Circuit With One Series and Three Parallel Resistors.

The following circuit and values as shown in figure 1 below are provided. We want to solve this circuit. In particular, the currents in the parallel resistors have to be determined.

FIG. 1: Circuit with One Series and Three Parallel Resistors.

The following values are given: V1=10V, R1=200Ω, R2=300Ω, R3=400Ω, and R4=100Ω.
Let's start with resistors R1, R2, and R3 and determine an equivalent value ReqR1R2R3 for them.
Since R1, R2, and R3 are in parallel we can say:
1/ ReqR1R2R3  = 1/ R1 + 1/ R2 + 1/ R3
                       = 1/200Ω + 1/300Ω + 1/400Ω
                       = 5.0x10^-3(1/Ω) + 3.33x10^-3(1/Ω) + 2.50x10^-3(1/Ω)
                       = 10.83x10^-3(1/Ω)
Thus ReqR1R2R3 = 92.31Ω.
We can now redraw the circuit by replacing R1, R2, and R3 with ReqR1R2R3 as follows:

FIG. 2: Circuit with R₁, R₂, and R₃ Replaced with R_eqR1R2R3.

As R₄ and R_eqR1R2R3 are in series to determine the total resistance of the circuit R_tot we simply add R₄ and R_eqR1R2R3.

Thus, it can be said that:
Rtot = R4 + ReqR1R2R3
       = 100Ω + 92.31Ω
Rtot = 192.31Ω

Now to calculate the total current in the circuit Itot it can be said that:
Itot = V1/Rtot
      = 10V/192.31Ω
      = 52.00x10^-3A
Or it can be said that Itot = 52mA.

 

FIG. 3: Circuit with I_tot=52mA. 

The I_tot = 52mA also flows through resistor R_eqR1R2R3. So, we can calculate the voltage drop across R_eqR1R2R3, V_eqR1R2R3.

VeqR1R2R3 = Itot x ReqR1R2R3
                  = 52mA x 92.31Ω
                  = 52x10-3A x 92.31
                  = 4.8V
VeqR1R2R3 = 4.8V

But remember VeqR1R2R3 is also the voltage drop across resistors R1=200Ω, R2=300Ω, and R3=400Ω as they make up ReqR1R2R3 = 92.31Ω. So, the current in resistors in resistors R1, R2, and R3=400Ω can be calculated as follows:
IR1 = VeqR1R2R3/R1 = 4.8V/200Ω = 24x10-3A or = 24mA.
IR2 = VeqR1R2R3/R2 = 4.8V/300Ω = 16x10-3A or = 16mA.
IR3 = VeqR1R2R3/R3 = 4.8V/400Ω = 12x10-3A or = 12mA.

Itot = IR1 + IR1 + IR3
     = 24mA + 16mA + 12mA
     = 52mA.

This is correct and complies with our previous calculation of Itot=52mA.

FIG. 4: Circuit with I_tot=52mA, I_R1=24mA, I_R2=16mA, I_R3=12mA, and Veq_R1R2R3=4.8V.

The voltage drop across the resistor R4, VR4can now be calculated as follows:

VR4  =10V-VeqR1R2R3
        = 10V-4.8V
        = 5.2V
VR4 = 5.2V

This circuit was also implemented in LTspice. See figure 5 below:

FIG. 5: Circuit Implemented in LTspice.

The current measured in R₄ in LTspice is as shown in figure 6 below:

FIG. 6: Current in R_4, IR4=52mA Measured in LTspice.

Shown in figure 6, the current in resistor R₄, I_R4=52mA as measured in LTspice. This is the same as I_tot=52mA calculated above.

The current measured in resistor R₁ in LTspice is shown in figure 7 below:

FIG. 7: Current in R_1, IR1=24mA Measured in LTspice.

Shown in figure 7, the current in R₁, I_R1=24mA as measured in LTspice. This is the same as I_R1=24mA calculated above.

The current measured in R₂ in LTspice is shown in figure 8 below:

FIG. 8: Current in R2, IR2=16mA Measured in LTspice.

Shown in figure 8, the current in R₂, I_R2=16mA as measured in LTspice. This is the same as I_R2=16mA calculated above.

The current measured in R₃ in LTspice is shown in figure 9 below:

FIG. 9: Current in R_3, IR3=12mA Measured in LTspice.

Shown in figure 9, the current in R₃, I_R3=12mA as measured in LTspice. This is the same as I_R3=12mA calculated above.

Voltage measured across R₁, R₂, and R₃, Veq_R1R2R3 in LTspice shown in figure 10 below:

FIG. 10: Voltage across R₁, R₂, and  R₃, Veq_R1R2R3=4.8V Measured in LTspice.

 Shown in figure 10 above, the voltage across R₁, R₂, and  R₃, Veq_R1R2R3=4.8V as measured in LTspice. It is the same as Veq_R1R2R3=4.8V calculated previously.

 

Circuit With Series and Parallel Resistors

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.