Solving Circuit Having a Current Source with Kirchhoff’s Law

 We are given the particulars in the circuit below in FIG. 1 and asked to calculate E₂.

FIG. 1: Circuit Having a Current Source

The voltage drop across resistor R₁=600Ω is given as 80V. Thus, using Ohm’s law of I=V/R the current flowing in R₁ can be calculated. The current flowing through R₁ is chosen to be I₂ in the circuit above.

I₂ = 80V/600Ω

   = 0,13A

Kirchhoff’s current law (KCL) which states:

“…total current or charge entering a junction or node is exactly equal to the charge leaving the node as it has no other place to go except to leave, as no charge is lost within the node.“

Using KCL and the currents chosen and indicated in the circuit it can be said that:

I₃ + I₂ = I₁

I₃ = I₁ – I₂, I₂ was calculated as 0,16A and I₁ is given as 3A.

   = 3A-0,13A

   = 2,87A

I₃ = 2,87A

Kirchhoff’s voltage law (KVL) which states:

“…in any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loop,”

Using (KVL) it can be stated that:

E₂ = I₃R₂ + I₁R₃

     = (2,87A)(150Ω) + (3A)(300Ω)

      = 431V + 900V

      = 1 331V

Therefore E₂ = 1 331V.

E₁ = 80V-(I3)(R2)

E₁ = 80V - (I₃)(150Ω) + 1 331V

     = 80V - (2,87A)(150Ω) + 1 331V

= 80V - 431V + 1 331V

= 980V

Therefore E₁ = 980V.

What we will do next is implement our circuit in LTspice and check if it is correct. Shown in FIG. 2 below is FIG. 1 implemented in LTspice.

FIG. 2: LTspice Circuit

So, we are working backwards so to speak. We have set V1 which is our E₁. We calculated it to be = 980V so that V1=980V. We also have set V2 which is our E₂ which we calculated to be = 1331V so that V2=1331V. Now we are going to make measurements to see if it complies with what we were originally given.

FIG. 3: Voltage Across Resistor R1 = 79.19999V

As can be seen in FIG. 3 the voltage across according to LTspice is R1 = 79.199999V. This compares well with the 80V we were initially given. Let’s now measure the current in R1 according to LTspice shown in FIG. 4 below.

FIG. 4: Current in Resistor R1 = 132mA.

As can be seen in FIG. 4 the current in R4 according to LTspice is 132mA. This compares well with our previous calculation of I₂ = 80V/600Ω  = 0,13A. Let’s measure the current I₃ according to LTspice which is the current flowing through resistor R2.

FIG. 5: Current in Resistor R2 = 2.868A

As can be seen in FIG. 5 the current in resistor R2 according to the LTspice simulation is 2.868A. According to FIG. 1, the current is I₃. We calculated that to be 2.87A which compares well with LTspice’s calculations.