Transcript Physics 6B - University of California, Santa Barbara

Physics 6B
Electric Current
And DC Circuit Examples
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Assistance Services at UCSB
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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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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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:
V 20V
I 
 2Amps
R 10
10Ω
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB