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Potential energy is lost as charge moves through a circuit. ThisIs 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
- -- - - - - - - - - - - -- -- -- --
+ + + + + + + + + + + + + +
∆V = EDD
Volts Potential Energy
Nichrome
Voltage Amperes
∆V= E L
e
e
e
e
∆V I = 8.4 watts
Copper
Voltage Current
∆V=EL
∆V I =33.6 watts
Copper
Nichrome
Voltage Current
-
electrons
Go back and check it out!
Nichrome
Copper
Voltage Current
Ed=∆V
Ed =∆V
∆V I < ∆VI
I
I
I
E
Length
Copper
Nichrome
Electrons build up at the Cu/Nichrome junction and reduce theelectric field in copper while increasing the electric field innichrome UNTIL the two currents are the SAME. It takes less force (electric field) to push the current throughthe copper wire than through the nichrome wire.
EL= Volts
EL= Volts
VI=(volts)(3a)= cool VI=(volts)(3a)= HOT
Cu wire
Nichrome wire
Electric field in the Nichrome wire mustbe larger because of its larger resistance.El cu +El nic.=V 6v
6v2a
Cool hot
Have studentstouch the wires.
Electric field in aSeries Circuit:
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
Copper
Nichrome
Voltage Current
12 amps3 amps
close to 15
E
Length
∆V=EL
Copper Wire
Length
E∆V=EL
Nichrome Wire
2.8 volts 2.8 volts
Power= VI Which wire gets Hotter?
Power = (2.8 v)(12a)=34watts
The resistance of copper wireis small so the current should be larger than in the Nichrome.
∆VI= hotter
The resistance of nichrome
wire is large so the currentshould be smaller than Cu.
∆V I= cooler
Power=(2.8v)(3a)=8.4watts
In a parallel circuit the voltages across the resistances areequal. (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 =.
Copper
Nichrome
Copper
Voltage Current
+ -
3.0
E
copper
nichrome
copper
Volts lost
Voltslost
Volts lost
L cu = L nichrome
Electric force field in the copper wires is very small; as in theprevious example the field in the nichrome must be larger toproduce the same current throughout.
RExtension Cord Light bulb
Therefore, negligible heat is generated in the ext. cord
∆V I + ∆V I + ∆V I = total power
In “Ohmic” devices…..that follow ohm’s law the current is proportional to the voltage. V α I therefore the ratio of V to Iis a constant V/I = constant This constant is called the resistance. Therefore V/I = R or V = IR
Graphing Ohm’sLaw: A demoon the black-board.
VII
VV
Variable resistor: VaryR and recordV and I. Graph.
V
I
R=4
R=2http://webphysics.ph.msstate.edu/jc/library/18-2/ohmslaw.htm
2 4
48
In “Ohmic” devices…..that follow ohm’s law the current is proportional to the voltage. V α I therefore the ratio of V to Iis 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 addresistors in series??
Variable resistor: VaryR and recordV and I. Graph.
V
I
R=4
R=2
2 4
48 .. a
b
What happens when you add anotherresistance in series? Is the resistance going “up” staying the “same” or “going down” ??
If you compare slides 4,5,and 7 youcan see that adding wires in series increases the resistance and lowersthe current. Adding two 2 ohm resistors in series will make the currenthalf and therefore must double thetotal resistance. Look at the graph frompoint “a” to point “b”.
Resistors placed in series have their values added to find the total resistance.
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