Complex Circuits
Presentation 2003 R. McDermott
Complex DC Circuits
•
A power supply with an internal resistance (lowers
terminal voltage):
•
Adds an additional resistor in series with the power
supply
•
Reduces the voltage at the terminals
•
Terminal voltage will depend on the external circuit
•
Example: A 12

volt battery has an internal resistance of
5.0 ohms. What is the terminal voltage of the battery when
it is connected in series to a 50 ohm resistor? A 25 ohm
resistor?
Complex Circuit #1
•
Example: A 12

volt battery has an internal resistance
of 5.0 ohms. What is the terminal voltage of the
battery when it is connected in series to a 50 ohm
resistor?
•
Rc
=
5.0
+
50
=
55
•
Ic
=
(12 v)
/
(55
)
=
.22 a
Internal drop
=
5.0
X .22 a
=
1.1 v
Terminal voltage
=
12 v
–
1.1 v
=
10.9 v
Complex Circuit #2
•
Example: A 12

volt battery has an internal resistance
of 5.0 ohms. What is the terminal voltage of the
battery when it is connected in series to a 25 ohm
resistor?
•
Rc
=
5.0
+
25
=
30
•
Ic
=
(12 v)
/
(30
)
=
.40 a
Internal drop
=
5.0
X .40 a
=
2.0 v
Terminal voltage
=
12 v
–
2.0 v
=
10 v
Complex Circuit #3
•
Multiple batteries in a single loop:
–
Voltages add if the batteries are positive to negative
–
Voltages subtract if the batteries are positive to
positive.
Circuit #3
•
The batteries are connected positive to
positive, so their voltages are subtracted
(they act against one another).
•
Vnet = 18v
–
12v = 6v
•
The current in the circuit is then:
•
Ic = Vc/Rc = 6v/6
=
1a
Complex Circuit #4
•
Compound series and parallel circuit:
•
Both sets of rules apply on various parts of the
circuit
•
Ratios become very important as time savers
Find the current and voltage drops for each resistor.
Circuit #4
•
The parallel combination has a total resistance of:
1
/
Rp
=
1
/
12
+
1
/
4
=
1
/
3
Rp
=
3 ohms
•
The parallel combination is in series with R
1
so:
Rc
=
6
+
3
=
9
•
The circuit current can now be found:
Ic
=
Vc
/
Rc
=
18v
/
9
=
2a
Circuit #4 (cont)
•
Resistor 1 gets the full circuit current, so its voltage drop
is:
V
1
=
I
1
R
1
=
(2a)(6
)
=
12 v
•
The parallel combination gets the remaining voltage:
V
P
=
Vc

V
1
=
18v

12v
=
6v
•
Each resistor in parallel gets the full 6 volts, so currents
are:
I
2
=
V
P
/
R
2
=
6v
/
4
=
1.5a
I
3
=
V
P
/
R
3
=
6v
/
12
=
0.5a
Complex Circuit #5
•
Multiple loops with multiple batteries:
Kirchoff’s junction current rule: I
IN
= I
OUT
.
Kirchoff’s loop voltage rule:
V
LOOP
= 0
Circuit #5
•
Kirchoff’s junction current rule: I
IN
=
I
OUT
•
Let’s pick junction A and assume all currents are
downward. Let the far left wire be I
1
, the middle
I
2
, and the far right be I
3
. All current is exiting the
junction, so
zero
current enters (our assumption):
•
I
1
+
I
2
+
I
3
=
0
•
We now have one equation which will allow us to
solve for one unknown
Circuit #5 (cont.)
•
Kirchoff’s loop voltage rule:
V
LOOP
=
0
•
We need two more equations, so we choose two of
the three loops in our circuit (we have one on the
left, one on the right, and the large outside loop)
•
Let’s take the left and right loops and arbitrarily
pick a clockwise direction to follow
•
If the positive side of a battery pushes current with
our loop, we assign the voltage a positive value. If
it works against our loop, we assign a negative
value
Sample #5
•
In the left loop, the 12v battery works with us (
+
)
and the 6v works against us (
–
)
•
In the right loop, both batteries work with us (
+
)
•
Current flowing through a resistor in the same
direction as we are looping produces a negative
voltage. Current flowing opposite to our loop
produces a positive voltage
•
I
1
flows through R
1
opposite to our loop, so I
1
R
1
is
positive. I
2
flows through R
2
in the same direction
as our loop, so I
2
R
2
is negative
#5 (cont.)
•
In the left loop, then:
•
12
+
I
1
R
1
–
I
2
R
2
–
6
=
0
and,
12
+
10I
1
–
5 I
2
–
6
=
0
•
This last is our second equation. We now have:
I
1
+
I
2
+
I
3
=
0
12
+
10I
1
–
5 I
2
–
6
=
0
#5
•
For the right loop:
•
6
+
I
2
R
2
–
I
3
R
3
+
22
=
0
and,
6
+
5I
2
–
20I
3
+
22
=
0
•
This is our third equation. We now have:
I
1
+
I
2
+
I
3
=
0
12
+
10I
1
–
5 I
2
–
6
=
0
6
+
5I
2
–
20I
3
+
22
=
0
This is it!
•
Solving:
•
I
1
=
–
0.657a
I
2
=
–
0.114a
I
3
=
0.771a
•
We were wrong about the direction of I
1
and I
2
(both up), but correct for I
3
(down)
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