In applying Kirchhoff’s law to specific problems, particular attention
should be paid to the algebraic signs of voltage drops and e.m.fs, otherwise
the result will come out wrong. The following sign convention is suggested.
should be paid to the algebraic signs of voltage drops and e.m.fs, otherwise
the result will come out wrong. The following sign convention is suggested.
a.
Sign of battery e.m.f
Sign of battery e.m.f
A rise in voltage should be given a positive sign and a fall in voltage
should be given a negative sign. Keeping this in mind, it is clear that as we
go from the negative terminal of a batter to its positive terminal, as shown in
figure 1 below, there is a rise in potential; hence this voltage should be
given a positive sign. If on the other hand, we go from a positive terminal to
the negative terminal, there is a fall in potential, so the voltage should be
given a negative sign. It is important to note that the sign of the battery
e.m.f is independent of the direction of current through that branch.
should be given a negative sign. Keeping this in mind, it is clear that as we
go from the negative terminal of a batter to its positive terminal, as shown in
figure 1 below, there is a rise in potential; hence this voltage should be
given a positive sign. If on the other hand, we go from a positive terminal to
the negative terminal, there is a fall in potential, so the voltage should be
given a negative sign. It is important to note that the sign of the battery
e.m.f is independent of the direction of current through that branch.
In analysis,
the IR drops across a mesh, if we go through a resistor in the same direction
as the current in Figure 2a above, there is a fall in potential because current
flows from a higher to a lower potential. Hence, their will be a fall in
voltage and it should be given a negative sign. However, if you go through the
resistors in a direction in opposition to that of the current flowing through
it, there is a rise in potential, and then the voltage drop should be given a
positive sign.
the IR drops across a mesh, if we go through a resistor in the same direction
as the current in Figure 2a above, there is a fall in potential because current
flows from a higher to a lower potential. Hence, their will be a fall in
voltage and it should be given a negative sign. However, if you go through the
resistors in a direction in opposition to that of the current flowing through
it, there is a rise in potential, and then the voltage drop should be given a
positive sign.
From the
voltage drop illustrations, it is clear that the sign of voltage drop across a
resistor is dependent on the direction of flow of current through that resistor
but independent of the polarity of any other source of e.m.f in the circuit
under consideration.
voltage drop illustrations, it is clear that the sign of voltage drop across a
resistor is dependent on the direction of flow of current through that resistor
but independent of the polarity of any other source of e.m.f in the circuit
under consideration.
Assuming the direction
of current when applying Kirchhoff’s laws
of current when applying Kirchhoff’s laws
In the
application of Kirchhoff’s laws to electrical networks, there have always been
questions on the proper direction of current to assume. The direction of
current assured should not be a thing to be worried about. The direction of
flow of current can be assumed to be either clockwise or anti-clockwise. If the
assured direction of current is not the actual direction, then on solving the
question, the current will be found to have a minus sign. If the answer is
positive, then the assumed current is the same as the actual direction.
However, the important point is that once a particular direction has been
assumed, the same should be used throughout the solution of the question.
application of Kirchhoff’s laws to electrical networks, there have always been
questions on the proper direction of current to assume. The direction of
current assured should not be a thing to be worried about. The direction of
flow of current can be assumed to be either clockwise or anti-clockwise. If the
assured direction of current is not the actual direction, then on solving the
question, the current will be found to have a minus sign. If the answer is
positive, then the assumed current is the same as the actual direction.
However, the important point is that once a particular direction has been
assumed, the same should be used throughout the solution of the question.
Note: it
should be noted that Kirchhoff’s law are applicable both to d.c and a.c
voltages and currents. However, in the case of alternating currents, any e.m.f
of self inductance or that existing across a capacitor should also be taken
into account
should be noted that Kirchhoff’s law are applicable both to d.c and a.c
voltages and currents. However, in the case of alternating currents, any e.m.f
of self inductance or that existing across a capacitor should also be taken
into account
Example
Calculate the
current in the 7Ω resistor in the figure shown below. All values of resistors
are in ohms
current in the 7Ω resistor in the figure shown below. All values of resistors
are in ohms
Solution
Applying KVL to each loop, we have
At loop 1,
(5 + 4 + 7)I1 – 7I2
– 4I3 = -5
– 4I3 = -5
16I1 – 7I2 – 4I3
= -5 – – – – – – 1
= -5 – – – – – – 1
At loop 2,
(7 + 2 + 1)I2 – 7I1
– I3 = 0
– I3 = 0
-7I1 + 10I2 – I3
= 0 – – – – – – 2
= 0 – – – – – – 2
At loop 3
(4 + 3 + 1)I3 – 4I1
– I2 = 5
– I2 = 5
-4I1 – I2 + 8I3
= 5 – – – – – – – 3
= 5 – – – – – – – 3
Equation 1, 2, and 3 can be solved
simultaneously using matrix method
simultaneously using matrix method
To solve for i1 and i2,
we use Crammer’s rule
we use Crammer’s rule
Note: The negative
sign in the value of current flowing through I7 means that the
actual direction of current flow is in opposition to that assumed in analysing
the mesh.
sign in the value of current flowing through I7 means that the
actual direction of current flow is in opposition to that assumed in analysing
the mesh.
