Transcript + - V

Today’s agenda:
Resistors in Series and Parallel.
You must be able to calculate currents and voltages in circuit components in series and in
parallel.
Kirchoff’s Rules.
You must be able to use Kirchoff’s Rules to calculate currents and voltages in circuit
components that are not simply in series or in parallel.
Resistances in Circuits
There are “two” ways to connect circuit elements.
Series:
A
B
Put your finger on the wire at A. If you can move along the
wires to B without ever having a choice of which wire to
follow, the circuit components are connected in series.
Truth in advertising: it is possible to have circuit elements that are connected
neither in series nor in parallel. See problem 24.73 in the 12th edition
of our text for an example with capacitors.
Parallel:
A
???
B
Put your finger on the wire at A. If in moving along the
wires to B you ever have a choice of which wire to follow,
the circuit components are connected in parallel.*
*Truth in advertising: actually, the circuit components are
not connected in series, and may be connected in parallel.
Are these resistors in series or parallel?
+ -
V
parallel
It matters where you put the source of emf.
Are these resistors in series or parallel?
+
V
series
It matters where you put the source of emf.
If resistors “see” the same potential difference, they are in
parallel. If resistors “see” the same current, they are in series.
V
+ -
+
I
V
V
parallel
series
It’s difficult to come up with a simple one- or two-sentence rule for series/parallel.
Here’s a circuit with three resistors and a battery:
I
I
I
R1
R2
R3
V1
V2
V3
+ -
I
V
Current flows…
…in the steady state, the same current flows through all
resistors…
…there is a potential difference (voltage drop) across each
resistor.
Applying conservation of energy allows us to calculate the
equivalent resistance of the series resistors.
I am including the derivation in these notes, for the benefit of
students who want to look at it.
In lecture, I will skip ahead past the derivation.
I
I
I
R1
R2
R3
V1
V2
V3
+ -
I
V
An electric charge q is given a potential energy qV by the
battery.
As it moves through the circuit, the charge loses potential
energy qV1 as it passes through R1, etc.
The charge ends up where it started, so the total energy lost
must equal the initial potential energy input:
qV = qV1 + qV2 + qV3 .
I
I
I
R1
R2
R3
V1
V2
V3
+ -
I
V
qV = qV1 + qV2 + qV3
V = V1 + V2 + V3
V = IR1 + IR2 + IR3
Now imagine replacing the three resistors by a single resistor,
having a resistance R such that it draws the same current as
the three resistors in series.
I
Req
V
+ -
I
As above:
From before:
Combining:
V
V = IReq
V = IR1 + IR2 + IR3
IReq = IR1 + IR2 + IR3
Req = R1 + R2 + R3
For resistors in series, the total resistance is the sum of the
separate resistances.
We can generalize this to any number of resistors:
R eq   R i
(resistors in series)
a consequence of
conservation of energy
i
R1
R2
R3
+ -
V
Note: for resistors in parallel, Req is always greater than any of the Ri.
Here’s another circuit with
three resistors and a battery.
I1
R1
V
I2
Current flows…
R2
V
…different currents flows through
different resistors…
R3
I3
V
…but the voltage drop across
each resistor is the same.
+ -
I
V
Applying conservation of charge allows us to calculate the
equivalent resistance of the parallel resistors.
I am including the derivation in these notes, for the benefit of
students who want to look at it.
In lecture, I will skip ahead past the derivation.
I1
In the steady state, the
current I “splits” into I1, I2,
and I3 at point A.
V
I2
A
I1, I2, and I3 “recombine” to
make a current I at point B.
B
R3
I3
I2 =
V
+ V
I
Because the voltage drop across
each resistor is V:
V
R1
R2
V
Therefore, the net current
flowing out of A and into B is I
= I1 + I2 + I3 .
I1 =
R1
V
R2
I3 =
V
R3
I
I
Now imagine replacing the
three resistors by a single
resistor, having a resistance R
such that it draws the same
current as the three resistors in
parallel.
Req
A
B
V
+ -
From above, I = I1 + I2 + I3, and
V
I1 =
R1
So that
V
I2 =
R2
V
V
V
V
= + + .
R eq R1 R 2 R 3
I
V
V
I3 =
.
R3
I
Dividing both sides by V gives
1
1
1
1
= + + .
R eq R1 R 2 R 3
We can generalize this to any number of resistors:
1
1

R eq
i Ri
(resistors in parallel)
a consequence of
conservation of charge
Note: for resistors in parallel, Req is always less than any of the Ri.
Summary:
Series
B
A
i
same I, V’s add
“just like” capacitors NOT
“just like” capacitors
Parallel
A
B
same V, I’s add
“just like” capacitors
R eq   R i
1
1

R eq
i Ri
“just like” capacitors NOT