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G482
ELECTRONS, PHOTONS AND
WAVES MODULE 3: DC Circuits
Answer Booklet
©2011 science-spark.co.uk
RAB Plymstock School
Lesson 15 Questions – Kirchoff’s Laws
1)a) State Kirchoff’s Second Law
……The sum of the E.M.F.s around a loop of a circuit equals the sum of the P.D.s around
a circuit.……………………………………………………………………………………
………………………………………………………………………………………… (2)
Fig.1.1 shows a circuit:
fig 1.1
b)
Use Kirchoff’s second law to determine the p.d. V.
…………4.5-1.5=V+0.2…………………………………………………………………
……………V=2.8V………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………… (2)
Total [4]
2)a) Kirchoff’s first law is based on the conservation of an electrical quantity. State the
law and the quantity conserved.
………The sum of the currents entering a point will equal the sum of the currents exiting
that point. ……………………………………………………………
………Conservation of charge……………………………………………………
………………………………………………………………………………………… (2)
b)
Fig 2.1 shows a potential divider circuit.
fig 2.1
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i)
Show that the voltmeter reading is 1.5V
…………………6=I x 1200 + I x 400……………………………………………………
…………………6=1600I…………………………………………………………………
…………………I=3.75x10-3…
So V=3.75x10-3 x 400 = 1.5V…………………………………………………… (2)
Total [4]
3)
Fig 3.1 shows part of an electric circuit.
fig 3.1
a)
Name the component marked X
…………Thermistor…………………………………………………………………… (1)
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 16 questions – resistors in series and parallel
1
Calculate the total resistance of the following resistance combinations:
a)
4Ω
12Ω
……………Total Resistance = R1 + R2 …………………………………………………
………………………………= 4 + 12………………………………………
………………………………= 16 Ω ………………………………………………… (1)
b)
4Ω
12Ω
………………RTot = R1xR2/(R1+R2) …………………………………………
……………………= (4 x 12)/(4 + 12)…………………………………………………
……………………= 48/16
=3Ω …………………………………………………………… (2)
c)
3Ω
2Ω
6Ω
…………Parallel section Resistance total = R1xR2/(R1+R2) ……………………
……………………………………………= (3 x 6)/(3 + 6)………………………
……………………………………………= 18/9
……………………………………………=2Ω ………………………
…………Add this to other resistor in series with these ………………
Total resistance = 2Ω +2Ω = 4Ω …………………………………………………… (3)
Total [6]
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RAB Plymstock School
2)
Calculate the total resistance between the points X and Y for the circuit shown in
fig. 1.1
fig1.1
……180Ω and 120Ω are parallel to each other so:…………………………………………
………………… Parallel section Resistance total = R1xR2/(R1+R2) ……………………
……………………………………………= (180 x 120)/(180 + 120)……………………
……………………………………………= 21600/300
……………………………………………=72Ω
……………………………………………………………………………
……These are in series with 100Ω resistor so add this on, ………………………
Total resistance =172Ω ……………………………………………………………… (3)
Total [3]
3)
A circuit of a combination of resistors is shown in figure 3.1.
6V
3Ω
4Ω
6Ω
fig 3.1
Calculate:
a)
the combined resistance of the 3Ω resistor and the 6Ω resistor in parallel,
Parallel section Resistance total = R1xR2/(R1+R2) ……………………
……………………………………………= (3 x 6)/(3 + 6)………………………
……………………………………………= 18/9
……………………………………………=2Ω ……………………………………… (2)
b)
the total resistance of the circuit,
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RAB Plymstock School
…………………2Ω + 4Ω = 6Ω…………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………… (1)
c)
the battery current,
…………………I=V/R……………………………………………………………………
…………………=6/6……………………………………………………………………
…………………=1A………………………………………………………………… (2)
d)
the power supplied to the 4Ω resistor.
………………………………………………………………………………………………
………P=I2R………………………………………………………………………………
………=12x4 ………………………………………………………………………………
………=4W…………………………………………………………………………… (3)
Total [8]
©2011 science-spark.co.uk
RAB Plymstock School
Lesson 17 questions – Internal resistance
1)
Fig. 1.1 shows two soldered resistors connected across terminals of a cell.
fig 1.1
The cell has a resistance of 0.8Ω and e.m.f. 1.5V.
a)
Define e.m.f. in terms of energy transformed and electric charge.
………e.m.f is the total energy gained per unit charge………………………
………Energy transformed into electrical………………………………………………
………………………………………………………………………………………… (2)
b)
Suggest why a cell has internal resistance.
……………The chemicals in the cell……………………………………………… (1)
c)
Calculate the total resistance R of the circuit in Fig. 1.1.
………………R=R1+R2…………………………………………………………
…………………=0.8+0.6………………………………………………………………
…………………=1.4Ω……………………………………………………………………
………………………………………………………………………………………… (2)
d)
Hence calculate the current I in the circuit.
……………I=1.5/1.4………………………………………………………………………
……………=1.07A………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………… (2)
e)i)
Write an equation for the power dissipated by a current-carrying resistor.
……………P=VI=I2R=V2/R……………………………………………………………
………………………………………………………………………………………………
ii)
For the circuit in Fig. 1.1, calculate the ratio:
power dissipated by internal resistance__
power dissipated by total external resistance
I2(internal R)/I2(external R)= (internal R)/ (external R)=0.8/0.6=4/3
ratio=…1.33……(4)
Total [7]
©2011 science-spark.co.uk
RAB Plymstock School
2)
Fig. 2.1 shows a cell of e.m.f. E and internal resistance r connected to a variable
resistor.
fig 2.1
fig 2.2
Fig. 2.2 shows the variation of the p.d. V across the terminals of the cell with the current I
drawn from the cell.
a)
Explain how fig 2.2 shows that the e.m.f. E is 1.4V
………intercept with y axis (V=E-Ir, since at intercept I=0)………………………………
………………………………………………………………………………………… (1)
b)i)
Use fig. 2.2 to determine the maximum possible current that can be drawn from
the cell.
………Graph extrapolated = 2A (+/-0.1A)……………………………………………
………………………………………………………………………………………… (1)
ii)
Calculate the internal resistance r of the cell.
………gradient = r (ignore sign)……………………………………………
………………= 0.7(0)Ω………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………… (2)
Total [5]
3)
Fig. 3.1 shows the I/V characteristics for a tungsten filament lamp.
fig 3.11
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A 5.0Ω resistor and the tungsten filament lamp are connected in series to a d.c. power
supply of e.m.f. 24V. The current drawn from the power supply is 2.0 A.
a)
Calculate the total power delivered by the supply.
…………P=VI……………………………………………………………………………
………… =24x2…………………………………………………………………………
…………=48W………………………………………………………………………… (2)
b)
Use fig 3.1 to determine the resistance of the filament lamp when the current in it
is 2.0 A.
…………R=V/I……………………………………………………………………………
……………=12/2…………………………………………………………………………
……………=6Ω……………………………………………………………………… (2)
c)
Calculate the total resistance of the series combination of the filament lamp and
the resistor.
…………R=R1+R2………………………………………………………………………
…………=6+5……………………………………………………………………………
…………=11Ω………………………………………………………………………… (1)
d)
Calculate the internal resistance of the supply.
…………E=IR+Ir…………………………………………………………………………
…………r=(E=IR)/I………………………………………………………………………
…………=(24-(2x11))/2 =1.0Ω……………………………………………………… (2)
Total [7]
4)
Fig 4.1 shows a car battery of e.m.f 12V and internal resistance 0.014Ω connected
to the starter motor of a car. When the car engine is being started, the car battery provides
a currnt of 160A to the starter motor.
fig 4.1
a)
Show that the p.d. across the internal resistance is about 2.2V.
…………………V=0.014x160……………………………………………………………
……………………=2.24V………………………………………………………………
………………………………………………………………………………………… (1)
b)
Determine the terminal p.d. across the battery.
……………………V=12-2.24……………………………………………………………
………………………=9.76V…………………………………………………………
………………………………………………………………………………………… (1)
Total [2]
©2011 science-spark.co.uk
RAB Plymstock School
Lesson 14 questions – Potential dividers
1)
A potential divider circuit based on a light-dependent resistor (LDR) is shown in
fig 1.1. The supply has negligible internal resistance.
fig 1.1
a)
The light intensity falling upon the LDR is increased. State:
i)
how the resistance of the LDR changes,
………………………Goes down……………………………………………………… (1)
ii)
how the p.d. across the LDR changes.
……………………… Goes down …………………………………………………… (1)
b)
At a particular light intensity, the resistance of the LDR is 420Ω.
i)
Calculate the p.d. across the LDR.
………………V1=Vs(R1/(R1+R2))…………………………………………………………
……………….…=5 x 420/(420 + 1000)
= 1.5V……………………………………………… (3)
ii)
An electronic circuit of resistance 50Ω is connected between the terminals X and
Y.
1.
Show that the total resistance of the parallel combination of this electronic
circuit and the LDR is about 45Ω.
………………Rtot=R1R2/(R1+R2)…………………………………………………………
……………….…=(50x420)/(50+420)……………………………………………………
…………………=45.16~45Ω…………… (1)
2.
Calculate the p.d. across the electronic circuit.
……………… V1=Vs(R1/(R1+R2))………………………………………………….…
=5 x 45/(45 + 1000)
= 0.2V……………………………………………… (2)
Total [8]
©2011 science-spark.co.uk
RAB Plymstock School
2)
Fig 2.1 shows a potential divider circuit. The battery has negligible internal
resistance and the voltmeter has a very high resistance.
fig2.1
i)
Show that the voltmeter reads 1.5V.
………………… V1=Vs(R1/(R1+R2))……………………………………………………
……………………=6(400/(400+1200))………………………………………………
……………………=1.5V…………………………………………………………… (2)
ii)
An electric device rated at 1.5V, 0.1A is connected between the terminals X and
Y. The device has constant resistance. The voltmeter reading drops to a very low value
and the device fails to operate, even though the device itself is not faulty.
1.
Calculate the total resistance of the device and the 400Ω resistor in
parallel.
…………………… R=V/I
=1.5/0.1
=15Ω
Rtot=R1R2/(R1+R2)
………………………=(15x400)/(15+400)………………………………………………
………………………=14.5Ω………………………………………………………… (3)
2.
Calculate the p.d. across the device when it is connected between X and Y.
……………………… V1=Vs(R1/(R1+R2))…………………………………
……………………
……………………
=6(14.5/(14.5+1200))……………………………………………
=0.07V……………………………………………………… (2)
3.
Why does the device fail to operate?
…………………p.d. across device is small………………………………………………
(1)
Total [8]
3)a) Fig 3.1 shows the circuit symbol for a component.
fig 3.1
©2011 science-spark.co.uk
RAB Plymstock School
Name the component and state how its resistance changes as the temperature of the
component is increased.
……Thermistor……………………………………………………………………………
……Resistance goes down as temperature increases……………………………………
………………………………………………………………………………………………
………………………………………………………………………………………… (2)
b)
Fig 3.2 shows a potential-divider circuit. The battery has a negligible internal
resistance and the voltmeter has a very high resistance.
fig 3.2
i)
At a particular temperature, the resistance of component X is 4.2kΩ. Calculate the
voltmeter reading.
……………………… V1=Vs(R1/(R1+R2))…………………………………
……………………
=9.0(4200/(4200+1200))…………………………………………
……………………
=7.0V
………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………… (3)
ii)
State how your answer to (b)(ii) changes when the temperature of the component
X is increased.
………………………………Decreases…………………………………………………
………………………………………………………………………………………… (1)
Total [6]
©2011 science-spark.co.uk
RAB Plymstock School
Lesson 18 questions – Potential dividers
1)
A potential divider circuit based on a light-dependent resistor (LDR) is shown in
fig 1.1. The supply has negligible internal resistance.
fig 1.1
a)
The light intensity falling upon the LDR is increased. State:
i)
how the resistance of the LDR changes,
………………………Goes down……………………………………………………… (1)
ii)
how the p.d. across the LDR changes.
……………………… Goes down …………………………………………………… (1)
b)
At a particular light intensity, the resistance of the LDR is 420Ω.
i)
Calculate the p.d. across the LDR.
………………V1=Vs(R1/(R1+R2))…………………………………………………………
……………….…=5 x 420/(420 + 1000)
= 1.5V……………………………………………… (3)
ii)
An electronic circuit of resistance 50Ω is connected between the terminals X and
Y.
1.
Show that the total resistance of the parallel combination of this electronic
circuit and the LDR is about 45Ω.
………………Rtot=R1R2/(R1+R2)…………………………………………………………
……………….…=(50x420)/(50+420)……………………………………………………
…………………=45.16~45Ω…………… (1)
2.
Calculate the p.d. across the electronic circuit.
……………… V1=Vs(R1/(R1+R2))………………………………………………….…
=5 x 45/(45 + 1000)
= 0.2V……………………………………………… (2)
Total [8]
©2011 science-spark.co.uk
RAB Plymstock School
2)
Fig 2.1 shows a potential divider circuit. The battery has negligible internal
resistance and the voltmeter has a very high resistance.
fig2.1
i)
Show that the voltmeter reads 1.5V.
………………… V1=Vs(R1/(R1+R2))……………………………………………………
……………………=6(400/(400+1200))………………………………………………
……………………=1.5V…………………………………………………………… (2)
ii)
An electric device rated at 1.5V, 0.1A is connected between the terminals X and
Y. The device has constant resistance. The voltmeter reading drops to a very low value
and the device fails to operate, even though the device itself is not faulty.
1.
Calculate the total resistance of the device and the 400Ω resistor in
parallel.
…………………… R=V/I
=1.5/0.1
=15Ω
Rtot=R1R2/(R1+R2)
………………………=(15x400)/(15+400)………………………………………………
………………………=14.5Ω………………………………………………………… (3)
2.
Calculate the p.d. across the device when it is connected between X and Y.
……………………… V1=Vs(R1/(R1+R2))…………………………………
……………………
……………………
=6(14.5/(14.5+1200))……………………………………………
=0.07V……………………………………………………… (2)
3.
Why does the device fail to operate?
…………………p.d. across device is small………………………………………………
(1)
Total [8]
3)a) Fig 3.1 shows the circuit symbol for a component.
fig 3.1
©2011 science-spark.co.uk
RAB Plymstock School
Name the component and state how its resistance changes as the temperature of the
component is increased.
……Thermistor……………………………………………………………………………
……Resistance goes down as temperature increases……………………………………
………………………………………………………………………………………………
………………………………………………………………………………………… (2)
b)
Fig 3.2 shows a potential-divider circuit. The battery has a negligible internal
resistance and the voltmeter has a very high resistance.
fig 3.2
i)
At a particular temperature, the resistance of component X is 4.2kΩ. Calculate the
voltmeter reading.
……………………… V1=Vs(R1/(R1+R2))…………………………………
……………………
=9.0(4200/(4200+1200))…………………………………………
……………………
=7.0V
………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………… (3)
ii)
State how your answer to (b)(ii) changes when the temperature of the component
X is increased.
………………………………Decreases…………………………………………………
………………………………………………………………………………………… (1)
Total [6]
©2011 science-spark.co.uk
RAB Plymstock School