I. Loop-current (mesh) Analysis This technique is similar to nodal ...

I. Loop-current (mesh) Analysis
This technique is similar to nodal analysis. This time, the objective is to find loop currents (rather
than node voltages) and you will use KVL (rather than KCL) and Ohm’s Law to solve for loop
currents. The steps are:
1) Label loop currents
2) Write KVL in each loop (unless there is a current source in that loop)
3) Express voltages as current × resistance (V = IR)
Example. Find V by loop analysis.
First label 3 currents in each loop. We generally choose a clockwise direction. I called them I 1 ,
I 2 , and I 3 .
Writing KVL in the first loop, notice that the 25 Ω resistor is shared between loops 1 and 2. The
rule is that the voltage across a shared resistor is R × (difference in shared currents). The current
that comes first is the loop current for the loop you are in. Keeping that in mind…
KVL in loop 1:
-100 + 25(I 1 -I 2 ) = 0
OR
25I 1 – 25I 2 = 100

KVL in loop 2:
25 (I2 – I1) + 10I2 + 50 (I2 – I3) = 0
*Here note 2 things:
Simplify this equation
a) I 2 comes first this time because I am in loop 2
b) For the 10 Ω resistor, the voltage is simply 10× I 2 because the 10 Ω resistor is
not shared between 2 loops.
-25I 1 + 85I 2 – 50I 3 = 0
In Loop 3 you have a current source so you already know I 3 . I 3 is flowing in a direction opposite
to the current source so I 3 = -5 A. No need to write KVL.
Substitute I 3 = -5 A into the equation above.
-25I 1 + 85I 2 = 50(-5)
-25I 1 + 85I 2 = -250
Now go back to the equation from loop 1,
25I 1 – 25I 2 = 100
Solve for I1 and I2
60I 2 = -150 or I 2 = -5/2 = -2.5 A.
Substitute that in and 25I 1 – 25(-2.5) = 100 or 25I 1 = 100-62.5 or I 1 = 1.5 A.
In this problem you were asked to solve for V, the voltage across the 50 Ω resistor (a resistor
shared between loops 2 and 3). You need an expression for V that involves a difference in
currents × 50 and this time the current that comes first will be the one that satisfies the passive
sign convention. The current entering the positive terminal of the resistor is I 2 so I 2 comes first.
V = 50(I 2 – I 3 ) = 50 × (-2.5 – (-5)) = 50 × 2.5 = + 125 V.

II.
More examples of Loop Analysis
Find I 2
Label currents I 0 and I 1 (This time I purposely did not name either of the loop currents I 2 because
that is the current I am trying to find flowing up the middle branch).
KVL in loop 1: 8I 0 + 2 I 0 + 2(I 0 – I 1 ) + 24 = 0
12I 0 – 2I 1 = -24
KVL in loop 2: -24 + 2(I 1 – I 0 ) + 4I 1 + 12 = 0
-2I 0 + 6I 1 = 12
Solve for loop currents. I 1 = 48/34 = 1.412 A and I 0 = -1.76 A. However, you are looking for I 2
and I 2 = I 1 – I 2 because it is a current flowing in a branch shared by the two loops.
I 2 = I 1 – I 2 = 1.412 – (-1.76) = 3.177 A.

Here is an example where a current source is present between two loops. In that case, the
difference in the two loop currents is known. However when KVL is written in the two loops,
you can not express the voltage as current × resistance. In this case, KVL is written around the
entire loop. Go back to the current source for an equation involving the source.
Example. Find V R by loop analysis.
Label loop currents, I 1 and I 2
Note that the difference between loop currents = 4 A. The one that comes first is the one flowing
in the direction of the current source so
I 2 – I 1 = 4 A.
Now write KVL in the outside loop (we are avoiding writing KVL around the current source
even though there is a voltage across it).
-12 + 4 I 1 – 2 + 2I 2 = 0
4 I 1 + 2I 2 = 14
Now you have 2 equations, 2 unknowns
Solve for I 1 , I 2
I 1 = 1 A and I 2 = 5 A. In that case, V R = 2I 2 = 2(5) = 10 V.
If you were also asked for the power associated with the 12 V source, you would find that by
multiplying the voltage by current or 12 (I 1 ) = 12(1) = 12 W.
What is the voltage across the 4A source (V CS ) Once you know the voltage across the 4 Ω
resistor (4I 1 ) or 4Ω(1A) = 4V you can write KVL in the inner loop: -12 + 4 +V CS = 0 so
V CS = 8V.