Question

What is a current source and how to solve using kcl and kvl

Answer 1

Answer:

For each of the $l$ independent loops in the circuit, define a loop current around the loop in clockwise (or counter clockwise) direction. These $l$ loop currents are the unknown variables to be obtained.

Apply KVL around each of the loops in the same clockwise direction to obtain $l$ equations. While calculating the voltage drop across each resistor shared by two loops, both loop currents (in opposite positions) should be considered.

Solve the equation system with $l$ equations for the $l$ unknown loop currents.

loopcurrentmethod.gif

Find currents $I_1$ from a to b, $I_2$ from c to b, and $I_3$ from b to d.

Assume two loop currents $I_a$ and $I_b$ around loops abda and bcdb and apply the KVL to them:  

$\displaystyle \sum V_{abda}$ $\textstyle =$ $\displaystyle -32+2I_a+8(I_a-I_b)=0$  

$\displaystyle \sum V_{bcdb}$ $\textstyle =$ $\displaystyle 8(I_b-I_a)+4I_b+20=0$

Explanation:

Answer 2

Answer:

The Kirchhoff's Laws are generally named as KCL (Kirchhoffs Current Law) and KVL (Kirchhoffs Voltage Law). The KVL states that the algebraic sum of the voltage at node in a closed circuit is equal to zero. ... For these kinds of calculations, we can use KVL and KCL.

Explanation:

Here in this simple single junction example, the current IT leaving the junction is the algebraic sum of the two currents, I1 and I2 entering the same junction. That is IT = I1 + I2.

Note that we could also write this correctly as the algebraic sum of: IT - (I1 + I2) = 0.

So if I1 equals 3 amperes and I2 is equal to 2 amperes, then the total current, IT leaving the junction will be 3 + 2 = 5 amperes, and we can use this basic law for any number of junctions or nodes as the sum of the currents both entering and leaving will be the same.

Also, if we reversed the directions of the currents, the resulting equations would still hold true for I1 or I2. As I1 = IT - I2 = 5 - 2 = 3 amps, and I2 = IT - I1 = 5 - 3 = 2 amps. Thus we can think of the currents entering the junction as being positive (+), while the ones leaving the junction as being negative (-).

Then we can see that the mathematical sum of the currents either entering or leaving the junction and in whatever direction will always be equal to zero, and this forms the basis of Kirchhoff’s Junction Rule, more commonly known as Kirchhoff’s Current Law, or (KCL).

Resistors in Parallel:

Let’s look how we could apply Kirchhoff’s current law to resistors in parallel, whether the resistances in those branches are equal or unequal. Consider the following circuit diagram:

Applying KCL to more complex circuits.

We can use Kirchhoff’s current law to find the currents flowing around more complex circuits. We hopefully know by now that the algebraic sum of all the currents at a node (junction point) is equal to zero and with this idea in mind, it is a simple case of determining the currents entering a node and those leaving the node. Consider the circuit below.