Physics Electricity

Question 1

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.

Answer 1

The LEDs only emit light when they are forward biased.
The change in polarity of the voltage changes the biasing.

Question 2

Determine the peak voltage of the output signal generator.

Answer 2

Vpeak= (no. of boxes * y-gain setting)

Question 3

Determine the frequency of the output of the signal generator.

Answer 3

f= 1/T
T= No. of boxes for a complete wave

Question 4

The power dissipated in a 120 Ω resistor is 4.8 W. What is the current in the resistor

Answer 4

P= I²R
I²= P/R
I = √P/R > √4.8/120
I = 0.2 A

Question 5

State what is meant by the term electromotive force (e.m.f.)

Answer 5

The energy given to each coulomb of charge passing through the battery.

Question 6

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.

Answer 6

The electrons in the n-type conduction band do not gain enough energy to move towards the conduction band of the p-type.

Question 7

Answer 7

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

Question 8

Answer 8

r = - gradient
r = - (y2 - y1/ x2 - x1)

Question 9

Answer 9

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Question 10

Answer 10

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

Question 11

Answer 11

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.

Question 12

Answer 12

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Question 13

Answer 13

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Question 14

Answer 14

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Question 15

Answer 15

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Question 16

Answer 16

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Question 17

Answer 17

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Question 18

Answer 18

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Question 19

Answer 19

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Question 20

Answer 20

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Question 21

The resistance of the variable resistor is now increased. State what happens to the reading of the voltmeter. Justify your answer

Answer 21

The reading on the voltmeter increases because the current in the circuit decreases, reducing the voltage lost across the internal resistance of the cells.

Question 22

Answer 22

Adjust the variable resistor and take readings of V and I

Plot a graph of V against I

Gradient of the graph = - r

Question 23

Answer 23

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

Question 24

Answer 24

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

Question 25

Answer 25

D

Question 26

Answer 26

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

Question 27

Answer 27

19.) E 20.) B C = Q/V

Question 28

Answer 28

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

Question 29

Answer 29

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

Question 30

Answer 30

20.) C 21.) E

Question 31

Answer 31

22.) A 23.) A

Question 32

Answer 32

24.) E 25.) D

Question 33

Answer 33

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.

Question 34

Answer 34

22.) C 23.) D

Question 35

Answer 35

24.) B 25.) E

Question 36

Answer 36

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

Question 37

Answer 37

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

Question 38

Answer 38

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

Question 39

Answer 39

24.) C

Question 40

Answer 40

Photovoltaic effect

Question 41

Answer 41

P = IV P = (35x10-3)x2.1 P = 0.074W

Question 42

Answer 42

Greater number of photons strike the solar cell per second

Question 43

Answer 43

17.) D Q = C x V 18.) D E = 1/2CV(squared)

Question 44

Answer 44

19.) E

Question 45

Answer 45

2a.) P = V(squared)/R > 12(squared)/9.6 = 15W

Question 46

Answer 46

21.) D 1/6 + 1/6 = 3 3 + 8 = 11

Question 47

Answer 47

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

Question 48

Answer 48

25.) D

Question 49

Answer 49

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.

Question 50

Answer 50

10bi.) 12.6V ii.) gradient = -r gradient = (12.2 - 12.4) / (40 - 20) = -0.01 Internal resistance = 0.01 ohms

Question 51

Answer 51

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

Question 52

Answer 52

C = Q/V Q = CV Q = (64 x 10-6) x ( 2.5 x 10+3) Q = 0.16 C

Question 53

Answer 53

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

Question 54

Answer 54

ii.) the voltage decreases iii.) smaller initial current, time to reach 0A is longer

Question 55

Answer 55

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

Question 56

Answer 56

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

Question 57

Answer 57

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.

Question 58

Answer 58

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

Question 59

Answer 59

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

Question 60

Answer 60

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

Question 61

Answer 61

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

Question 62

Answer 62

Gradient = - r Gradient = ( 310x10-3 - 600x10-3 / 100x10-6 - 20x10-6) Gradient = -3625 r = 3625 ohms

Question 63

Answer 63

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

Question 64

Answer 64

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

Question 65

Answer 65

C = Q/V Q = CV Q = (47 x 10-6) x (6) Q = 2.82 x 10-4 C

Question 66

Answer 66

Question 67

Answer 67

As resistance decreases current increases. Lost volts increases, terminal potential difference decreases

Question 68

Answer 68

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

Question 69

Answer 69

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

Question 70

Answer 70

15.) material 2 Material 1 is an insulator Material 3 is a conductor

Question 71

Answer 71

X - insulator Y - semiconductor Z - conductor

Question 72

Answer 72

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

Question 73

Answer 73

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.

Question 74

Answer 74

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.