LECTURE 3 Circuits
18
Lecture3 Series, Parallel, and Series Parallel Circuit Neon S. de Jesus 06/21/2011
-
Upload
aira-krizelle -
Category
Documents
-
view
159 -
download
0
Transcript of LECTURE 3 Circuits
Lecture3 Series, Parallel, and Series Parallel Circuit
Neon S. de Jesus
06/21/2011
Internal Resistance of Voltage Sources
DE JESUS 2011
2
,4
max int L
L
RPE
RR
Parallel Elements
• Have two points in common
• The total resistance is always less than the value of the smallest resistor
• RT always decrease when additional element are added
DE JESUS 2011
two resistor
three resistor
equal resistor
Parallel Elements
• Example
• Solution
DE JESUS 2011
Parallel Circuits
DE JESUS 2011
1 2 3
1 2 3 I
T N
T N
V V
I
V V V
I I I
Parallel Circuits• Example Solution
– Determine R3
– Calculate E– Find Is
– Find I2
– Determine P2
DE JESUS 2011
Current Divider Rule
DE JESUS 2011
General Form
Two Resistor
Current Divider Rule
• Example
• Solution
DE JESUS 2011
Series-Parallel Circuits
• Example Solution
DE JESUS 2011
Series-Parallel Circuits
• Example
DE JESUS 2011
At path 1
At path 2
At path 3
Series-Parallel Circuits
• Exercise
DE JESUS 2011
Series-Parallel Circuits
• Exercise
DE JESUS 2011
Series-Parallel Circuits
• Exercise
DE JESUS 2011
Wye (Y) to Delta (Δ) and Delta to WyeConversion (T-π and π -T)
DE JESUS 2011
If resistors value are equal
Wye (Y) to Delta (Δ) and Delta to WyeConversion (T-π and π -T)
• Example
DE JESUS 2011
Wye (Y) to Delta (Δ) and Delta to WyeConversion (T-π and π -T)
DE JESUS 2011
• Example
a. Convert Delta to wye
Wye (Y) to Delta (Δ) and Delta to WyeConversion (T-π and π -T)
DE JESUS 2011
• Example
b. Convert wyeto delta
Homework (Due: Jun 28, 2011)
• Derivation of wye to delta and delta to wye conversion formula
• Solve for RT in the figure
• Quiz on Jun 28, 2011– Practice solving
problem in circuit
DE JESUS 2011
