Question 1.26: Using the circuit of Figure 1.49, assume that R1 = 6 Ω, R2 =...
Using the circuit of Figure 1.49, assume that R_1 = 6 Ω, R_2 = 3 Ω, R_3 = 10 Ω, V_1 = 5 V, and V_2 = 4 V. Write the KCL equation for node A and KVL equations for loops 1 and 2 as shown on the figure. For the reader’s convenience Figure 1.49 is repeated here and referred to as Figure 1.50.
By inspection of Figure 1.50 and applying KCL for node A,
I_1+I_2+=I_3 (1.171)
Applying KVL on loop 1,
V_1=I_1R_1+=I_3R_3 (1.172)
Applying KVL on loop 2,
V_2=I_2R_2+=I_3R_3 (1.173)
Equations (1.171) through (1.173) are three linearly independent equations, and since we have three unknowns which are I_1, I_2, and I_3, we obtain unique solutions for each branch current. Plugging the resistor and voltage sources values given into Equations (1.172) and (1.173) we obtain
5=I_16+=I_310 (1.174)
4=I_23+=I_310 (1.175)
I_3=I_1+I_2 (1.176)
The solving of the three simultaneous Equations (1.174) through (1.176) is left as an exercise to the reader. As promised on the Preface, this book covers hardware essentials, and it is not intended to be a math book. The solutions for the three currents are
I_1 = 0.2315 A (1.177)
I_2 = 0.1296 A (1.178)
I_3 = 0.3611 A (1.179)
All three numerical results were rounded to the fourth decimal place. Let us plug results given by Equations (1.177) through (1.179) into Equations (1.174) through (1.176):
5 = 0.2315×6 + 0.3611×10, (1.180)
4=0.1296×3 +0.3611×10, (1.181)
0.3611 = 0.1296 + 0.2315 . (1.182)
It is also easy to verify that the voltage at node A (Fig. 1.50) is
V_A = I_3R_3. (1.183)
V_A = 0.36111 A \times 10 Ω=3.6111 V. (1.184)
It is also instructive to realize that node voltage V_A also equals from Figure 1.50 using KVL:
V_A = V_1-I_1R_1. (1.185)
Using the values for V_1, I_1, and R_1 in Equation (1.185) we find that
V_A =5-0.2315\times 6 =3.6111 V. (1.186)
Applying KVL to loop 2 of the circuit of Figure 1.50,
V_A = V_2-I_2R_2. (1.187)
And using the values known for V_2, I_2, and R_2 yields
V_A =4-0.1296\times 3=3.6112 V. (1.188)
In summary, the nodal voltage V_A was found using one set of KVL Equations (1.174) and (1.175) and found to be the same when using an alternate set of KVL equations given by Equations (1.185) through (1.187). Keep in mind that due to the use of finite precision (4 decimal places in our example), the numbers may not be 100% exact. This is just due to numerical round-off errors and not to Kirchhoff’s laws.