## Q. 2.7

Using the KVL method, determine the current flows,$I_1$ and $I_2$, in the circuit below.

## Verified Solution

Current $I_1$ and $I_2$ are two of the five unknown parameters in the circuit above. The other unknown parameters are $V_1$ Ω, $V_2$ Ω, and $V_4$ Ω. One approach for determining the values of $I_1$ and $I_2$ would be to formulate two equations, using the KVL, such that each equation includes the same two unknown variables, $I_1$ and $I_2$. Then, by applying the simultaneous equation technique to the two-equation system, with two unknowns, we can determine the values of $I_1$ and $I_2$.
Application of KVL requires that voltage around each circuit element be defined in terms of known values and the unknown variables. In addition, the sign or polar-ity of each voltage must be assigned. See the diagram below.

Rules for assuming the current direction and assigning voltage polarities to various loads (resistors) and sources (voltage sources) are as follows:
1. As shown in the figure above, by convention, the currents are assumed to be emanating from the positive pole (positive electrode or anode) of the voltage source and are assumed to be terminating into the negative pole (negative electrode or cathode) of the voltage source.
2. As shown in the figure above, the end or side of the resistor or load that the current enters from is labeled as positive.
3. The polarity for a voltage source is assumed as encountered in the direc-tion of current flow

Examination of the circuit above reveals that there are three loops in the given circuit. The left loop will be referred to as loop 1, the loop on the right segment of the circuit is loop 2. The third loop in this circuit is formed by the outer perimeter. We will focus on the first two loops to derive two equations for the determination of the two unknown currents. Assume that $I_1$ is greater than $I_2$. Conventionally, it is acceptable to make such assumptions as long as the assumptions are, strictly, adhered to in deriving all equations necessary for the solution.
“Walking” loop 1, beginning at the negative terminal of the 24 V DC source, yields the following equation:

$-24 V+V_{1Ω}+V_{2Ω}+12 V=0$      (2.9)

“Walking” loop 2, beginning at the negative terminal of the 12 V DC source, yields the following equation:

$-12V-V_{2Ω}+V_{4Ω}=0$    (2.10)

Based on Ohm’s law:

$V_{1Ω}=(I_1).(1 Ω)=I_1$      (2.11)

$V_{4Ω}=(I_2).(4 Ω)=4I_2$      (2.12)

$V_{2Ω}=(I_1-I_2).(2 Ω)=2.(I_1-I_2)$     (2.13)

Then, by substituting Eqs. 2.11–2.13 into Eqs. 2.9 and 2.10, we get:

$-24+I_1+2.(I_1-I_2)+12=0$

$3I_1-2I_2=12$      (2.14)

$-12-2.(I_1-I_2)+4I_2=0$

$-2I_1+6I_2=12$      (2.15)

Equations 2.14 and 2.15 represent the two simultaneous equations that were needed to solve for currents $I_1$ and $I_2$. These equations will be solved simultane-ously to determine the values of $I_1$ and $I_2$.

For simultaneous equation solution, multiply left- and right-hand sides of Eq. 2.14 by 3 and add it to Eq. 2.15:

$\begin{matrix}9I_1-6I_2=36\\\underline{-2I_1+6I_2=12}\\7I_1=48\end{matrix}$

$∴I_1=6.86 A$

Then, by substituting this value of $I_1$ into Eq. 2.15 yields:

$-2(6.86)+6I_2=12$

Or

$I_2=4.29 A$

Note: The values of unknown currents $I_1$ and $I_2$ can also be determined by apply-ing Cramer’s Rule to Eqs. 2.14 and 2.15 in matrix format and linear algebra.