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ELECTRIC CURRENT 2. Ohm’s law shows the relationship between current, potential, and voltage. We need a few more rules to make predictions about current flow through circuits. Rule 1: Conservation of Charge. - PowerPoint PPT Presentation
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ELECTRIC CURRENT 2
Ohm’s law shows the relationship between current, potential, and voltage.
We need a few more rules to make predictions about current flow through circuits.
Rule 1: Conservation of Charge
The number of charges flowing into a point is the same as the number of charges flowing out of a point
5 amps
3 amps
?
5 amps
3 amps
2 amps
Rule 2: Conservation of Energy
The total drop in potential energy of a circuit is equal to the total voltage of the circuit
SERIES CIRCUITS
The voltage drops across each resistor, but the total voltage drop is still 60 volts.
So, V1 + V2 = 60 volts…
V1
V2
And V=IR
R1
R2
V1
V2
R1
R2
So, IR1 + IR2 = IRtotal
But the current is the same through R1 and R2.
V1
V2
R1
R2
IRtotal = I (R1 +_R2)
Rtotal = R1 +_R2
or
Summary:
For series circuits the total resistance is equal to the sum of all the resistors in series.
Functionally, this is the same as increasing the length of a resistor.
As L increases, so does R.
A
LR
From the Reference Tables
V1
V2
R1
R2
So, back to our problem!
R1+ R2 = 30 Ω = total resistance
Therefore the total current = V/R = 2 amps
2 A
V1= 40 volts
Remember, IR1 + IR2 = IRtotal
So: (2 amps x 20 ohms) = 40 voltsand
(2 amps x 10 ohms) = 20 volts
V2 = 20 volts
PARALLEL CIRCUITS
The voltage drop across R1 and R2 is the sameand
R1 R2It
I1 I2
It = I1 + I2 R
VI
R1 R2It
I1 I2
21 R
V
R
V
R
V
t
R1 R2It
I1 I2
So:
21
111
RRRt
Summary:For parallel circuits the reciprocal of the total resistance is equal to the sum of the reciprocal of each resistance.
Functionally, this is the same as increasing the area of a resistor.
As A increases, R decreases.
A
LR
From the Reference Tables
R1 R2It
I1 I2
Back to our problem
5
255
5
1
5
11 11
tR
R1 R2It
I1 I2
5.22
5
5
21 1ansorRR t
t
Notice that the total resistance is less that either one
R1 R2It
I1 I2
amps6.05.2
5.1
V
R
VI
tt
We can now calculate the total current…
R1 R2
It= 0.6 amps
I1 I2
amps3.05
5.121
VII
…and the current through each resistor.
amps6.0amps3.0amps3.021 III t
Finally,
R1 R2
It= 0.6 amps
I2= 0.3 amps I2= 0.3 amps
