Resistance and Ohm’s Law:
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Resistance and Ohm’s Law:
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Resistance and Ohm’s Law:
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Resistance and Ohm’s Law:
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Kirchoff’s Laws:
Success Criteria
I can state Kirchhoff's laws and Ohm's law
and use them to explain, in quantitative
terms, direct current, potential difference,
and resistance in mixed circuit diagrams.
 I can construct real and simulated mixed
direct current circuits and analyse them in
quantitative terms to test Ohm's and
Kirchhoff's laws.

Kirchoff’s Laws and
Equivalent
Resistance
SPH4C
Kirchoff’s Current Law

At any junction point in an
electrical circuit, the total current
into the junction equals the total
current out of the junction.
Kirchoff’s Current Law
At any junction point in an
electrical circuit, the total current
into the junction equals the total
current out of the junction.
(“What goes in must come out.”)

In the diagram at right,
I1 + I2 = I3
Kirchoff’s Voltage Law

In any complete path in an electrical circuit, the
sum of the potential increases equals the sum
of the potential drops.
Kirchoff’s Voltage Law
In any complete path in an electrical circuit, the
sum of the voltage increases equals the sum of
the voltage drops.
(“What goes up must come down.”)

The Laws for a Series Circuit
The current is the same at all points in the circuit:
IT = I 1 = I 2 = . . .
The total voltage supplied to the circuit is equal
to the sum of the voltage drops across the
individual loads:
VT = V1 + V2 + . . .
Equivalent Resistance in Series
Given
VT = V1 + V2 + . . .
Equivalent Resistance in Series
Given
VT = V1 + V2 + . . .
From Ohm’s Law, V = IR
ITRT = I1R1 + I2R2 + . . .
Equivalent Resistance in Series
Given
VT = V1 + V2 + . . .
From Ohm’s Law, V = IR
ITRT = I1R1 + I2R2 + . . .
Since IT = I1 = I2 = . . .
IRT = IR1 + IR2 + . . .
Equivalent Resistance in Series
Given
VT = V1 + V2 + . . .
From Ohm’s Law, V = IR
ITRT = I1R1 + I2R2 + . . .
Since IT = I1 = I2 = . . .
IRT = IR1 + IR2 + . . .
Divide all terms by I and the equivalent resistance
is the sum of the individual resistances:
Req or RT = R1 + R2 + . . .
Series Resistance Example
Series Resistance Example
Req = R1 + R2 + R3
Series Resistance Example
Req = R1 + R2 + R3
Req = 17 W + 12 W + 11 W = 40 W
Series Resistance Example
IT = VT/Req
Series Resistance Example
IT = VT/Req
IT = 60 V/40 W = 1.5 A
Series Resistance Example
IT = VT/Req
IT = 60 V/40 W = 1.5 A
And I1 = I2 = I3 = IT = 1.5 A
Series Resistance Example
V1 = I1R1 = (1.5 A)(17 W) = 25.5 V
V2 = I2R2 = (1.5 A)(12 W) = 18 V
V3 = I3R3 = (1.5 A)(11 W) = 16.5 V
The Laws for a Parallel Circuit
At a junction:
IT = I 1 + I 2 + . . .
But the total voltage across each
of the branches is the same:
VT = V1 = V2 = . . .
Equivalent Resistance in Parallel
Given
IT = I 1 + I 2 + . . .
From Ohm’s Law, V = IR or I = V/R
VT/RT = V1/R1 + V2/R2 + . . .
Since VT = V1 = V2 = . . .
V/RT = V/R1 + V/R2 + . . .
Divide all terms by V and the reciprocal of the
equivalent resistance is the sum of the
reciprocals of the individual resistances:
1/Req or 1/RT = 1/R1 + 1/R2 + . . .
Equivalent Resistance in Parallel
1/Req = 1/R1 + 1/R2 + 1/R3
1/Req = 1/(12 W) + 1/(12 W) + 1/(12 W)
Using a calculator ...
1/Req = 0.25 W-1
Req = 1 / (0.25 W-1)
Req = 4.0 W
Equivalent Resistance in Parallel
1/Req = 1/R1 + 1/R2 + 1/R3
Equivalent Resistance in Parallel
1/Req = 1/R1 + 1/R2 + 1/R3
1/Req = 1/(5.0 W) + 1/(7.0 W) + 1/(12 W)
1/Req = 0.42619 W-1
Req = 1 / (0.42619 W-1)
Req = 2.3 W
Combination Circuits
What do we do if a circuit has both series and
parallel loads?
Find the equivalent resistance of the loads in
parallel and continue the analysis. E.g.:
More Practice

Equivalent Resistance Lab Activity