electrical circuits (topic 2) Flashcards
calc the pd across the lamp (pd = current x resistance) current is 0.02A
0.02 x 150 = 3V
calc the pd across the resistor, pd of lamp is 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
state the reading on the voltmeter
10
state the voltage across the 300Ω resistor
10
calculate the current through the 200Ω resistor, current = pd/resistance
10/200 = 0.05A
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)
- total circuit current = current through 200 Ω + current through 300 Ω
- total circuit current = 0.05 + 0.0333 = 0.0833 A
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)
- variable resistor allows the circuit resistance to vary
- varying the circuit resistance allows the current to be varied
describe what happens to the resistance of the lamp as the voltage is increased from 0V to 10V
- 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
pd = current x resistance
- 12 - 5
- 7/350 = 0.02A
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)
- 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
explain what is meant by a D.C. power supply (2)
- direct current
- current flows continuously around circuit in the same direction
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-
- 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
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)
- 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
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)
- 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
- ammeter reading = total circuit current - current through top branch
- ammeter reading = 2.66 - 0.66 = 2A
the student opens the switch, describe what happens to the reading on the ammeter and the voltmeter
- ammeter reading drops to zero as no current flows through the bottom branch
- voltmeter reading is unchanged
calculate the reading on the voltmeter
- voltage = current x resistance
- v = 0.66 x 5 = 3.33
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)
- 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
define current
the rate of flow of charge
a filament lamp has a current of 275 mA flowing through it for 2 hours, calculate the charge that flows through the filament lamp
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)
- current through S = total circuit current - current through R
- current through S = 325 - 200 = 125 mA
(current flowing through S is 125mA, 0.125A)
calculate the potential difference across T (pd across S = 2.5)
- potential difference across T + potential difference across S = 10 V
- potential difference across T = 10 - 2.5 = 7.5 V
calculate the resistance of T at this temperature (voltage = 7.5)
the student moves the circuit into a warm room, describe what happens to the potential difference across T
- thermistor resistance decreases
- current through S and T increases
- higher potential difference across S
- lower potential difference across T
describe the shape of the I-V graph (3)
- no current when a negative voltage is applied
- diode begins to conduct after 0.6 V
- above 0.6 V current increases linearly
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 ?????
all turned off
describe the changes the student would need to make to the switches to turn on bulbs 3 and 4 (1)
both switches C and D would need to be closed
explain the advantage of using parallel circuits for lighting control (2)
- if the bulb blows in one branch, the other branch will still operate normally
- light bulbs can be switched on/off independently
- 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)
- 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
