2. What I expected you to know
The role of the battery: To maintain potential difference between its
terminals
Potential difference creates electric fields between its terminals at a speed
approximately that of light
Electric field becomes stronger when directed from thicker to a thinner
conductor.
When the field is strong, charges in that region move faster because the
field are proportional to the force in that region
When moving faster, their kinetic energy/temperature increases energy is
dissipated as heat and light.
State and apply Kirchoff’s rules
3. Circuits
Series Parallel
Implications:
𝑃 = 𝐼 𝑉 (𝐼 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)
𝑃 = 𝐼2
𝑅 (𝐼 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)
𝑃 𝛼 𝑅 (𝐼 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)
Same current
Higher resistor will have more power
𝑃 𝛼 𝑏𝑟𝑖𝑔ℎ𝑡𝑛𝑒𝑠𝑠
more power more brightness
Same Potential difference V
𝑃 = 𝐼 𝑉 (𝑉 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)
𝑃 𝛼
1
𝑅
(𝑉 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)
𝑃 =
𝑉2
𝑅
(𝑉 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)
Higher resistor will have less power
𝑃 𝛼 𝑏𝑟𝑖𝑔ℎ𝑡𝑛𝑒𝑠𝑠
Higher resistor: less power less brightness
Implications:
4. 3 bulb question
The circuit above shows three identical light bulbs attached to an ideal
battery. If the bulb#2 burns out, which of the following will occur?
a) Bulbs 1 and 3 are unaffected. The total light emitted by the circuit decreases.
b) Bulbs 1 and 3 get brighter. The total light emitted by the circuit is unchanged.
c) Bulbs 1 and 3 get dimmer. The total light emitted by the circuit decreases.
d) Bulb 1 gets dimmer, but bulb 3 gets brighter. The total light emitted by the circuit is
unchanged.
e) Bulb 1 gets brighter, but bulb 3 gets dimmer. The total light emitted by the circuit is
unchanged.
f) Bulb 1 gets dimmer, but bulb 3 gets brighter. The total light emitted by the circuit
decreases.
5. When the bulb #2 is not burnt
out:
R
2
3
2
R
RReq
R3
V2
R
V
I
2
31
RIP,Power 2
R
V
I
For Bulb #1
For Bulb #2
For Bulb #3
R
V
44.
R9
V4
RIP
22
2
11
R3
V
2
I
I 1
2
R
V
11.
R9
V
RIP
22
2
22
R3
V
2
I
I 1
3 R
V
11.
R9
V
RIP
22
2
33
1I
2
I
I 1
2 2
I
I 1
3
Conclusion:
Bulb 1 brighter, bulbs 2 & 3 equally dim
6. R2RRReq
R2
V
I1
RIP,Power 2
R
V
I
For Bulb #1
For Bulb #2
For Bulb #3
R
V
25.
R4
V
RIP
22
2
11
0I2 0RIP 2
22
R2
V
II 13
R
V
25.
R4
V
RIP
22
2
33
1I
13 II
So, Bulb #1 gets dimmer and bulb #3 gets
brighter. And the total power decreases.
f) is the answer.
Before total power was
R
V
66.
R
V
R
V
P
2
2
3
2
eq
2
b
R
V
50.
R2
V
R
V
P
22
eq
2
a After total power is
When the bulb #2 is burnt out:
7. Problem 65 Chapter 20
Information to use
• When two or more resistors are in series, the equivalent resistance is given by
Rs R1 R2 R3 . . .
• When resistors are in parallel, the expression to be solved to find the equivalent resistance is given by
1
Rp
1
R1
1
R2
1
R3
...
.
We will successively apply these to the individual resistors in the figure in the text
beginning with the resistors on the right side of the figure.
• Since the 4.0- and the 6.0- resistors are in series, the equivalent resistance of the combination of those two resistors is 10.0 .
• The 9.0- and 8.0- resistors are in parallel; their equivalent resistance is 4.24 .
• The equivalent resistances of the parallel combination (9.0 and 8.0 ) and the series combination (4.0 and the 6.0 ) are in parallel;
• therefore, their equivalent resistance is 2.98 .
• The 2.98- combination is in series with the 3.0- resistor, so that equivalent resistance is 5.98 .
• Finally, the 5.98- combination and the 20.0- resistor are in parallel, so the equivalent resistance between the points A and B is 4 6.
.
Related problem no 64 page 632
8. Applying Kirchhoff's rules
For a resistor: Apply V=IR
(a) the potential difference is negative(a decrease) if your chosen
direction is the same as the chosen current direction through the
resistor
(b) the potential difference is positive(an increase) if your chosen
direction is opposite the chosen current direction through the
resistor
For a battery:
(c) The potential difference is positive if your loop direction is from the
negative terminal toward the positive terminal
(d) The potential difference is negative if your loop direction is from
the positive terminal toward the negative terminal
9. Element Analysis Direction Current Direction Voltage Drop
iR
iR
iR
iR
Vemf
Vemf
Vemf
Vemf
10. Problem 85 Chapter 20
Step 1: Choose a junction and directions of currents and allocate signs on resistors
Step 2: Apply the junction rule :
𝑰 𝟏 + 𝑰 𝟑 = 𝑰 𝟐 … … … … … … (𝟏)
Step 3: Applying the loop rule
Loop ABCD Clockwise:
−𝟐 𝑰 𝟏 + 𝟔𝑽 + 𝟒 𝑰 𝟑 + 𝟑𝑽 = 𝟎 … … … … … 𝟐
Loop BEFC Clockwise:
−𝟖 𝑰 𝟐 − 𝟗 𝑽 − 𝟒 𝑰 𝟑 − 𝟔𝑽 = 𝟎 … … … … … 𝟑
Solving simultaneous equations We get
𝑰 𝟑 = −𝟏. 𝟖𝟐 𝑨
Significance of minus sign:
B
C
6.00 V
A
D
E
F
2.00 Ω
I1 I2
8.00 Ω
I3 4.00 Ω
9.00 V3.00 V
+ + +
+ +
+
The minus sign indicates that the current in the 4.00- resistor is directed downward , rather
than upward as selected arbitrarily in the drawing.
11. 𝐼1
𝐼2
𝐼3
Step 1: Choose a junction and directions of currents
Step 2: Apply the junction rule
𝑰 𝟏 + 𝑰 𝟑 = 𝑰 𝟐 … … … … … … (𝟏)
Step 3: Applying the loop rule
-I1 + -2.0 I2 +1.0 = 0 ……(2)
Top loop clockwise
Bottom loop clockwise
3.0 I3 + 2.0 I2 -1.0- 4.0 = 0….(3)
Solving Equations (1) , (2) and (3) simultaneously, we find I2 = 0.73 A
The positive sign shows that the assumed direction is correct.
That is, to the LEFT
