Kirchhoff’s Circuit Law and Circuit Analysis

circuisKirchhoff’s Circuit Law and Circuit Analysis

When we deal with the complex electric circuits (like T-circuits), we use some mathematical techniques to find the current and voltage drops in the circuit. It’s not as difficult as it may seem at the first sight. In this post, we will tell you how to analyze the circuit with the help of the Kirchhoff’s Circuit Law.

There are two simple rules about the voltage and the current in the closed electric circuit. The first one, called Kirchhoff’s Current Law, states that the sum of the currents flowing into a point (or node) in a circuit equals the sum of the currents flowing out. The second, Kirchhoff’s Voltage Law, states that the sum of the voltage drops around any closed circuit is zero. Let’s now figure out what a branch, a node and a loop mean. The branch is simply the component of the circuit such as the resistor or a battery. The node is the point of connection between two or more branches. And, finally, the loop is any closed path in the circuit.

Let’s consider, for example, T-circuit in the scheme below:

kirch

Let the first voltage source (or the battery) be 10\ V and the second one 15\ V. And let the resistances be R_1=15\ \Omega, R_2=22\ \Omega and R_3=30\ \Omega. Our task is to analyze the circuit and find the currents I_1, I_2, I_3 and the voltage drops on resistances R_1, R_2, R_3.

As we can see in the scheme above, this circuit has three branches, two nodes (designated as 1 and 2) and three loops (designated as A, B and C). So, firstly, applying Kirchhoff’s Current Law to the nodes 1 and 2, we get:

I_3 = I_1 + I_2

Then, we can apply Kirchhoff’s Voltage Law to the loops A, B and C:

V_1 = I_1R_1 + I_3R_3 (1)

V_2 = I_2R_2 + I_3R_3 (2)

V_1 – V_2 = I_1R_1 – I_2R_2 (3)

Let’s substitute I_3 in equations (1)-(2) to the sum of currents I_1 and I_2:

V_1 = I_1(R_1+R_3) + I_2R_3

V_2 = I_2(R_2+R_3) + I_1R_3

After substituting the numbers, we get the system of equations:

45I_1 + 30I_2 = 10 (4)

30I_1 + 52I_2 = 15 (5)

Let’s express I_1 from the equation (5) via I_2 and substitute it into the equation (4):

I_1=(15-52I_2)/30

I_2=0.26\ A

As we know I_2, we can find I_1:

I_1=0.05\ A

Finally, we can find I_3:

I_3=I_1+I_2 = 0.31\ A

Then, the voltage drops on resistances R_1, R_2, R_3 will be:

I_1R_1 = 0.05\ A \times 15\ \Omega = 0.75\ V

I_2R_2 = 0.26\ A \times 22\ \Omega = 5.72\ V

I_3R_3 = 0.31\ A \times 30\ \Omega = 9.3\ V

So, as you may see, it is easier than you expected. We are glad to help you with this and other problems in physics, just let us know about them!

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