Resistors in series and parallel circuits were discussed. In a series circuit, the total resistance is equal to the sum of the individual resistances. In a parallel circuit, the total resistance is lower than the individual resistances. Examples of circuit analysis problems involving calculating equivalent resistances and currents/voltages in various components were provided. The effects of the internal resistances of ammeters, voltmeters and cells on circuit measurements were also explained.

2. Resistors in series and in parallel
R1
R2
R3
V
• •
R1 R2 R3
V

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