 2.4
P = 15 W
C = Q/V
Q = CV
Q = (47 x 10-6) x (6)
Q = 2.82 x 10-4 C
As resistance decreases current increases.
Lost volts increases, terminal potential difference decreases
14a.) V = IR
R = V/R
R = 12 / 3 x 10-5
R = 400000 ohms
b.) Q = It
Q = (3 x 10-5) x (25)
Q = 7.5 x 10-4 C
bii.) C = Q/V
V = Q/C
V = 7.5 x 10-4 / 220 x 10-6
V = 3.4 V
Potential difference = 12 - 3.4
Potential difference = 8.6 V
15.) material 2
Material 1 is an insulator
Material 3 is a conductor
X - insulator
Y - semiconductor
Z - conductor
14b.) the band gap between the valance and conduction bands is small.
Some electrons have enough energy to move from the valance to the conduction band.
c.) increases conductivity
d.) (2.3 x 10+3) / (1.7 x 10-8)
= 1.35 x 10+11
Resistivity of silicon is 11 orders of magnitude greater
9a.) a semiconductor that has specific impurities added
b.) the voltage applied causes the electrons to move from the n-type conduction band to the p-type conduction band.
The electrons then drop to the valance band of the p-type and photons are emitted.
3di.) photovoltaic effect
ii.) electrons absorb energy from photons.
Electrons move from the valance band to the conduction band.
Electrons move towards the n-type semiconductor producing a potential difference.
