09 Ohm_s law problems

09 Ohm_s law problems - Ohm's
law
problems
 
...

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Unformatted text preview: Ohm's
law
problems
 1/13/10
 Ohm’s
Law
for
Resistor
 Ganago’s
problems
on
Ohm’s
law

 Master
these
concepts
for
 success
on
homework
and
 exams
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 1
 •  For
a
resistor,
 current
 (through)
 
 
 
is
directly
 proporMonal
to
 voltage
 
(across)

 ©
2009
A.
Ganago
 Voltage‐current

 CharacterisMc
 Problems
on
Ohm's
law
 2
 Independent
Voltage
Source
 •  Regardless
of
 anything
else
 in
the
circuit,
 provides
 voltage
VS
 between
its
 terminals

 ©
2009
A.
Ganago
 Independent
Current
Source
 •  Regardless
of
 anything
else
 in
the
circuit,
 pumps
 current
IS
 through
its
 terminals

 ©
2009
A.
Ganago
 Voltage‐current

 CharacterisMc
 Voltage‐current

 CharacterisMc
 Problems
on
Ohm's
law
 3
 Problems
on
Ohm's
law
 4
 How
much
power
does
the
 voltage
source
absorb?
 3A 2Ω Recall
KCL,
KVL,
Ohm’s
Law,
and

 the
Passive
Sign
ConvenMon
 I4 3A 2Ω 4V Choices:
 +
12
W

 +
4
W

 4V ‐
4
W

 ‐
8
W

 ‐
12
W
 5
 The
absorbed
power
=
P4V
=
(4
V)(I4)
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 6
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 ©
2010
A.
Ganago
 1
 Ohm's
law
problems
 1/13/10
 By
KCL
at
the
top
node,

 3
A
=
I2
+
I4

 3A I2 2Ω I4 4V By
KVL
(right
loop)
the
voltage
 across
2
Ω
must
equal
4
V
 3A I2 2Ω I4 4V By
Ohm’s
law,
I2
=
(4
V)/(2
Ω)
=
2
A
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 7
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 8
 By
KCL
(top
node),
I4
=
(3
A)
–
I2

 I4
=
(3
A)
–
(2
A)
=
1
A

 3A I2 2Ω I4 4V The
absorbed
power

 P4V
=
(4
V)(1
A)
=
4
W
 I2 2Ω I4 4V 3A ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 9
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 10
 How
much
power
does
the
 current
source
absorb?
 3A 2Ω Recall
KCL,
KVL,
Ohm’s
Law,
and

 the
Passive
Sign
ConvenMon
 3A Choices:
 +
12
W

 +
4
W

 4V ‐
4
W

 ‐
8
W

 ‐
12
W
 11
 ©
2009
A.
Ganago
 I2
 3A 4V 2Ω 4V I4
 4V ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 Problems
on
Ohm's
law
 12
 ©
2010
A.
Ganago
 2
 Ohm's
law
problems
 1/13/10
 Power
P3
=
(4
V)(‐
3
A)
=
‐
12
W
 Note
the
negaMve
sign!
 3A Check
the
Power
Balance
 3A I2
 3A 4V 2Ω 4V I4
 4V 3A 4V I2
 4V 2Ω I4
 4V P3A
+
P2Ω
+
P4V
=
‐
12W
+
(2A)2(2Ω)+4
W
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 13
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 14
 How
much
power
does
the
 voltage
source
absorb?
 3A 2Ω Recall
KCL,
KVL,
Ohm’s
Law,
and

 the
Passive
Sign
ConvenMon
 3A 2Ω I4N Subscript N refers to 4V new problem Choices:
 +
20
W

 +
4
W

 4V Zero
 ‐
4
W

 ‐
20
W

 15
 The
absorbed
power
=
P4VN
=
(4
V)(I4N)
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 16
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 By
KCL
at
the
top
node,

 3
A
+
I2N
+
I4N
=
0


 I4N 3A 2Ω I2N 4V By
KVL
(right
loop)
the
voltage
 across
2
Ω
must
equal
4
V
 I4N 3A 2Ω I2N 4V By
Ohm’s
law,
I2N
=
(4
V)/(2
Ω)
=
2
A
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 17
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 18
 ©
2010
A.
Ganago
 3
 Ohm's
law
problems
 1/13/10
 By
KCL
(top
node),

 (3
A)
+
I4N
+
I2N

=
0

 I4N
=
–
(3
A)
–
(2
A)
=
–
5
A

 I4N 3A 2Ω I2N 4V The
absorbed
power

 P4VN
=
(4
V)(‐5
A)
=
‐
20
W
 I4N 3A 2Ω I2N 4V ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 19
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 20
 How
much
power
does
the
 current
source
absorb?
 3A 2Ω Two
ways
to
use
KVL
in
order
to
 find
voltage
across
the
3‐A
source:
 3A 3A 4V 2Ω I2N 4V Choices:
 +
12
W

 +
4
W

 4V ‐
4
W

 ‐
8
W

 ‐
12
W
 21
 4V (1)  For the left loop (2)  For the large loop (whole circuit) ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 22
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 Recall
KCL,
KVL,
Ohm’s
Law,
and

 the
Passive
Sign
ConvenMon
 3A 3A 4V 2Ω I2N 4V Check
the
Power
Balance
 3A 3A 4V 2Ω I2N I4N 4V 4V 4V P3AN
=
(3
A)(4
V)
=
12
W
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 23
 P3AN+P2Ω+P4VN
=
+12W
+
(2A)2(2Ω)‐20
W
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 24
 ©
2010
A.
Ganago
 4
 Ohm's
law
problems
 1/13/10
 Equivalent
Resistance
 •  If
voltage
VS
 makes
current
I
 flow
through
 the
circuit,
the
 equivalent
 resistance
 equals
Req=VS/I

 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 A 20 20 B Req
=
?
 Choices:
 13.75
 17.50
 19.375
 21.25
 25.00
 26
 Req
 5 D 10 15 C All
Resistances
are
in
Ω
 25
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 20 20 5 D 20 5 20 10 15 C D 10 15 C ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 27
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 28
 10 5 D 15 10 15 C D 10 15 C ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 29
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 30
 ©
2010
A.
Ganago
 5
 Ohm's
law
problems
 1/13/10
 Answer:
17.50
Ω

 D 10 7.5 C D 17.5 C ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 31
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 32
 The
source
is
connected
to
a
and
b
 The
1Ω
resistor
absorbs
1
W
 P3Ω

=
?
 Choices:
 a
 1Ω 3
W
 9
W
 27
W
 Not
enough
info
 2Ω None
of
the
above
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 Recall
KCL,
KVL,
and
Ohm’s
Law

 1
W
=
(I1)2(1
Ω)
thus
I1
=
1
A

 Ohm’s
law:
 (1
A)
(1
Ω)
=
 a
 =
1
V

 1Ω 3Ω 3Ω b
 b
 2Ω 33
 ©
2009
A.
Ganago
 Problems
on
Ohm's
law
 34
 Recall
KCL,
KVL,
and
Ohm’s
Law

 KVL:
(1
A)
(1
Ω)
=
(I2)
(2
Ω)
=
1
V


 Ohm’s
law:
 I2
=
(1V)/(2Ω)
 a
 =
0.5
A
 KCL:I3
=
I1
+
I2
 I3
=
1.
5
A
 ©
2009
A.
Ganago
 Recall
KCL,
KVL,
and
Ohm’s
Law

 Power
P3
=
(1.5
A)2
(3
Ω)
=
6.75
W


 a
 b
 Answer:
 None
of
the
above
 ©
2009
A.
Ganago
 1Ω 3Ω 1Ω 3Ω b
 2Ω 36
 2Ω Problems
on
Ohm's
law
 35
 Problems
on
Ohm's
law
 ©
2010
A.
Ganago
 6
 ...
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