2013 Checkpoints Chapter 8
ELECTRONICS BASICS
Question 296
This is not a very good question, all they actually want you to do is to read the scale and work out what the value
of the line is.
1.2 Volt is a good approximation.
Question 297
C
This is because it is an AC signal that has been translated up a constant amount. The translation is due to the
DC.
Question 298
This is the simplest case of a voltage divider, the two resistors are the same, so the voltage is split equally
across both. So Vout = 12V
Question 299
The resistors in parallel effectively half the resistance in that part of the circuit. So the resistance across the
output is half the resistance in the other section of the circuit. This means that the potential drop across the
output will be half the potential drop across the other resistor. With a total of 24 V around the circuit. 16V will be
lost across the top resistor and 8 Volt will be lost across the resistors in parallel.
Question 300
The voltage across the single resistor will be 16 Volt. (See previous question)
Question 301
The resistors are now all in series, so there will be a potential drop of 8 Volt across each resistor. This means the
potential drop across the output will be 16 Volt.
Question 302
With a 'short circuit' as one option for the current and a resistor in the other path, all the current will flow through
the short circuit. This means that there will not be a potential difference across the output resistor ∴ Vout = 0
Question 303
With a 'short circuit' as one option for the current and a resistor in the other path, all the current will flow through
the short circuit. This means that there will not be a potential difference across the upper resistor. ∴All the
potential will drop across the output.
∴ Vout = 24V
Question 304
6 Volt
6 Volt
A
1.2 V
5 kΩ resistor
0 Volt
This circuit needs to lose 4.8 Volt across the thermistor.
This is 4 times the potential drop across the 5 kΩ resistor, so the resistance of the thermistor = 4 × 5 kΩ
R = 20 kΩ
Ω
(ANS)
Question 305
The voltage drop across L2 must be 12.0 – 3.0 = 9V.
As the voltage drop is 3 times larger, the resistance must also be three times larger, because the current in both
is the same.
∴ 45 Ω (ANS)
Question 306
From the graph the resistance is 1 000Ω
Question 307
Vthermistor = 8.0 V because the Vvariable = 4.0 V.
∴ Vthermistor = 2 × Vvariable
0
At 5 C Vthermistor = 4000Ω (from graph)
∴Rvariable = 2000Ω (ANS)
Question 308
The refrigeration needs to turn on at a lower temperature.
0
For example, turn on at 4 C. The resistance of the thermistor is now ≈ 5000Ω.
The voltage across the variable resistor still needs to be 4 V.
∴the variable resistor needs still to be ½ the voltage of the thermistor.
∴ variable resistance is now 2500Ω
So R needs to increase (ANS)
Question 309
To find the potential difference across R2 use:
V2
R
= 2
VS RT
V2
30
=
10 70
V2 = 4.3 V (ANS)
Question 310
First step: Find the current through the resistor
V
I= s
RT
10
70
I = 0.1429 A
Then use the power equation
P = I2R
P = 0.1429 2 × 40 (Don’t round off)
I=
P = 0.82 W (ANS)
Question 311
First step: find the effective resistance of the 40 Ω resistor and the 20 Ω resistor in parallel.
1
1
1
=
+
RT R1 R2
1
1
1
=
+
RT
40 20
R T = 13.3 Ω
Second step: Find the overall resistance of the circuit.
R T = 13.3 + 30
R T = 43.3 Ω
Third step: calculate the resistance.
V
I= s
RT
10
43.3
I = 0.23 A (ANS)
I=
Question 312
0
The resistance is read straight from the graph, at 5 C the resistance is 5000 Ω.
Question 313
The voltage across the variable resistor needs to be 8.0 V, this means that the voltage across the thermistor is
0
4.0 V. At 25 C the resistance of the thermistor will be 500 Ω as read from the graph.
Vthemistor Rthermistor
=
Vvariable
Rvariable
4.0 500
=
8.0 R2
∴ R2 = 1000 Ω (ANS)
Question 314
Vthemistor 4.0
=
since the 8.0 volts is what is required to turn on the air
Vvariable
8.0
conditioner. If the air conditioner is to come on at a lower temperature the resistance of the thermistor will be
greater.
The ratio needs to be the same so increasing the thermistor the variable resistor needs to increase to ensure
that the ratio remains constant.
The ratio of the voltages will still be
Question 315
The diode has a maximum voltage drop across it of 1 V. This means that the voltage drop across the 500Ω
resistor will be 6V – 1V = 5V.
Using v = iR gives 5 = i x 500
∴ i = 10 mA. (ANS)
This will be the current in the ammeter, because it is in series with both the diode and the resistor. You need to
double check that this current fits on the I V curve for the diode.
Question 316
The 200Ω resistor and the diode are in parallel. This means that the voltage drop across both these elements
will be the same. The voltage drop across the diode is limited to 1V. This means that the voltage drop across the
500Ω resistor will be 5.0 V.
∴ the current remains at 10 mA
(ANS)
Question 317
It is often best to try to redraw the diagram so that the elements in series and parallel are more obvious.
2Ω
0.50Ω
0.50Ω
2Ω
24 V
The effective resistance of two 2 Ω resistors in parallel is 1 Ω.
(
1
R total
=
1
2
+
1
2
)
This will lead to the circuit looking like this.
0.50Ω
1Ω
0.50Ω
24 V
This means the potential drops around the circuit will be 6 V across each resistor and 12 V across both globes.
Question 318
The total resistance of the circuit is 0.5 + 1 + 0.5 = 2Ω. Using V = iR, gives i = 24/2.
∴ 12 A
(ANS)
Question 319
The voltage drop across the two resistors in parallel will be half the voltage drop across the single resistor. This
means that the drop across the parallel elements will be 4 V and across the single element it will be 8 V.
Question 320
The circuit can be redrawn as
A
The resistance of the parallel components is given by
1
1
1
=
+
RParallel
2
4
∴ Rparallel =
C
B
4
D
3
∴ RTOTAL = 2 + 1.33
= 3.33 Ω
(ANS)
Question 321
10
= 3 A.
3.3
Therefore the current in B will be 2 A and the current in both C and D will be 1 amp.
∴2A
(ANS)
Since the total resistance is 3.3 Ω, the current in the circuit is
Question 322
Use V = iR
where I = 3 A and R = 2 Ω
∴V=6V
(ANS)
Question 323
Use P = VI, where V = 2V and I = 1 A
∴ P = 2W
(ANS)
Since the voltage drop across A is 6 V, the voltage across C + D, must be 4 V combined. ∴ VD = 2V
Question 324
From the graph 1500 Ω (ANS)
Question 325
fan
4500Ω
Switching
circuit