# Kirchoff`s Laws and Equiv Resistance Jan 2016 ```Kirchoff’s Laws and
Equivalent
Resistance
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
VT = V1 + V2 + . . .
Ohm’s Law
Equivalent Resistance in Series

Equivalent (Total) Resistance in series:
RT

= R1 + R2 + R3…RN
Equivalent (Total) Resistance in parallel:
R T
= 1 / (1/R1 + 1/R2 + 1/R3…1/RN)
Equivalent Resistance in Parallel

Equivalent (Total) Resistance in parallel:
R T
= 1 / (1/R1 + 1/R2 + 1/R3…1/RN)
Equivalent Capacitance in Series

Equivalent (Total) Capacitance in parallel:
CT
= C1 + C2 + C3…CN
Equivalent Capacitance in Parallel

Equivalent (Total) Capacitance in series:
C T
= 1 / (1/C1 + 1/C2 + 1/C3…1/CN)
Series Resistance Example
RT =
I=
Vdrop R1 =
Vdrop R2 =
Vdrop R3 =
RT = 40 
I = V/R = 60/40 = 1.5 A
Vdrop R1 = 60 x (17/40) = 25.5 V
Vdrop R2 = 60 x (12/40) = 18 V
Vdrop R3 = 60 x (11/40) = 16.5 V
25.5 + 18 + 16.5 = 60 Vtotal
Series Resistance Example
RT =
I=
Vdrop R1 =
Vdrop R2 =
Vdrop R3 =
RT = 40 
I = V/R = 60/40 = 1.5 A
Vdrop @ R1 = 60 x (17/40) = 25.5 V
Vdrop @ R2 = 60 x (12/40) = 18 V
Vdrop @ R3 = 60 x (11/40) = 16.5 V
25.5 + 18 + 16.5 = 60 Vtotal
Series Resistance Example
If you are more comfortable using Ohm’s Law to solve
the Vdrop over each load, that will work too!
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
Req = 1/ (1/R1 + 1/R2 + 1/R3 )
Req = 1/ (1/12  + 1/12  + 1/12 )
Req = 1/ 0.083 + 0.083 + 0.083
Req = 1/ .249
Req = 4.02 
OR
Req = 1/ (1/12  + 1/12  + 1/12 )
Req = 1/(3/12) 
Req = 1 / 0.25 
Req = 4.0 
Equivalent Resistance in Parallel
Req = 1/ (1/R1 + 1/R2 + 1/R3 )
Req = 1/ (1/5  + 1/7  + 1/12 )
Req = 1/ 0.2 + 0.143 + 0.083
Req = 1/ 0.426
Req = 2.35 
Combination Circuits
What do we do if a circuit has both series and
Find the equivalent resistance of the loads in
parallel and continue the analysis. E.g.:
Combination Circuits
So…How would we find the current in this complex (combination) circuit??
Req = 1/ (1/R1 + 1/R2 + 1/R3 )
Req = 1/ (1/4 W + 1/4 W )
Req = 1/ (2/4)
Req = 1/ (.5)
Req = 2 
First, find the resistance
in the parallel branch.
Combination Circuits
Req = R1 + Req2+3 + R4
Req = 2 + 2 + 6
Req = 10 
Then find the
overall resistance
in the circuit.
Combination Circuits
I = V/R
I = 12/10
I = 1.2 A
Then use Ohm’s Law
to find the current in
the circuit.
Combination Circuits
What is the
voltage drop
here?
What is the
voltage drop
here?
I = 1.2 A
Req = 10 
V = 12 V
What is the voltage
drop here? (Across
ENTIRE branch.)
Current will split here
into two (2) equal paths
of 0.6 A each. Why
equal parts?
Now you should be able to describe many
Combination Circuits
V= I x R
V = 1.2 x 2
V = 2.4
V= I x R
V = 1.2 x 6
V = 7.2
Vtot = 2.4 + 2.4 + 7.2
Vtot = 12 V
I = 1.2 A
Req = 10 
V = 12 V
V= I x R
V = 1.2 x 2
V = 2.4
If we measure the voltage
drop across each of the
individual 4  resistors
inside the parallel branch,
what will we find?
```