ELECTRICITY Flashcards

1
Q

Potential Divider:

A

A combination of resistors in series connected by a voltage source.(To produce a required pd.)

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2
Q

Series circuit with a Variable Resistor(R) and a thermistor with a voltmeter connected across the terminals of the thermistor.
State and explain what happens to the voltmeter reading when the resistance of R is
increased while the temperature is kept constant. (3 marks)

A
  • As R increases, more pd is drawn to R so pd across R increases
  • Supply pd=Pd across R+ Pd across T
  • So pd across T decreases and therefore voltmeter reading decreases.
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3
Q

Series circuit with a Variable Resistor(R) and a thermistor with a voltmeter connected across the terminals of the thermistor.
State and explain what happens to the ammeter reading when the temperature of the
thermistor increases. (2 marks)

A
  • As temperature of the thermistor increases, resistance of thermistor decreases
  • Therefore circuit resistance is lower as a whole, I=V/R So current reading at ammeter increases
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4
Q

Is the value of the current through one component the same as the overall current in the circuit, AND the same value of current through other components?

A

Yes.
Example:
Circuit of 12v emf. A thermistor has a resistance of 2.5x10^3 Ohms when 3v is across it.
I=V/R =3/2.5x10^3 =1.2x10^-3A
Second resistor must have pd of 9v, and has a resistance of 7.5x10^3 .
Overall current: I=V/R =12/(2.5x10^3+7.5x10^3)=1.2x10^-3A

Checking current of resistor I=9/7.5x10^3=1.2x10^-3A
So in a series circuit, CURRENT IS THE SAME IN EACH COMPONENT.

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5
Q

The battery has an emf of 12.0 V. At a temperature of 0 °C the resistance of the
thermistor is 2.5 × 103 Ω.
The voltmeter is replaced by an alarm that sounds when the voltage across it exceeds
3.0 V.
State one change that you would make to the circuit so that instead of the alarm
coming on when the temperature falls, it comes on when the temperature rises above
a certain value. (1 Mark)

A

-Connect the alarm across variable resistor (R) rather than Thermistor (T).

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6
Q

CIRCUIT RULES:

A

1)Voltage is divided out in a series circuit (Potential divider)
2)Voltage is equal in each branch in a paralell circuit (12v emf battery, 12v each branch)
3)Voltage gets ‘used up’ (boiler analogy)
4)Current is the same at each point in a series circuit
5)Current is split in paralell
ALWAYS USE SIMPLEST METHOD

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7
Q

A voltmeter is attached across a 8 ohm resistor, and the circuit also contains a 4 ohm resistor. Battery has emf of 3v.
The voltmeter reads 1.8v. Calculate the resistance of the voltmeter.

A

-Treat the voltmeter and the component it is attached to as parallel
-Use the 4 ohm resistor to work out current.
3-1.8= 1.2v for 4 ohm resistor
I=V/R = 1.2/4 = 0.3A
Total resistance of parallel voltmeter and resistor: R=V/I = 1.8/0.3 = 6 ohms.
1/Rt=1/8 + 1/Rv
1/6=1/8 + 1/Rv
Rv=24 Ohms.

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8
Q

Branched Resistors in parallel In a series circuit:

A

-Current splits along the branches
(eg 2A in, one branch gets 1.3A the other 0.7A
-Calculations of individual resistors in that branch use THE INDIVIDUAL RESISTORS RESISTANCE (eg 4v across the whole branch, resistor Y has resistance of 3 Ohms work out current. I=4/3 = 1.37A)

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9
Q

State and explain one reason why it is an advantage for a rechargeable battery to have a low internal resistance. (2 marks)

A
  • Internal resistance limits current

- Hence can provide higher current

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10
Q

EMF:

A

Work done per unit charge.

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11
Q

Current

A

The rate of flow of charge

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12
Q

State and explain why it is important for car batteries to have a very low internal resistance. (2 Marks)

A
  • A large current is needed to start the car

- Internal resistance limits the flow of current therefore needs to be low

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13
Q

When does internal resistance act?

A

When a current is flowing - Internal resistance acts

When no current is flowing (eg switch is open)- No Internal resistance, therefore emf value

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14
Q

Time period:

A

The time for one cycle of a wave to complete (eg time for two half wavelengths to complete) (Think AC question)

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