Download Application of electric field to currents in conductors. pp

Potential energy is lost as charge moves through a circuit. This
Is measured in Volts. As an electron moves toward the (+)
Terminal it will lose potential energy which is usually
Converted into heat. The symbol (V) actually means V.
V=ED or EL means Energy lost/charge = Force/charge x Length
V = ED or EL
Work = Force x Distance
Volts Potential Energy
- -- - - - - - - - - - - -- -- -- -D
∆V = ED
+++++++ ++++++ +
Voltage
Amperes
∆V I = 8.4 watts
e
∆V= E L
e
e
Nichrome
e
Voltage
∆V
Current
I =33.6 watts
∆V=EL
Copper
Voltage
Current
-
Copper
Nichrome
Go back and check it out!
Voltage
∆V I <
Current
∆VI
Ed=∆V
Copper
I
Nichrome
Ed =∆V
I
EL= Volts
Nichrome
E
EL= Volts
Copper
Length
Electrons build up at the Cu/Nichrome junction and reduce the
electric field in copper while increasing the electric field in
nichrome UNTIL the two currents are the SAME.
It takes less force (electric field) to push the current through
the copper wire than through the nichrome wire.
VI=(volts)(3a)= cool
VI=(volts)(3a)= HOT
2a
6v
Electric field in a
Series Circuit:
Cu wire
Have students
touch the wires.
Nichrome wire
Electric field in the
Nichrome wire must
be larger because of
its larger resistance.
El cu +El nic.=V 6v
Cool hot
Conclusion about Series Circuits:
1. All the currents must be the same!
2. The resistor with the most resistance must have the
largest electric field in it.
3. Therefore, The resistor with the most volts lost
(greatest resistance) must get the hottest.
∆V I = watts ∆V = E d
Voltage
Current
close to 15
Nichrome
3 amps
12 amps
Copper
Power= VI Which wire gets Hotter?
Copper Wire
E
Nichrome Wire
E
∆V=EL
2.8 volts
Length
The resistance of copper wire
is small so the current should
be larger than in the Nichrome.
I= hotter
∆V
Power = (2.8 v)(12a)=34watts
∆V=EL
2.8 volts
Length
The resistance of nichrome
wire is large so the current
should be smaller than Cu.
∆V I= cooler
Power=(2.8v)(3a)=8.4watts
In a parallel circuit the voltages across the resistances are
equal. (Logic tells us this must be true)
The currents add up to the total current.
The electric fields which push the electrons around are
equal in each resistor…since ∆V=Ed and the d’s are =.
Voltage
-
Current
+
Copper
Copper
Nichrome
R
Extension Cord
∆V I + ∆
VI
Light bulb
+ ∆V I = total power
L cu = L nichrome
nichrome
Volts
lost
E
copper
Volts lost
copper
Volts lost
Electric force field in the copper wires is very small; as in the
previous example the field in the nichrome must be larger to
produce the same current throughout.
Therefore, negligible heat is generated in the ext. cord
In “Ohmic” devices…..that follow ohm’s law the current is
proportional to the voltage. V α I therefore the ratio of V to I
is a constant V/I = constant This constant is called the
resistance. Therefore V/I = R or V = IR
Graphing Ohm’s
Law: A demo
on the blackboard.
I
Variable
resistor: Vary
R and record
V and I. Graph.
V
R=4
V
http://web
physics.ph.
msstate.ed
u/jc/library
/182/ohmslaw.
htm
R=2
V
8
4
2
I4
In “Ohmic” devices…..that follow ohm’s law the current is
proportional to the voltage. V α I therefore the ratio of V to I
is a constant V/I = constant This constant is called the
resistance. Therefore V/I = R or V = IR
What happens to the total resistance of a circuit when you add
resistors in series??
What happens when you add another
resistance in series?
Is the resistance going “up” staying
the “same” or “going down” ??
If you compare slides 4,5,and 7 you
can see that adding wires in series
increases the resistance and lowers
the current. Adding two 2 ohm
resistors in series will make the current
half and therefore must double the
total resistance. Look at the graph from
point “a” to point “b”.
Resistors placed in series have their
values added to find the total resistance.
Variable
resistor: Vary
R and record
V and I. Graph.
R=4
V
..
b
8
4
2
I4
R=2
a
Conclusion about Series Circuits:
1. All the currents must be the same!
2. The resistor with the most resistance must have the
largest electric field in it.
3. Therefore, The resistor with the most volts lost
(greatest resistance) must get the hottest.
∆V I = watts ∆V = E d
4. R total = R 1 + R 2 + R 3 etc