This document discusses the superposition theorem, which allows the effects of multiple independent sources in an electrical circuit to be calculated separately and then summed. It explains that each source can be analyzed individually by setting all other sources to zero. For a circuit with two sources, the document calculates the current and voltage contributions of each source separately, then sums the results to find the total response. It provides an example calculation using two sources (a voltage source and current source) in a circuit with two resistors to demonstrate how to apply the superposition theorem.

1. (Series of Lectures for DAE / B-Tech / BS
Electrical)
Presented by. Engr. Fazal Ur Rehman
Lecturer Electrical GPI Karak
Former HOD Electrical RPI, Rwp

2.  NEED.
 Calculations of KVL and KCL are complicated for
circuit having more than one sources.
 It makes calculation easy and understandable.
 It can be used to find out the voltage and current
of a circuit;
 For series Circuits
 Parallel Circuits
 Complex Circuits

3.  The effects of multiple sources in a circuit
are superimposed on each other. Therefore
the overall effect (current or Voltage) is due
to combined effort of all the sources
connected in a circuit.
OR
 The response in any branch of a circuit
having multiple independent sources is equal
to algebraic sum of responses due to each
source.

5.  Select one source at a time for calculation
 Set all the sources to zero except the
selected one.
 Zeroed Current source is replaced by an open
circuit.
 Zeroed voltage source is replaced by short
circuit.
 Then select second source for calculation
and set all other to zero.
 Repeat the above three steps for all the
sources
 At the end calculate the result from the
algebraic sum of all the sources responses.

8.  In final step we will find the resultant
algebraic sum for the combined effect of two
sources (i.e. V1 and V2 as shown in previous
figures).

10.  For the first voltage source we have the
circuit where current source is zerod
(replaced by an open circuit)
 According to ohm's law, I= V/R= 10V/(1KΩ +
100Ω)= 10V/1.1KΩ= 9.1mA.
 Calculating the voltage drop across the 100Ω
resistor is calculated through ohm's law,
V=IR= (9.1mA)(100Ω)= 0.91V.
 The voltage drop across the 1KΩ resistor is,
V=IR=(9.1mA)(1KΩ)= 9.1V.

11.  For the second source which is a current
source the first source is zeroed (replaced by
a short circuit)
 The current through the 100Ω resistor is,
IS(RT/RX)= 20mA(91Ω/100Ω)= 1.82Ω=
18.2mA.
 The voltage across the 100Ω resistor is, V=IR=
(18.2mA)(100Ω)= 1.82V.
 The voltage across the 1KΩ resistor is, V=IR=
(1.82mA)(1KΩ)= 1.82V.
 The voltage across the 100Ω resistor and the
1KΩ resistor are the same,

12.  The current flowing through the 100Ω resistor
is, 9.1mA + (-18.2mA)= -9.1mA, which means
that current is flowing toward the voltage
source.
 The current flowing through the 1KΩ resistor
is, 9.1mA + 1.82mA= 10.92mA.
 The voltage across the 100Ω resistor is,
0.91V - (1.82V) = -0.91V.
 The voltage across the 1KΩ resistor is,
9.1V + 1.82V= 10.92