3. Warm-up
1 Margaret connects 10
bulbs in a circuit and the
light of the bulbs is not
visible.
(a) increases with the number of bulbs,
(b) is equal to 10 times the resistance of a bulb?
Do you agree with Margaret that total resistance of the
circuit
4. Warm-up
2 Flow of charge in a conductor
vs. students walking on a road.
(a) If the road is narrow, how will the no. of
students walking on it be affected?
Fewer students can walk on it at a time.
To make the road wider.
(b) How should the road be changed to let more students walk on
it at a time?
5. 1 Bulbs in series and in parallel
The bulb glows
brightly.
bulb 1
bulb 2
The bulbs glow
dimmer.
The bulbs glow
brightly.
bulb 1 bulb 2
Bulbs are
connected in
series.
Bulbs are
connected in
parallel.
bulb 1
--
6. 1 Bulbs in series and in parallel
bulb 1
bulb 2
Vbattery = Vbulb 1 =
Vbulb 2
bulb 1 bulb 2
Remove bulb 1 ⇒
breaks circuit &
bulb 2 goes out
bulb 1
Remove bulb 1 ⇒
NOT break circuit &
bulb 2 glows
--
Vbattery = Vbulb 1 Vbattery = Vbulb 1 +
Vbulb 2
in series in parallel--
7. 1 Bulbs in series and in parallel
bulb 1
bulb 2bulb 1 bulb 2
bulb 1
Ibulb 1 = Ibulb 2 (=
0.5I0)
I0 Ibulb 1 = Ibulb 2
= I0
(= 0.5 × Itotal)
in series in parallel--
same current ⇒ same brightness
8. 2 Resistors in series
R1, R2 and R3 are connected in series.
R1 R2 R3
V
I = current through R1, R2 and R3
I
V = V1 + V2 + V3
V1
IR1
V2
IR2 IR3
V3
= I × (R1 + R2 + R3)
Equivalent resistance
9. 2 Resistors in series
R1 R2 R3
V
I.e.
I
R (= R1 + R2 + R3)
V
I
= I × (R1 + R2 + R3)
= IR
10. 2 Resistors in series
If 2 or more resistors are connected in series,
the equivalent resistance of resistors > resistance of each
resistor.
Analogy — joining wires...
...to give a longer
wire (of higher
resistance).
11. 3 Resistors in parallel
R1, R2 and R3 are connected in parallel.
R1
R2
R3
V
• •
V = voltage across R1, R2
and R3
I1
I2
I3
1
R1
1
R2
1
R3
= V × ( + + )
I = I1 + I2 + I3
V
R
1
V
R
2
V
R3
I
Equivalent resistance
= V
(R1
–1
+ R2
–1
+ R3
–1
)–1
12. 3 Resistors in parallel
I.e.
R1
R2
R3
V
• •
I1
I2
I3
I
R [= (R1
–1
+ R2
–1
+ R3
–1
)–1
]
V
I =
= I1 + I2 + I3
V
R
13. 3 Resistors in parallel
If 2 or more resistors are in parallel,
the equivalent resistance of resistors < resistance of each
resistor.
Analogy — putting wires side by side...
...to give a wider
wire (of lower
resistance).
14. Example 4
Equivalent resistance
(a) (i) What is the
equivalent resistance
of the circuit?
3-Ω and 6-Ω resistors are in series.
Equivalent resistance = 3 + 6 = 9 Ω
3 Ω
9 V
6 Ω
15. Example 4
Equivalent resistance
3 Ω
9 V
6 Ω
(a) (ii) What is the
voltage across the 6-Ω
resistor?
Current through each resistor = V
R
9
9
equivalent resistance
of the series circuit
equivalent
resistance = 9 Ω
= 1 A
Voltage across 6-Ω resistor = IR 1 × 66 V
16. Example 4
Equivalent resistance
(b) (i) What is the
equivalent resistance
of the circuit?
9 V
• •
3 Ω
6 Ω
3-Ω and 6-Ω resistors are in parallel.
Equivalent resistance = (R1
–1
+ R2
–1
)–1
3 6
= 2 Ω
17. Example 4
Equivalent resistance
(b) (ii) What is the
current passing the
6-Ω resistor?
9 V
• •
3 Ω
6 Ω
Voltage across each resistor = 9 V
Current passing 6-Ω resistor = V
R
9
6
=
= 1.5 A
18. Example 4
Equivalent resistance
(b) (iii)What is the
total current in the
main circuit?
9 V
• •
3 Ω
6 Ω
I6-Ω = 1.5 A
Current passing 3-Ω resistor = V
R
9
3
= = 3 A
Total current in the main circuit
=
I3-Ω + I6-Ω = 3 + 1.5 = 4.5 A
19. Example 5
Circuit analysis
(a) What is the total resistance
of the circuit?
Equivalent resistance of B and C
=
Total resistance = RA + RB // C
= 8 + 4
= 4 Ω
A = 8 Ω
12 V
• •
B = 6 Ω
C = 12 Ω
(R1
–1
+ R2
–1
)–1
6 12
= 12 Ω
20. Example 5
Circuit analysis
(b) What is the current
passing A?
A = 8 Ω
12 V
• •
B = 6 Ω
C = 12 Ω
total resistance = 12 Ω
Total current from battery = V
R
12
12
= = 1 A
⇒ Current passing A = 1 A
21. Example 5
Circuit analysis
(c) What is the voltage
across A?
A = 8 Ω
12 V
• •
B = 6 Ω
C = 12 Ω
Voltage across A = IR
I = 1 A
= 1 × 8 = 8 V
22. Example 5
Circuit analysis
(d) What is the voltage
across the parallel
combination of B and
C? A = 8 Ω
12 V
• •
B = 6 Ω
C = 12 Ω
Voltage across the combination of B & C
=
= 4 V12 – 8
8 V
23. Example 5
Circuit analysis
(e) What is the current
passing (i) B and (ii)
C?
A = 8 Ω
12 V
• •
B = 6 Ω
C = 12 Ω
4 V
(i) Current passing B = V
R
= = 0.667 A4
6
(ii) Current passing C = V
R
= = 0.333 A4
12
24. 4 Short circuit
When the key is open...
...the bulb lights.
• •
shorting key
When the key is closed...
Also, the bulb goes out.
...a large current passes the key (~ 0
Ω) and the wires become very hot.
The bulb is short-circuited.
25. 4 Short circuit
shorting key
In this case, the battery goes ‘flat’
quickly.
In a mains circuit, a short circuit
overheats the cables and may cause a
fire.
• •
26. Q1 Which circuit is NOT...
Which circuit is NOT equivalent to the others?
A B
C D
27. Q2 Find I1 and I2.
Find I1 and I2.
6 V 1 Ω 2 Ω
I1 =____ I2 =____6 A 3 A
28. Q3 Fill in the values of...
Fill in the values of resistance in the blanks.
12 V
6 Ω
____ Ω
A
B
6
1 A
29. Q3 Fill in the values of...
Fill in the values of resistance in the blanks.
12 V
6 Ω
A
C D
___ Ω ___ Ω1212
1 A equivalent to
B (6 Ω); RC = RD
30. Q3 Fill in the values of...
Fill in the values of resistance in the blanks.
12 V
6 Ω
A
C
E
12 Ω
1 A
___ Ω
___ Ω6
6
equivalent to
D (12 Ω); RE = RF
F
31. Q3 Fill in the values of...
Fill in the values of resistance in the blanks.
12 V
6 Ω
A
C
G
12 Ω
1 A
___ Ω
___ Ω4
8
equivalent to
D (12 Ω); RG = 2RH
H
32. Q3 Fill in the values of...
Fill in the values of resistance in the blanks.
12 V
6 Ω
A
C
G
12 Ω
1 A
RK = ___ Ω, RL = ___ Ω88
equivalent to
H (4 Ω);
RK = RL
LK
33. Q4 Which circuit in Q3...
Which circuit in Q3 has the highest equivalent resistance?
All circuits in Q3 have the same equivalent resistance.
34. 5 Resistance of ammeters,
voltmeters and cells
Ammeter connected in series to
a resistor gives the current
passing the resistor.
Rcircuit = Rammeter + Rresistor
A
⇒ I in the circuit < I without ammeter
Resistance of ammeter should be very low.
a Resistance of ammeter
I
35. 5 Resistance of ammeters,
voltmeters and cells
b Resistance of voltmeter
Voltmeter connected in parallel
to a resistor gives the voltage
across the resistor.
V
• •
R// branch = (Rvoltmeter
–1
+ Rresistor
–1
)–1
⇒ V across resistor ≠ V without voltmeter
Resistance of voltmeter should be very high.
I
V
36. 5 Resistance of ammeters,
voltmeters and cells
c Resistance of cells
All power supplies (batteries, power packs,
etc.) have some resistance — internal
resistance.
• •
V
voltage
= 3 V
voltage
= 2.8 V
E.g.
• •
V
closed
< 3 V
37. 5 Resistance of ammeters,
voltmeters and cells
c Resistance of cells
‘Lost volt’ is due to the resistance of the cells.
In a complete circuit, a battery is like...
3 V lost
volt
battery
voltage supplied in a
complete circuit < 3 V
Internal resistance of a cell is
usually neglected if not
otherwise indicated.
I
38. Example 6
Applying Ohm’s law
A student uses this set-up to
investigate Ohm’s law.
1. Ammeter and voltmeter should be interchanged.
2. Rheostat is connected up as a fixed resistor, not a
variable resistor.
(a) (i) What are the 2
mistakes in the
circuit?
ammeter
voltmeter
39. A student uses this set-up to
investigate Ohm’s law.
(a) (ii) Draw a correct circuit
diagram for this set-up.
ammeter
voltmeter
Example 6
Applying Ohm’s law
40. Example 6
Applying Ohm’s law
After fixing mistakes, the student obtains...
(b) (i) What is the resistance of the wire when V is (1) 1.8
V (2) 4.8 V?
Voltage V / V 0.6 1.2 1.8 2.4 3.2 4.8
Current I / A 0.2 0.4 0.6 0.8 1.0 1.2
V
I
R = (1) R = 1.8
0.6
= 3.0 Ω
(2) R = 4.8
1.2
= 4.0 Ω
41. Example 6
Applying Ohm’s law
After fixing mistakes, the student obtains...
(b) (ii) Plot V-I graph. V / V
I / A
0 0.2 0.4 0.6 0.8 1.0 1.2
1
2
3
4
5
Voltage V / V 0.6 1.2 1.8 2.4 3.2 4.8
Current I / A 0.2 0.4 0.6 0.8 1.0 1.2
42. V / V
I / A
0 0.2 0.4 0.6 0.8 1.0 1.2
1
2
3
4
5
Example 6
Applying Ohm’s law
After fixing mistakes, the student obtains...
(b) (iii) When is the voltage
not ∝ current?
Voltage V / V 0.6 1.2 1.8 2.4 3.2 4.8
Current I / A 0.2 0.4 0.6 0.8 1.0 1.2
When V > 2.4 V
43. V / V
I / A
0 0.2 0.4 0.6 0.8 1.0 1.2
1
2
3
4
5
Example 6
Applying Ohm’s law
After fixing mistakes, the student obtains...
(b) (iv) Why is the proportion
not held there?
Voltage V / V 0.6 1.2 1.8 2.4 3.2 4.8
Current I / A 0.2 0.4 0.6 0.8 1.0 1.2
The wire is heated up by the
current and its resistance ↑.
44. 1 (a) What would I and V be if
resistance of ammeter R
were...
1 V
3 Ω
6 Ω
Vbulb
Ibulb
The effect of the resistance of ammeter, voltmeter and cell
on the circuit.
ideal R = 6 ΩR = 6 kΩR = 6 MΩ
I bulb
Vbulb
111 mA
0.667 V
66.7 mA 0.166 mA 0.167 µA
0.4 V 0.999 mV 1.00 µV
A
45. 1 (a) 1 V
3 Ω
6 Ω
Vbulb
Ibulb
The effect of the resistance of ammeter, voltmeter and cell
on the circuit.
ideal R = 6 ΩR = 6 kΩR = 6 MΩ
I bulb
Vbulb
111 mA
0.667 V
66.7 mA 0.166 mA 0.167 µA
0.4 V 0.999 mV 1.00 µV
(b) A good ammeter should have a __________
resistance.low
A
46. 2 (a) What would I and V be if
resistance of voltmeter R were...
The effect of the resistance of ammeter, voltmeter and cell
on the circuit.
1 V
3 Ω
6 Ω
Vbulb
Ibulb
ideal R = 6 ΩR = 6 kΩR = 6 MΩ
I bulb
Vbulb
0.111 A
0.667 V
83.3 mA 0.111 A 0.111 A
0.5 V 0.666 V 0.667 V
V
47. 2 (a)
The effect of the resistance of ammeter, voltmeter and cell
on the circuit.
1 V
3 Ω
6 Ω
Vbulb
Ibulb
ideal R = 6 ΩR = 6 kΩR = 6 MΩ
I bulb
Vbulb
0.111 A
0.667 V
88.3 mA 0.111 A
0.5 V 0.666 V
0.111 A
0.667 V
(b) A good voltmeter should have a
__________ resistance.large
V
48. 3 (a) What would I and V be if
resistance of battery R were...
The effect of the resistance of ammeter, voltmeter and cell
on the circuit.
EE
1
V
3 Ω
6 Ω
Vbulb
Ibulb
ideal R = 6 ΩR = 6 kΩR = 6 MΩ
I bulb
Vbulb
111 mA
0.667 V
66.7 mA 0.166 mA 0.167 µA
0.4 V 0.999 mV 1.00 µV
49. 3 (a)
3 Ω
6 Ω
Vbulb
Ibulb
The effect of the resistance of ammeter, voltmeter and cell
on the circuit.
ideal R = 6 ΩR = 6 kΩR = 6 MΩ
I bulb
Vbulb
111 mA
0.667 V
66.7 mA 0.166 mA 0.167 µA
0.4 V 0.999 mV 1.00 µV
(b) A good battery should have a __________
resistance.low
1
V
