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.