Network Analysis II
Using Method of Branch Currents to
solve for unknowns
Circuit overview
Using the method of
branch currents,
solve for the
unknown values of
voltage and current
in the figure shown.
To do this, complete
steps a through m.
The assumed
direction of all
currents is shown in
the figure.
Listing of steps
a. Using Kirchhoff’s current law, write an equation
for the currents I1, I2 and I3 at point C.
b. Specify the current I3 in terms of I1 and I2.
c. Write a KVL equation for the loop ABDCA, going
counterclockwise from point A, using the terms
V1, VR1 and VR3. This loop will be called Loop 1.
d. Write a KVL equation for the loop FECDF, going
counterclockwise from point F, using the terms
V2, VR2 and VR3. This loop will be called Loop 2.
Listing of steps (cont.)
e. Specify each resistor voltage drop as an IR
product using the actual resistor values for R1, R2
and R3.
f. Rewrite the KVL equation for Loop 1 in step c
using the IR voltage values for VR1 and VR2
specified in step e.
g. Rewrite the KVL equation for Loop 2 in step c
using the IR voltage values for VR2 and VR3
specified in step e.
h. Reduce the Loop 1 and Loop 2 equations in step
f and step g to their simplest possible form.
Listing of steps (cont.)
i. Solve for the currents I1 and I2 using any of
the methods for the solution of simultaneous
equations. Next, solve for I3.
j. In the figure under consideration, were the
assumed directions of all currents correct?
How do you know?
k. Using the actual values of I1, I2 and I3,
calculate the individual voltage drops.
Listing of steps (cont.)
l. Rewrite the KVL loop equations for both Loops 1
and 2 using actual voltage values. Go
counterclockwise around both loops when
adding voltages. (Be sure that the resistor
voltage drops all have the correct polarity based
upon the actual directions for I1, I2 and I3).
m. Based upon the actual directions for I1, I2 and I3,
write a KCL equation for the currents at point C.
Step a
a. Using Kirchhoff’s
current law, write an
equation for the
currents I1, I2 and I3 at
point C.
• I1 + I3 – I2 = 0
Step b
b. Specify the current I3
in terms of I1 and I2.
• I3 = I2 – I1
Step c
c.
Write a KVL equation for
the loop ABDCA, going
counterclockwise from
point A, using the terms
V1, VR1 and VR3. This loop
will be called Loop 1.
• –V1 – VR3 + VR1 = 0
Step d
d. Write a KVL equation for
the loop FECDF, going
counterclockwise from
point F, using the terms V2,
VR2 and VR3. This loop will
be called Loop 2.
• –V2 + VR2 + VR3 = 0
Step e
e. Specify each resistor
voltage drop as an IR
product using the
actual resistor values
for R1, R2 and R3.
• VR1 = I1R1 = I1*12 = 12I1
• VR2 = I2R2 = I2*24 = 24I2
• VR3 = I3R3 = (I2 – I1)12 =
12(I2 – I1)
Step f
f. Rewrite the KVL
equation for Loop 1 in
step c using the IR
voltage values for VR1
and VR2 specified in
step e.
• –V1 – 12(I2 – I1) + 12I1 = 0
Step g
g. Rewrite the KVL
equation for Loop 2 in
step c using the IR
voltage values for VR2
and VR3 specified in
step e.
• –V2 + 24I2 +12(I2 – I1) = 0
Step h
h. Reduce the Loop 1 and Loop 2 equations in step f and
step g to their simplest possible form.
• –V1 – 12(I2 – I1) + 12I1 = 0
(distribute the 12) =>
• –V2 – 12I2 + 12I1 + 12I1 = 0
(combine like terms) =>
• –V1 – 12I2 + 24I1 = 0
(substitute in value for V1) =>
• –24 – 12I2 +24I1 = 0
(factor out common multiple of 12) =>
• –2 – I2 + I1 = 0 ∴ –I2 + 2I1 = 2V
Step h (cont.)
• –V2 + 24I2 +12(I2 – I1) = 0
(distribute the 12) =>
• –V2 + 24I2 + 12I2 – 12I1 = 0
(combine like terms) =>
• –V2 + 36I2 – 12I1 = 0
(substitute in value for V2) =>
• –12 + 36I2 – 12I1 = 0
(factor out common multiple of 12) =>
• –1 + 3I2 – I1 = 0 ∴ 3I2 – I1 = 1V
Step i
i. Solve for the currents I1 and I2 using any of the
methods for the solution of simultaneous
equations. Next, solve for I3.
(reorder first formula to solve for I2) =>
• 2I1 – 2 = I2
(substitute value for I2 into second formula) =>
• 3(2I1 – 2) – I1 = 1
(distribute the 3) =>
• 6I1 – 6 – I1 = 1
Step i (cont.)
(combine like terms) =>
• 5I1 – 6 = 1
(move all constants to one side [right]) =>
• 5I1 = 7
(solve for I1 by dividing out the 5 on both sides)
=>
• I1 = 7/5 = 1.4A
Step i (cont.)
(substitute value for I1 back into formula) =>
• 2(1.4) – 2 = I2
(multiply the two values) =>
• 2.8 – 2 = I2 ∴ I2 = 0.8A or 800 mA
(substitute both values for I1 and I2 into formula
found in step b) =>
• I3 = I2 – I1 = 0.8A – 1.4A = –600 mA
Step j
j.
In the figure under
consideration, were the
assumed directions of all
currents correct? How
do you know?
• No. I3 should be a positive
value, which indicates
that I1 should set the
polarity for VR3 and the
direction of I3, not I2 as
indicated.
Step k
R1
A
C
12Ω
V1
24 V
R2
E
24Ω
R3
12Ω
V2
12 V
D
B
F
k. Using the actual values
of I1, I2 and I3, calculate
the individual voltage
drops.
• VR1 = I1R1 = 1.4(12) =
16.8V
• VR2 = I2R2 = 0.8(24) =
19.2V
• VR3 = I3R3 = 0.6(12) =
7.2V
Step l
l.
Rewrite the KVL loop equations
for both Loops 1 and 2 using
actual voltage values. Go
counterclockwise around both
loops when adding voltages. (Be
sure that the resistor voltage
drops all have the correct
polarity based upon the actual
directions for I1, I2 and I3).
• –24 + 7.2 + 16.8 = 0
(Loop 1 [ABDCA])
• –12 + 19.2 – 7.2 = 0
(Loop 2 [FECDF])
Step m
m. Based upon the actual
directions for I1, I2 and
I3, write a KCL
equation for the
currents at point C.
• I1 – I2 – I3 = 0
• 1.4A – 0.8A – 0.6A = 0