Transcript Physics 6B Electric Current - UCSB Campus Learning Assistance

Physics 6B
Electric Current
And DC Circuit Examples
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Electric Current is the RATE at which charge flows(usually through a wire).
We can define it with a formula:
I
Q
t
Units are Coulombs/second, or Amperes (A)
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Resistance and Ohm’s Law
Under normal circumstances, wires present
some resistance to the motion of electrons.
Ohm’s law relates the voltage to the current:
Be careful – Ohm’s law is not a universal law
and is only useful for certain materials
(which include most metallic conductors).
Resistance and Ohm’s Law
The units of resistance, volts per ampere,
are called ohms:
Two wires of the same dimensions will have different
resistances if they are made of different materials. This
property of a material is called the resistivity.
Resistance and Ohm’s Law
Example: 2 cylindrical resistors are made from the same
material. The length of resistor B is twice the length of A, and
the radius of B is 3 times the radius of A.
The ratio of their resistances, RA/RB, is:
a) 1/2
b) 2/3
c) 3/2
d) 9/2
Resistance and Ohm’s Law
Example: 2 cylindrical resistors are made from the same
material. The length of resistor B is twice the length of A, and
the radius of B is 3 times the radius of A.
The ratio of their resistances, RA/RB, is:
a) 1/2
b) 2/3
c) 3/2
d) 9/2
Energy and Power in Electric Circuits
In materials for which Ohm’s law holds, the
power can also be written:
This power mostly becomes heat inside the
resistive material.
Resistors in Series and Parallel
Resistors connected end to end are said to be in
series. They can be replaced by a single
equivalent resistance without changing the
current in the circuit.
Resistors in Series and Parallel
Since the current through the series resistors
must be the same in each, and the total potential
difference is the sum of the potential differences
across each resistor, we find that the equivalent
resistance is:
Resistors in Series and Parallel
Resistors are in parallel
when they are across the
same potential
difference; they can
again be replaced by a
single equivalent
resistance:
Resistors in Series and Parallel
Using the fact that the potential difference across each
resistor is the same, and the total current is the sum of
the currents in each resistor, we find:
If you have a pair of resistors in parallel you can use a
quick shortcut:
R1  R2
R eq 
R1  R2
Example: Find the current through, and power used by each
resistor in this circuit.
Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω
20V
R1
R3
R2
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Example: Find the current through, and power used by each
resistor in this circuit.
Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω
20V
6Ω
Before we can calculate the individual currents, we need
to know how much current is supplied by the battery. So
we need to find the equivalent resistance for the circuit.
6Ω
12Ω
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Example: Find the current through, and power used by each
resistor in this circuit.
Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω
Before we can calculate the individual currents, we need
to know how much current is supplied by the battery. So
we need to find the equivalent resistance for the circuit.
20V
6Ω
6Ω
Parallel – Req=4Ω
12Ω
Notice that R1 and R2 are in parallel. We can combine
them into one single resistor:
1
1
1
R R


 Req  1 2
Req R1 R2
R1  R2
Req 
6  12
 4
6  12
This shortcut formula
will work for any pair
of parallel resistors.
20V
4Ω
6Ω
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Example: Find the current through, and power used by each
resistor in this circuit.
Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω
Before we can calculate the individual currents, we need
to know how much current is supplied by the battery. So
we need to find the equivalent resistance for the circuit.
20V
6Ω
6Ω
Parallel – Req=4Ω
12Ω
Notice that R1 and R2 are in parallel. We can combine
them into one single resistor:
1
1
1
R R


 Req  1 2
Req R1 R2
R1  R2
Req 
This shortcut formula
will work for any pair
of parallel resistors.
6  12
 4
6  12
20V
Series – Req=10Ω
4Ω
6Ω
The next step is to combine the remaining resistors, which
are in series. The formula is simple – just add them together.
20V
10Ω
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Example: Find the current through, and power used by each
resistor in this circuit.
Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω
Before we can calculate the individual currents, we need
to know how much current is supplied by the battery. So
we need to find the equivalent resistance for the circuit.
20V
6Ω
6Ω
Parallel – Req=4Ω
12Ω
Notice that R1 and R2 are in parallel. We can combine
them into one single resistor:
1
1
1
R R


 Req  1 2
Req R1 R2
R1  R2
Req 
This shortcut formula
will work for any pair
of parallel resistors.
6  12
 4
6  12
20V
Series – Req=10Ω
The next step is to combine the remaining resistors, which
are in series. The formula is simple – just add them together.
4Ω
6Ω
2 Amps
20V
Now that we finally have our circuit simplified down to a single resistor
we can use Ohm’s Law to compute the current supplied by the battery:
I
V 20V

 2Amps
R 10
10Ω
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Example: Find the current through, and power used by each
resistor in this circuit.
Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω
20V
6Ω
Now that we have the total current, we have to find the
individual currents. It might help to make a table like this:
Current (I)
Voltage (V)
Power (P)
6Ω
Parallel – Req=4Ω
12Ω
R1=6Ω
R2=12Ω
R3=6Ω
20V
Battery
Series – Req=10Ω
4Ω
6Ω
2 Amps
20V
10Ω
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Example: Find the current through, and power used by each
resistor in this circuit.
Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω
20V
6Ω
Now that we have the total current, we have to find the
individual currents. It might help to make a table like this:
Current (I)
Voltage (V)
Power (P)
6Ω
Parallel – Req=4Ω
12Ω
R1=6Ω
R2=12Ω
R3=6Ω
2 Amps
Battery
2 Amps
20V
20 volts
Take a look at the original circuit. Notice that all the
current has to go through R3. So I3 = 2 Amps. We can
also fill in the information for the battery – we know its
voltage and we already found the total current, which has
to come from the battery.
Series – Req=10Ω
4Ω
6Ω
2 Amps
20V
10Ω
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Example: Find the current through, and power used by each
resistor in this circuit.
Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω
20V
6Ω
Now that we have the total current, we have to find the
individual currents. It might help to make a table like this:
Current (I)
Voltage (V)
R3=6Ω
2 Amps
12 volts
Battery
2 Amps
20 volts
Power (P)
6Ω
Parallel – Req=4Ω
12Ω
R1=6Ω
R2=12Ω
Take a look at the original circuit. Notice that all the
current has to go through R3. So I3 = 2 Amps. We can
also fill in the information for the battery – we know its
voltage and we already found the total current, which has
to come from the battery.
Now that we have the Current for resistor #3, we can use
Ohm’s Law to find the voltage drop:
V  I  R  2A   6  12V
20V
Series – Req=10Ω
4Ω
6Ω
2 Amps
20V
10Ω
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Example: Find the current through, and power used by each
resistor in this circuit.
Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω
20V
6Ω
Now that we have the total current, we have to find the
individual currents. It might help to make a table like this:
Current (I)
Voltage (V)
R3=6Ω
2 Amps
12 volts
Battery
2 Amps
20 volts
Power (P)
6Ω
Parallel – Req=4Ω
12Ω
R1=6Ω
R2=12Ω
The current for R1 and R2 will be easier to find if we
calculate the voltage drops first (they have to be the same
voltage because they are in parallel – make sure you
understand why!).
20V
Series – Req=10Ω
4Ω
6Ω
2 Amps
20V
10Ω
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Example: Find the current through, and power used by each
resistor in this circuit.
Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω
20V
6Ω
Now that we have the total current, we have to find the
individual currents. It might help to make a table like this:
Current (I)
Voltage (V)
R1=6Ω
8 volts
R2=12Ω
8 volts
R3=6Ω
2 Amps
12 volts
Battery
2 Amps
20 volts
Power (P)
The current for R1 and R2 will be easier to find if we
calculate the voltage drops first (they have to be the same
voltage because they are in parallel – make sure you
understand why!).
Either use Ohm’s law (use the 2nd diagram with the
combined resistance of 4Ω), or notice that the total voltage is
20V, and R3 uses 12V, so there is 8V left over for R1 or R2.
Now we can use Ohm’s Law again for the individual
resistors to find the current through each:
6Ω
Parallel – Req=4Ω
12Ω
20V
Series – Req=10Ω
4Ω
6Ω
2 Amps
20V
10Ω
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Example: Find the current through, and power used by each
resistor in this circuit.
Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω
20V
6Ω
Now that we have the total current, we have to find the
individual currents. It might help to make a table like this:
Current (I)
Voltage (V)
R1=6Ω
4/3 Amps
8 volts
R2=12Ω
2/3 Amps
8 volts
R3=6Ω
2 Amps
12 volts
Battery
2 Amps
20 volts
Power (P)
The current for R1 and R2 will be easier to find if we
calculate the voltage drops first (they have to be the same
voltage because they are in parallel – make sure you
understand why!).
Either use Ohm’s law (use the 2nd diagram with the
combined resistance of 4Ω), or notice that the total voltage is
20V, and R3 uses 12V, so there is 8V left over for R1 or R2.
Now we can use Ohm’s Law again for the individual
resistors to find the current through each:
8V 4
 Amps
6 3
8V
2
I2 
 Amps
12 3
6Ω
Parallel – Req=4Ω
12Ω
20V
Series – Req=10Ω
4Ω
6Ω
2 Amps
20V
10Ω
I1 
Total = 2 Amps
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Example: Find the current through, and power used by each
resistor in this circuit.
Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω
20V
6Ω
Now that we have the total current, we have to find the
individual currents. It might help to make a table like this:
Current (I)
Voltage (V)
R1=6Ω
4/3 Amps
8 volts
R2=12Ω
2/3 Amps
8 volts
R3=6Ω
2 Amps
12 volts
Battery
2 Amps
20 volts
Power (P)
Finally, we can calculate the power for each circuit
element. You have your choice of formulas:
V2
P  I V  I R 
R
6Ω
Parallel – Req=4Ω
12Ω
20V
Series – Req=10Ω
2
4Ω
6Ω
2 Amps
20V
10Ω
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Example: Find the current through, and power used by each
resistor in this circuit.
Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω
20V
6Ω
Now that we have the total current, we have to find the
individual currents. It might help to make a table like this:
Current (I)
Voltage (V)
Power (P)
R1=6Ω
4/3 Amps
8 volts
32/3 Watts
R2=12Ω
2/3 Amps
8 volts
16/3 Watts
R3=6Ω
2 Amps
12 volts
24 Watts
Battery
2 Amps
20 volts
40 Watts
Finally, we can calculate the power for each circuit
element. You have your choice of formulas:
V2
P  I V  I R 
R
12Ω
20V
Series – Req=10Ω
2
I suggest using the simplest one. Plus it’s easy to
remember because you probably live there…
6Ω
Parallel – Req=4Ω
4Ω
6Ω
2 Amps
20V
10Ω
As a final check you can add the powers to make sure
they come out to the total power supplied by the battery.
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Kirchhoff’s Rules
More complex circuits cannot be broken down
into series and parallel pieces.
For these circuits, Kirchhoff’s rules are useful.
The junction rule is a consequence of charge
conservation; the loop rule is a consequence
of energy conservation.
Kirchhoff’s Rules
The junction rule: At any junction, the current
entering the junction must equal the current
leaving it.
Kirchhoff’s Rules
The loop rule: The algebraic sum of the potential
differences around a closed loop must be zero (it
must return to its original value at the original
point).
Kirchhoff’s Rules
Using Kirchhoff’s rules:
• The variables for which you are solving are the
currents through the resistors.
• You need as many independent equations as
you have variables to solve for.
• You will need both loop and junction rules.
Example – Find the current through each resistor, and find the
power used by each.
Example – Find the current through each resistor.
We can start by writing a formula for each
loop shown. The direction of current has
already been chosen for us, but that is an
important first step.
Example – Find the current through each resistor.
We can start by writing a formula for each
loop shown. The direction of current has
already been chosen for us, but that is an
important first step.
Loop 1: +15V -100I3 -100I1 = 0
Loop 2: -9V -100I2 +100I3 = 0
Example – Find the current through each resistor.
We can start by writing a formula for each
loop shown. The direction of current has
already been chosen for us, but that is an
important first step.
Loop 1: +15V -100I3 -100I1 = 0
Loop 2: -9V -100I2 +100I3 = 0
We also need to write down a junction formula:
At point A we have
+I1-I2-I3 = 0
Example – Find the current through each resistor.
We can start by writing a formula for each
loop shown. The direction of current has
already been chosen for us, but that is an
important first step.
Loop 1: +15V -100I3 -100I1 = 0
Loop 2: -9V -100I2 +100I3 = 0
We also need to write down a junction formula:
At point A we have
+I1-I2-I3 = 0
The rest is just algebra to solve for the 3
unknown currents.
Example – Find the current through each resistor.
We can start by writing a formula for each
loop shown. The direction of current has
already been chosen for us, but that is an
important first step.
Loop 1: +15V -100I3 -100I1 = 0
Loop 2: -9V -100I2 +100I3 = 0
We also need to write down a junction formula:
At point A we have
+I1-I2-I3 = 0
The rest is just algebra to solve for the 3
unknown currents.
Answers: I1=0.07A, I2=-0.01A, I3=0.08A
Notice that I2 came out negative. This is
no problem – it just means that the
current actually flows the opposite way
from what was chosen in the picture.
Circuits Containing Capacitors
Capacitors can also be connected in series or in
parallel.
When capacitors are
connected in parallel,
the potential difference
across each one is the
same.
Circuits Containing Capacitors
Therefore, the equivalent capacitance is the
sum of the individual capacitances:
Circuits Containing Capacitors
Capacitors connected in
series do not have the
same potential difference
across them, but they do
all carry the same charge.
The total potential
difference is the sum of the
potential differences
across each one.
Circuits Containing Capacitors
Therefore, the equivalent capacitance is
Note that this equation gives you the inverse of
the capacitance, not the capacitance itself!
Capacitors in series combine like resistors in
parallel, and vice versa.
RC Circuits
In a circuit containing
only batteries and
capacitors, charge
appears almost
instantaneously on the
capacitors when the
circuit is connected.
However, if the circuit
contains resistors as
well, this is not the case.
RC Circuits
Using calculus, it can be shown that the charge
on the capacitor increases as:
Here, τ is the time constant of the circuit:
And
is the final charge on the capacitor, Q.
RC Circuits
Here is the charge vs. time for an RC circuit:
RC Circuits
It can be shown that the current in the circuit
has a related behavior:
RC Circuits
Interactive example: Charging a capacitor