Lesson 3 – Current and charge - science

advertisement
science-spark.co.uk
G482
ELECTRONS, PHOTONS AND
WAVES MODULE 1: ELECTRIC
CURRENT
Answer Booklet
©2011 science-spark.co.uk
RAB Plymstock School
Lesson 2 – Current and charge. ΔQ=IΔt.
1. Fig.1.1 shows a lightning strike between a cloud and the ground.
a)
i)
The current in the lightning strike is 7800A. The strike lasts for a time of 230ms.
Calculate
1
the charge flowing between the cloud and the ground
………
ΔQ=IΔt ……………
=7800 x .23……………………………………………………
…………………=1794C ~1.8kC…………………………………………
………………………………………………………………………………………… (3)
2
the number of electrons transferred to the ground.
……………Total charge = 1.8kC, charge per electron = 1.6 x 10-19C
……………so, number of electrons = charge/charge per electron
……………1800 / 1.6 x 10-19 = 1.1x1022 electrons…………… (2)
Total [5]
2)a) Name an instrument used to measure
i)
Electric current
………………Ammeter……………………………………………………………… (1)
ii)
potential difference
………………Voltmeter………………………………………………… (1)
b)
The electric charge ΔQ passing a point in a circuit is given by the equation
ΔQ=IΔt
©2011 science-spark.co.uk
RAB Plymstock School
State what is represented by the other symbols I and Δt.
I: ………Current………………………………………………………… (1)
Δt: ………Change in time……………………………………… (1)
c)
A water heater is switched on for 1500s. During this time, a charge of 7.5x103 C
passes. Calculate the electric current,
ΔQ=IΔt, I = ΔQ/Δt………………………………………………………………………
……………= 7.5/1.5 = 5A …………………………………………………
………………………………………………………………………………………… (2)
Total [6]
3)a)i) State what is meant by electric current
…………Rate of change of charge …………………………………………………… (1)
ii)
A mobile phone is connected to a charger for 600s. The charger delivers a
constant current 350mA during this interval. Calculate the total charge supplied to the
mobile phone.
…………… ΔQ=IΔt …………………
= 0.35 x 600………………………………………………………………
…………………= 210 C ……………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………… (3)
Total [4]
©2011 science-spark.co.uk
RAB Plymstock School
Lesson 3 Questions – Kirchoff’s first Law
1)
i)
Kirchoff’s first law is based on the conservation of an electrical quantity.
State the law and the quantity conserved.
………Kirchoff’s 1st law says that the sum of the currents entering a point equal the sum
of the current leaving a point.
Since current is the charge in a given time, charge must be conserved.… (2)
ii)
Determine the current I in each of the circuits below.
a)………………2+1=3A………………………………………………………………
………………………………………………………………………………………………
b)………………0.5+0.7=1.2A……………………………………………………………
………………………………………………………………………………………………
c)………………I + 2.0 = 4.0 + 0.5 …………………
I=2.5A………………………………………………………………………
Total [6]
2)
Fig 2.1 shows part of an electric circuit.
fig 2.1
a)
Name the component marked X
………………Thermistor……………………………………………………………… (1)
©2011 science-spark.co.uk
RAB Plymstock School
b)
Determine the magnitude of the currents I1, I2 and I3.
I1 =……51………………mA
I2 =……9..………………mA
I3 =……29………………mA
(3)
Total [4]
©2011 science-spark.co.uk
RAB Plymstock School
Lesson 4 questions - Current Equation
1
(a) Explain in terms of band theory what is meant by a free electron.
....................... An electron in the conduction band (of the copper);................................
........................ able to take part in conduction....................................................[2]
(b) Describe in detail the motion of free electrons in a copper wire
when there is no current in the wire
when there is a current in the wire.
Your answer should include the meanings of root-mean-square speed and drift velocity
and the factors that determine them.
With no current in the wire:
r.m.s.speed is square root of mean of squares of the speed of free electrons; (1)
r.m.s. speed depends on / increases with increasing temperature; (1)
free electrons move (fast) in random directions; (1)
colliding with atoms (in the lattice); (1)......................................................................
.......................................................................................................................................
With current in the wire:
free electrons move in opposite direction to current / electric field; (1)
free electrons accelerate between collisions with atoms; (1)
this motion is superimposed on the random motion; (1)
Drift velocity is the mean value of electrons’ velocities due to this motion; (1)
value depending on current, cross-section of wire, free electron concentration
and electron charge / I, A, n and e. (1) max [7].....................................
.......................................................................................................................................
...................................................................................................................................[7]
(c) A current of 0.75 A is carried in a copper wire of cross-sectional area
4.0 x10–7 m2.
The drift velocity of free electrons in the wire is 1.4 x 10–4 m s–1.
(i)
Calculate n, the number of free electrons per unit volume in
copper.
n = I/Aev (1)
-7
-19
-4
28
= 0.75 / (4.0 x 10 x 1.6 x 10 x 1.4 x 10 ) = 8.4 x 10 (1)
(ii)
Calculate the new drift velocity when
1
the current is changed to 0.25 A in the same wire
-5
drift velocity = ........ 4.7 x 10 ....... m s–1 [1]
2
a current of 0.75 A is carried in a copper wire of twice the diameter.
-5
drift velocity = ........ 3.5 x 10 ...... m s–1 [1]
©2011 science-spark.co.uk
RAB Plymstock School
Download