Lecture (01)
Ohm’s and Kirchhoff’s laws
Dr. Ahmed M. ElShafee
1
Ideal basic circuit element
1. Has only 2 terminal
2. Described mathematically in terms of voltage and current
3. It can’ t be divided.
+
i
1
v
-
2
2
Passive sign convention theory
• Passive sign convention:
if the current reference direction is in the direction of the
voltage
lt
d
drop
across the
th element,
l
t +ev sign
i in
i any expression
i
relates voltage to the current, otherwise use a –ve sign.
• Note: polarity reference is independent on function of the
basic element, or its interconnection.
+
i
1
v
-
3
2
Power and Energy
• Power and energy calculations are so important in electric
circuit analysis because
1 The
1.
Th useful
f l output
t t off these
th
systems
t
often
ft nonelectrical
l ti l
(heat, light, movement, ….).
2 All practical devices have limitations on the amount of
2.
power that can handle.
4
Power and Energy (2)
• Passive sign convention
if the current flow is in the same direction of voltage drop,
th
then
P=i×v
dissipating, absorbing power
if not
P ‐i×v
P=
generating, losing power
5
Power and Energy (3)
+
i
1
v
+
2
i
2
-
P= vii
P= -vii
1
v
-
i
1
v
2
+
P= -vi
6
1
v
-
-
i
2
+
P= vi
Power and Energy (4)
• Example: in the following model
if i = 4 A and v = ‐10V
then P = ‐ (‐10)(4) = 40 W
The element is absorbing (dissipating) 40 watts
7
Power and Energy (5)
• Example: in the following model
if i = ‐4 A and v = 10V
then P = ‐ (10)(‐4) = 40 W
The element is absorbing (dissipating) 40 watts
8
Example 01
Assume that a 10V drop occurs across an element from terminal
2 to terminal 1 and that a current of 4A enters terminal 2.
a)) Specify
S if th
the values
l
off v and
d i for
f the
th polarity
l it
references shown in figures.
b) SState whether
h h the
h circuit
i i iinside
id the
h box
b
is absorbing or delivering power.
c)) How much power is the circuit absorbing?
9
Answer 01
• Start with sketching the circuit model
• For the first model
v= ‐20v
i=‐4A;
P=(‐20)x(‐40)=80W, absorbing
• For the second model
v= ‐20v
i=4A;
P=‐(‐20)x(4)=80W, absorbing
10
Answer 01(2)
• For the 3rdd model
v= 20v
i=‐4A;
P=‐(20)x(4)=80W, absorbing
• For the 4th model
v= 20v
i=4A;
P=(20)x(4)=80W, absorbing
11
Ohm’ss law
Ohm
The voltage v across a resistor is directly proportional to the
current I flowing through the resistor.
1 Ohm
h = 1 V/A
/
12
Ohm’ss law (2)
Ohm
Passive sign convention:
There are two possible reference choices for the current and
voltage
lt
13
Ohm’ss law (3)
Ohm
Conductance is the opposite of resistance:
Symbol: G
Units: Siemens (S) or mho (
14
)
Ohm’ss law (4)
Ohm
Power dissipated by a resistor:
15
Example 09
For each figure, calculate;
1. V, and I.
2. Power.
+
50V
+
1A Vg
-
+
0.2S
‐
+
‐
16
8Ω
1A Vg
-
20Ω
50V
25Ω
Answer 09
I = 1A
V = I.R = 8V
P = V2/R = 8W
I = 1A
V = ‐I.R = ‐20V
P = V2/R = 20W
17
+
1A Vg
-
+
1A Vg
-
8Ω
20Ω
Answer 09 (2)
V = 50V
I = V.G = 10A
P = I2/G = 8W
V = 50V
I = V/R = 2A
P = I2.R = 100W
+
0.2S
50V
25Ω
‐
+
‐
18
50V
Example 10
a. Vg=1KV, Ig=5mA, find R&Presistor.
b. Ig=75mA, Psource=‐3W, find Vg, R, Presistor.
c. R=300Ω, Presistor=480mW, find Ig, Vg.
+
‐
19
Vg
R
Answer 10
a) Vg=1KV, Ig=5mA. find R&Presistor.
+
R= Vg/Ig = 103/5x10‐3 = 0.2x106=
200KΩ.
P= V.I = 5W.
b) Ig=75mA, Psource=‐3W. find
f d Vg, R, Presistor.
Vg=‐P/Ig = ‐( ‐3/(75x10‐3)) = 40V.
R=Vg/Ig= 40/(75x10‐3)=533.33Ω.
Presistor = 3W.
20
‐
Vg
R
Answer 10 (2)
R=300Ω, Presistor=480mW, find Ig, Vg.
+
Ig=(P/R)0.5=((480x10‐3)/300)0.5=
0.04A = 40mA.
Vg=(PxR)0.5=((480x10‐3)x300) 0.5=12V.
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‐
Vg
R
Kirchhoff’ss laws
Kirchhoff
• Simple circuit
– Voltage
– Current
– Power
+
+
Vs
R
‐
-
Kirchhoff’ss laws (2)
Kirchhoff
• More Complex circuit,
circuit
– Circuit contains 2 loops 1, 2.
– bd branch shared between
the two loops.
V1
– Nodes b, d are shared
between the two loops.
a
R1
b
R3
+
c
+
R2
‐
‐
1st
loop
2nd
loop
d
V2
Kirchhoff’ss laws (3)
Kirchhoff
Kirchhoff's Voltage Law (KVL):
“The algebraic sum of all voltage drops around any closed loop is
zero””
M: number
b off voltages
l
in the
h loop
l
Vm: mth element voltage.
– Satisfied around all loops of a circuit.
– The law is about energy conservation
Kirchhoff’ss laws (4)
Kirchhoff
• KVL example:
• 1st loop:
‐Vg1 + V1 + V2 = 0
• 2nd loop:
‐Vg2 + V3 + V2 = 0
a
b
R1
+
+
++
+
Vg1
-
-
c
R3
+
R2
‐
1st
loop
‐
- -
2nd
loop
• Passive sign convention:
d
Voltage get +ve if the current flow in side component is in the
same direction of voltage drop, otherwise voltage get –ve
sign.
Vg2
Kirchhoff’ss laws (5)
Kirchhoff
• Kirchhoff's current Law (KCL):
“The algebraic sum of currents around a node is zero”
N: number of branches connected to the node
in: h current entering the node
– Satisfied at all nodes of a circuit
– The law is about charge conservation
Kirchhoff’ss laws (6)
Kirchhoff
• KCL example:
node a: ‐Ig1+I1 = 0
Node b: ‐I1‐I3+I2=0
Node c: ‐Ig2+I3=0
Node d: ‐I2+Ig1+Ig2=0
So: I2=Ig1+Ig2.
R1
a
+
-
-
c
+
++
+
Vg1
R3
b
+
R2
‐
1st
loop
‐
- -
2nd
loop
d
• Passive sign convention:
Current gets +ve if the current leave the node is while gets –ve
sign if the current entering the node.
Vg2
Example 3
use KL and Ohm’s law to
1. find Io
2. Verify that total dissipated
power – total generated power
10Ω
+
120V
‐
Io
50Ω
6A
Example 3 solution
KCL:
‐Io‐6+I50=0
I50= Io+6 ………………..(1)
KVL:
‐120
120 + (10xIo)+ (50xI50) =0
0
……….…………………………(2)
Submit 2 in 1
(10xIo)+ (50x(Io+6))=120
Io=‐3A
10Ω
+
+
+
120V
‐-
Io a
-
+
50Ω
-
6A
Example 3 solution (2)
I50 = 3A
P10=(io)2x10=9x10=90W
P50=9x50=450W
P120V=‐ 120 x –i120V =360W
Total dissipated power= 360+90+450=
900W
V50=50x3=150V
P6A=‐V50x6=‐150x6=‐900W (generated)
10Ω
+
+
+
120V
‐-
Io a
-
+
50Ω
-
6A
Example 4
For the shown circuit;
Find
Is.
V3.
V2.
V7
Power delivered by source.
Example 4 solution
KVL:
‐24 + 3xIs + 7xIs + 2xIs = 0
12xIs = 24
Is = 2A
‐‐‐>
Ohm:
V3Ω = Is x 3 = 6v
V7Ω = Is x 7 = 14v
V2Ω = Is x 2 = 4V
Ps = ‐Vs x Is = ‐24 x 2 = ‐48V (generated)
32
Example 5
Find R
Example 5 solution
KVL@1:
‐200 + VR + 120 =0
VR = 80V
KVL@2:
‐120 + 8xI8 =0
I8 = 120/8 = 15A
Ohm:
I24 = 120/24 = 5A
KCL:
IR = I24 + I8
IR = 20A
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Example 5 solution (2)
Ohm:
R = VR / IR = 80/20 = 4Ω
Thanks,…
See you next week (ISA),…
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