electrical circuits (topic 2)

Question 1

Answer 1

Question 2

calc the pd across the lamp (pd = current x resistance) current is 0.02A

Answer 2

0.02 x 150 = 3V

Question 3

calc the pd across the resistor, pd of lamp is 3

Answer 3

• potential difference across lamp + potential difference across resistor = 12 V
• 3 (voltage of lamp) + voltage across resistor = 12 V
• voltage across resistor = 12 - 3 = 9V

Question 4

state the reading on the voltmeter

Answer 4

10

Question 5

state the voltage across the 300Ω resistor

Answer 5

10

Question 6

calculate the current through the 200Ω resistor, current = pd/resistance

Answer 6

10/200 = 0.05A

Question 7

the 300 Ω resistor has a current of 33.3 mA flowing through it, calculate the total circuit current
(0.05 is the current through the 200 resistor)

Answer 7

• total circuit current = current through 200 Ω + current through 300 Ω
• total circuit current = 0.05 + 0.0333 = 0.0833 A

Question 8

Answer 8

Question 9

explain why the student uses a variable resistor and not a fixed resistor (2) (a student is investigating the current-voltage characteristic of a filament lamp)

Answer 9

• variable resistor allows the circuit resistance to vary
• varying the circuit resistance allows the current to be varied

Question 10

describe what happens to the resistance of the lamp as the voltage is increased from 0V to 10V

Answer 10

• between 0 V and 4 V, the current increases linearly
• constant resistance between 0 V and 4 V
• between 4 V and 10 V, the current increases at a decreasing rate
• resistance increases between 4 V and 10 V

Question 11

pd = current x resistance

Answer 11

• 12 - 5
• 7/350 = 0.02A

Question 12

the student moves the circuit into a warmer room, describe what happens to the voltage across the lamp, you may assume the resistance of the lamp stays constant (3)

Answer 12

• resistance of thermistor decreases (resistance of a thermistor decreases as temp increases)
• circuit current increases
• potential difference across lamp increases
• potential difference across thermistor decreases

Question 13

Answer 13

Question 14

Answer 14

Question 15

explain what is meant by a D.C. power supply (2)

Answer 15

• direct current
• current flows continuously around circuit in the same direction

Question 16

a student is investigating how the resistance of a wire varies with length, explain why the student must turn the power supply off between readings (3-

Answer 16

• when a current flows through the wire it causes it to heat up
• turning off the power supply between readings reduces the heating effect of the current
• an increase in temperature would increase the resistance of the wire

Question 17

a student is investigating how the resistance of a wire varies with length, describe how the student could use the circuit to identify the relationship between length and resistance (6)

Answer 17

• using a ruler and crocodile clips, set the length of wire to 10 cm
• turn on power supply and record voltage and current values
• turn off power supply and increase length of wire by 5 cm
• repeat experiment and calculate average values
• resistance = voltage/current
• plot a graph of length against resistance

Question 18

the student concludes that resistance is directly proportional to the length of the wire, using data from the graph, show that resistance is directly proportional to length (3)

Answer 18

• resistance/length = constant
• use of one set of values to calculate constant = 2 eg 60/30
• use of second set of values to calculate constant = 2 eg 90/45
• constant is the same, therefore resistance is directly proportional to length

Question 19

Answer 19

• ammeter reading = total circuit current - current through top branch
• ammeter reading = 2.66 - 0.66 = 2A

Question 20

the student opens the switch, describe what happens to the reading on the ammeter and the voltmeter

Answer 20

• ammeter reading drops to zero as no current flows through the bottom branch
• voltmeter reading is unchanged

Question 21

calculate the reading on the voltmeter

Answer 21

• voltage = current x resistance
• v = 0.66 x 5 = 3.33

Question 22

the 10 Ω resistor is replaced with a thermistor, the student places the circuit in a cold room, describe how the voltmeter and ammeter reading will change (4)

Answer 22

• ammeter reading will remain the same
• resistance of thermistor increases
• lower current through top branch
• potential difference across the 5 Ω resistor decreases/voltmeter reading decreases

Question 23

define current

Answer 23

the rate of flow of charge

Question 24

a filament lamp has a current of 275 mA flowing through it for 2 hours, calculate the charge that flows through the filament lamp

Answer 24

Question 25

the filament lamp is connected to a 10 V cell, calculate the energy transferred to the filament lamp in 2 hours (charge is 1980 C)

Answer 25

Question 26

Answer 26

- current through S = total circuit current - current through R - current through S = 325 - 200 = 125 mA

Question 27

(current flowing through S is 125mA, 0.125A)

Answer 27

Question 28

calculate the potential difference across T (pd across S = 2.5)

Answer 28

- potential difference across T + potential difference across S = 10 V - potential difference across T = 10 - 2.5 = 7.5 V

Question 29

calculate the resistance of T at this temperature (voltage = 7.5)

Answer 29

Question 30

the student moves the circuit into a warm room, describe what happens to the potential difference across T

Answer 30

- thermistor resistance decreases - current through S and T increases - higher potential difference across S - lower potential difference across T

Question 31

Answer 31

Question 32

describe the shape of the I-V graph (3)

Answer 32

- no current when a negative voltage is applied - diode begins to conduct after 0.6 V - above 0.6 V current increases linearly

Question 33

when the voltage across the diode is 1.2 V, the current flowing through it is 17.6 mA, alculate the resistance of the resistor ?????

Answer 33

Question 34

Answer 34

all turned off

Question 35

describe the changes the student would need to make to the switches to turn on bulbs 3 and 4 (1)

Answer 35

both switches C and D would need to be closed

Question 36

explain the advantage of using parallel circuits for lighting control (2)

Answer 36

- if the bulb blows in one branch, the other branch will still operate normally - light bulbs can be switched on/off independently

Question 37

- the student closes switch A and B - after a short period of time the bulbs stop working - the student suspects that only one of the bulbs has blown - describe how the student could identify which bulb has blown (4)

Answer 37

- close switches C and D and ensure both bulbs light - open switch C and replace bulb 3 or 4 with bulb 1 - close switch C. If bulb 1 does not light, it has blown - repeat for bulb 2