Kirchhoff’s Law and Superposition Theorem

Question

For the given circuit I found the current and the voltage at R1, R2, R3, using the superposition theorem. 1.I replaced the VS2 with a short then calculated: I1= 0.57mA, I2=0.33mA, I3=0.2mA, V1= 2.68V, V2=2.02V, V3=2.97V. 2. Then I replaced the VS1 with a short and then calculated: I1=-0.99mA, I2=-0.58mA, I3=0.4mA, V1=3.19V, V2=6.7V, V3=3.2V

So, the total current and voltage for each resistor is : I1=0.26mA, I2=-0.65mA, I3=0.91mA V1= 5.86V, V2=9.16V, V3=5.5V

I have to prove the Kirchhoff’s voltage and current Law are valid in the circuit given. Can someone help me with this?

Answer 1

^{simulate this circuit – Schematic created using CircuitLab}

First of all, mark all currents on all branches with their magnitude and direction as you calculated.

To verify Kirchoff's current law:

• For node E, consider currents entering and leaving the node - \$I_{R1}, I_{R2}, I_{R3} \$

• Similarly for node G, consider currents - \$I_{R3}, I_{DG}, I_{GB} \$

• Prove that the algebraic sum of currents leaving the node = algebraic sum of currents entering the node.

To verify Kirchoff's voltage law:

• Consider all 3 loops in the circuit - AEGB, ECDG, ACDB.

• Prove that the sum of all voltage drops and voltage sources = 0 in each loop.

Answer 2

I may have wrongly understood the question but... for Kirchhoff’s Laws.

Assign nodes to the given circuit Before you conduct KCL and KVL, assign nodes to the given circuit. This is necessary.

Apply Kirchhoff’s Current Law Utilize KCL at the middle node (bounded by the 10,000-ohm resistor and the ground). Let us label the node between the two X and Y where X is the top node and Y is the ground. Assign the directions of the current at your will but I will share mine at my own will.

As I have said, apply KCL at node X. By doing so, we have

$$i_1 = i_2 + i_3$$

where

\$i_1\$ is the current entering node X

\$i_2\$ and \$i_3\$ are the currents leaving the node X.

Apply Kirchhoff’s Voltage Law Assign loops to the given circuit. If there is an element that separates the given circuit into two planes, it will have two loops. The direction is assigned by free will.

Let 1 and 2 be the loops on the given circuit. We take the sign conventions into account both for sources and passive elements. A current passing from the negative polarity to the positive polarity of a voltage source is considered a voltage rise while a current that passes from the positive polarity to negative polarity is called voltage drop. On a side note; I have defined both loops to be clockwise direction.

For passive elements, when the direction of the loop is the same as the direction of the current, the passive element takes a voltage drop and when the direction of the loop opposes the direction of the current, the passive element takes a voltage rise.

Let us formulate another set of equations.

At Loop 1:

$$ 5 - 4,700i_1 - 10,000i_2 = 0 $$ $$ 4,700i_1 + 10,000i_2 = 5$$

At Loop 2: $$ -6,800i_3 - 10 + 10,000i_2 = 0$$ $$ 10,000i_2 - 6,800i_3 = 10$$

Solve the system of equations Arrange the equations obtained from KCL and KVL respectively. Solve by any means.

Check whether the answers agree You first did the Principle of Superposition. Double check if the answers are the same.

Answer 3

I have just come across this question and nearly 3 years late but wish to show the asked for equivalence.

Labelling the circuit using the [passive sign convention](Passive sign convention).

Applying Kirchhoff's voltage law for loop BACDB

$$-5 +xR_1+yR_2+10 =0$$

Now let the subscript for currents produced by the 5 volt supply be labelled 5 and those by the 10 volt supply be labelled 10.

$$-5 +(x_5+x_{10})R_1+(y_5+y_{10})R_2+10 =0$$

$$\left[-5+x_5R_1+y_5R_2 \right]+ \left[x_{10}R_1+y_{10}R_2 +10\right] = 0$$

Now note that the terms in the square brackets are those that you would have started with when solving the problem via superposition.