The following circuit shown in figure 1 below has the following values: R1=6.5Ω, R2=4Ω, R3=8Ω, R4=5Ω, and R5=6Ω. The voltage is given as V1=9V. We want to solve the circuit and determine the voltage drop across R5 as well as the current flowing in R5.
FIG. 1: Circuit to be Solved.
Let’s start with resistors R2 and R3 and determine an equivalent value ReqR2R3 for resistors R2 and R3. Since R2 and R3 are in parallel we can say:
1/ ReqR2R3 = 1/R2+1/R3
= 1/4Ω+1/8Ω
= 0,25(1/Ω) + 0,125(1/Ω)
1/ ReqR2R3 = 0,375(1/Ω).
Rounded off to first two digits: ReqR2R3 = 2,67Ω.
We can now redraw the circuit by replacing R2 and R3 with ReqR2R3 as follows as shown in figure 2 below:
FIG. 2: Circuit Redraw with ReqR2R3.
As ReqR2R3 and R4 are in series we can determine the equivalent resistance for ReqR2R3 and R4, which is indicated as ReqR2R3R4. Thus for resistors ReqR2R3 and R4 in series:
ReqR2R3R4 = ReqR2R3 + R4
= 2,67Ω + 5Ω
= 7,67Ω
ReqR2R3R4 = 7,67Ω.
The circuit can thus be redrawn by replacing ReqR2R3 and R4 with ReqR2R3R4 as shown in figure 3 below:
FIG. 3: Circuit Redraw with ReqR2R3R4.
Now what must be obtained is the equivalent resistor ReqR2R3R4R5 for resistors ReqR2R3R4 and R5.
As they are in parallel it can be said that:
1/ReqR2R3R4R5 = 1/ ReqR2R3R4 + 1/R5
= 1/(7,67Ω) + 1/(6Ω)
= 1/(7,67)(1/Ω) + 1/6(1/Ω)
= (1/(7,67) + 1/6)(1/Ω)
= (0,13 +0,17) (1/Ω)
= 0,30(1/Ω).
Or we can say ReqR2R3R4R5 = 3,33Ω.
We can now once again redraw the circuit by replacing resistors ReqR2R3 and R5 with equivalent resistor ReqR2R3R4R5 as shown in figure 4 below:
FIG. 4: Circuit Redraw with ReqR2R3R4R5.
We now have a much-simplified circuit with two series resistors ReqR2R3R4R5=3,33Ω, R1=6,5Ω and a voltage source V1=9V. To calculate the current in the circuit, shown in figure 4, Icircuit, and also flowing through ReqR2R3R4R5 and R1 we can say: Icircuit = V1/(ReqR2R3R4R5+R1 ) = 9V/(3,33Ω+6,5Ω) =9V/9,83Ω = 0,92A Therefore, Icircuit is equal to 0,92A. Now we can calculate the voltage across ReqR2R3R4R5 which is VReqR2R3R4R5. VReqR2R3R4R5 = (Icircuit)(ReqR2R3R4R5) = (0,92A)(3,33Ω) = 3,06V.
Thus VReqR2R3R4R5 = 3,06V Referring to figure 3 ReqR2R3R4R5 is in fact ReqR2R3R4 and R5 in parallel. Thus, VReqR2R3R4R5 is also the voltage across R5. Thus VReqR2R3R4R5 is also the voltage drop across R5 that being VR5, thus VR5=3,06V. The current through R5, IR5 can be calculated as follows: IR5 = VR5/(R5 ) = 3,06V/(6Ω ) = 0,51A
Thus IR5=0,51A.
Figure 1 was simulated on LTspice. Measurements were made during the simulation in LTspice of the voltage drop VR5 and current IR5 in resistor R5 and the results are shown in figure 5 below:
FIG. 5: Measurements of IR5 and VR5 Made During LTspice Simulation.
As can be seen in figure 5 the measurement of IR5=-0,5117A, indicated in blue, and VR5=3,0705V, indicated in green, during the simulation of the circuit in figure 1 in LTspice. Don't worry about the (-) in IR5. This is purely due to the chosen measurement convention. This compares well with the calculated values of IR5=0,51A and VR5=3,06V.
