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Physics > Electric Circuits > Flashcards

Electric Circuits Flashcards

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1
Q

Define P.D

A

Energy transferred per unit charge (V = E/Q)

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2
Q

Define EMF

A

Work done per unit charge by the power supply (converting chemical energy into electrical potential energy of the charges)

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3
Q

Define Current

A

Rate of flow of charge

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4
Q

Define resistance

A

Ratio of P.D to current

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5
Q

P.D divider derivation

A
  1. I is same
  2. V1 + V2 + V3 = VT
  3. I = V/R
  4. VT/RT = V1/R1 + V2/R2
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6
Q

Parallel current derivation

A
  1. V is same
  2. V = IR
  3. I1R1 = I2R2
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7
Q

Series resistor derivation

A
  1. I is same
  2. V1 + V2 + V3 = VT
  3. I = V/R
  4. RT = R1 + R2 + R3
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8
Q

Parallel resistor derivation

A
  1. I is same
  2. V1 + V2 + V3 = VT
  3. I = V/R
  4. 1/RT = 1/R1 + 1/R2 + 1/R3
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9
Q

P.D law

A

Sum of the P.D drops = sum of the emf’s around a closed loop within a circuit (conservation of energy)

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10
Q

Current Law

A

Sum of currents into a junction = sum of the currents out of the junction (conservation of charge)

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11
Q

EMF & Ir circuit

A
  • ammeter in series
  • voltmeter parallel to a variable resistor
  • (draw cell resistor)
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12
Q

Define lost volts

A

Energy per unit charge transferred to the internal resistance of a cell

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13
Q

Define power

A

Rate of energy transfer (P = E/t)

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14
Q

Derive P = I2R & P = V2/R

A
  1. P = IV
  2. I = V/R
  3. P = V2/R
  4. P = IV
  5. V = IR
  6. P = I2R
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15
Q

Brighter bulb in series (B) or parallel (A)

A

*P.D in each bulb in A is equal
*P = V2/R, lower resistance bulb has higher power
*So 20Ω bulb is brighter in A

*I in each bulb in B is equal
*Therefore, as P = I2R, the bulb with the higher resistance will have a higher power
*So the 40Ω bulb is brighter in B

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16
Q

no. electrons in a certain amount of charge calculation

A
  • Number of electrons = total charge / charge on one electron (n = Q/e)
  • N = Q/ 1.6 x10 -19
17
Q

I = nqvA

A

I = current
n = number density of free charge carriers (per unit vol.)
q/e = charge on 1 electron
v = (average) drift vel.
A = cross sectional area of current carrying material

18
Q

Derive I = nqvA

A
  • There are N electrons moving with a drift velocity v along a wire of length l and cross sectional area A
  • The current I = ∆Q/∆t
  • ∆Q = Nq where q is the charge on an electron
  • I = Nq/∆t
    *∆t = length/velocity
    *I = Nqv/l
    *N (no. of elctrons) = V (volume) * n (no. density) = Al (volume) * n
  • = Alnqv/l
  • = nqvA
19
Q

Determining resistance of a wire

A
  1. Wire connected to ohmmeter
  2. voltmeter parallel to wire and m ammeter in series, to cell.
20
Q

Wire resistivity practical

A
  1. measure resistance of wire using an ohmmeter for different lengths
  2. measure diameter of wire using micrometer, & average
  3. Calculate the cross sectional area
  4. Plot a graph of resistance against length, compare to equation
  5. resistivity = gradient * area
21
Q

Filament bulb IV relationship

A

*As I increases electrons collide more frequently with lattice resistor ions, transferring more energy/second OR as P.D increases, electrons gain & transfer energy to resistor ions
*Resistor ions to oscillate with greater amplitudes so the temperature of the resistor increases
*Frequent collisions between electrons & lattice ions, limiting drift velocity
*Current, I=nqvA increases by a smaller factor than the P.D (as n, q, A all fixed)
*Resistance is the ratio of pd to current, as pd increases by a greater factor than current does, the ratio R=V/I increases

22
Q

Temp. increase on thermistor resistance

A
  • Temp. increase electrons gain energy and
  • electrons are promoted to conductor
  • n increases
  • I increases by a greater factor than P.D
  • As R = ratio resistance decreases, as current increases by a greater factor than P.D
23
Q

Light increase on LDR resistance

A
  • electrons gain photon energy from light intensity increase
  • electrons are promoted into the conduction band, so n increase
  • Current increases
  • So, as resistance is the ratio of pd to current, if pd remains constant as current increases, resistance = V/I decreases
24
Q

Diode IV graph description

A
  • Diode
  • Allows current to only flow in one direction
  • diode conducts above a threshold voltage
  • resistance is high for negative P.D’s, but will conduct for very high -P.D’s
25
Q

Describe a P.D circuit

A

a circuit involving a fixed P.D supply, shared between 2 resistors

26
Q

Draw 2 P.D divider circuits

A
  1. resistor and variable resistor in series voltmeter parallel to variable resistor
  2. resistor in series and voltmeter in series to variable resistor parallel to resistor
27
Q

Voltmeter change of a decrease in variable resistor resistance in double resistor P.D divider

A
  • Fixed supply of P.D and 1 resistor has fixed resistance
  • When the variable resistor’s resistance increases, the voltmeter reading increases
  • As P.D is shared in the ratio of the resistances
28
Q

Voltmeter change of a decrease in variable resistor resistance in single resistor P.D divider

A
  • at the bottom of resistor voltmeter = 0
  • top of resistor voltmeter = terminal P.D
  • Potential difference is proportional to the length of the wire (as resistance is proportional to length for a constant cross sectional area and resistivity, R=ρl/A, and for a fixed current V is directly proportional to R as V=IR)
29
Q

Better circuit for IV graph

A
  • more parallel wouild ne better
  • P.D can vary from 0-max
  • current can vary from 0-max
  • less parallel circuit would recieve share of P.D & current never = 0
  • min P.D: V = EMF*Rcomponent
  • min current = EMF/RT
30
Q

Define resistivity

A

resistance of a unit cube

31
Q

Resistivity dependence

A

Tempurature

32
Q

Thermistor resistivity change due to temperature

A

Resistivity decreases, as temp increases, as n increases

33
Q

Filament resistivity change due to temp increase

A

resistivity increases, as v decreases due to more frequent collisions, so I decreases, which increases R

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