Electrical Circuit Analysis | Implementation of KCL and KVL...

# Circuit Analysis | Implementation of KCL and KVL | CS School

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Kirchhoff’s circuit laws are especially important for circuit analysis. There are two circuit laws both we have discussed in previous articles here in EE section. In this article we will talk about implementation of KCL, and KVL in circuit analysis or to solve circuit problem. Kirchhoff’s first law of circuit analysis is the current law. Which is applied to a node, or junction. And the second law is kirchhoff’s voltage law,  which is applied to a closed loop circuit. In a circuit analysis to solve circuit problem we need either current rule, or voltage rule, or both of them. Figure: Current $i_{1}$ passing through node A

KCL – the current rule stated that for current through a node, what goes in must come out. So, if $i_{1}$ current enter into node A (fig above), and $i_{2},$ $i_{3}$ current come out or leave the node. Then according to kirchhoff, $i_{1}=i_{2}+i_{3},$ or $\sum i_{a}=o$.

Similarly second postulate of Kirchhoff’s circuit rule is the voltage law. It stated that algebraic sum of voltage drop across all terminals around a closed loop circuit is always equal to zero.

If voltage source v cause i current to flow through the circuit (above), then voltage drop by $r_{1},$ $r_{2}$ resistors, $ir_{1},$ $ir_{2}$ respectively. Therefore according to Kirchhoff, $v+(-ir_{1})+(-ir_{2})=0$

Here, negative sign indicates the voltage fall, positive sign indicates voltage rise. v is the source that provide or rise voltage, and $ir_{1},$ $ir_{2}$ are the voltage fall, or dropped voltage by $r_{1},$ $r_{2}$ resistors.

Let, $ir_{1}=v_{1},$ $ir_{2}=v_{2},$ therefore, $v+(-v_{1})+(-v_{2})=0$.
Or simply, $\sum v_{i}=0$

Now let’s see how to apply them in a circuit analysis to solve a circuit problem. Suppose we have a circuit below. We need to determine the current through, voltage across, and power dissipated by the two resistors of the circuit.

## Circuit Analysis | Implementation of KCL and KVL

We generally start by labeling the nodes in the circuit. As we have done for the above circuit, nodes $N_{1},$ and $N_{2}$. Then we label the current flowing through the circuit. Here the direction of current is arbitrary. Because, at this point we don’t know the direction of current. After solving the circuit we will be able to choose the right direction. However, $i_{0}$ current entering into the node, $N_{1,}$  and $i_{1},$ $i_{2}$ leaving the node. Therefore $i_{0}=i_{1}+i_{2}$. Again, $i_{1},$ $i_{2}$ current entering to the node $N_{2},$ and $i_{0}$ leaving the node.

As total charge is conserved, therefore $i_{0}$ current come out from the battery and the same $i_{0}$ current get back into the battery. We also labeled loop A and B to loop rule or KVL around the loop.

Now applying KCL on node $N_{1}$ we get: $i_{0}=i_{1}+i_{2}\text{ (KCL)}$

Applying KVL to loop A we have: As we traveling the loop clockwise direction with the direction of current $i_{2}$, therefore voltage will drop across 50Ω resistor, $-50i_{2}$. Continuing the loop with clockwise direction our next component is the 1.5V battery, we traveling from lo to high ( to +), therefore voltage will rise. So for loop A we have:

$$-50i_{2}+1.5=0\text{ (KVL)}$$

Next applying loop rule (KVL) on loop B, gives: traveling clockwise, in loop B our first component is 3V battery, which we traveling high to low (+ to -), therefore voltage will fall, means $-3V$. Then we counter 25Ω resistor with the direction of $i_{1}$ current flow, therefore the voltage across 25Ω resistor will also fall. Then our next component is 50Ω resistor, as we’re traveling the loop clockwise direction with the opposite direction of current $i_{2}$ through the 50Ω resistor, therefore voltage will rise across 50Ω resistor. So from loop B we have:

$$-3-25i_{1}+50i_{2}=0\text{ (KVL)}$$

So, at this point we are done applying KCL, and KVL on the given circuit. Only thing we are left with some algebra.

From Node A, we find the $i_{2}$, so, $i_{2}=\frac{1.5}{50}=0.03A$

now plug in the value of $i_{2}$ to the equation we got from loop B, gives:

$$-3-25i_{1}+50*(0.03)=0\\\text{Or, }25i_{1}=1.5-3$$

$$\text{Or, }i_{1}=\frac{-1.5}{25}=-0.06A$$

Hence $i_{1}=-0.06$ amps. Here negative indicates that we assume the direction for $i_{1}$ is the wrong direction. So the right direction of $i_{1}$ will be the other way. Now, taking $i_{1},$ $i_{2}$ and plug them into the KCL equation will give the current $i_{0}$. So, $i_{0}=i_{1}+i_{2},$ or $i_{0}=-0.06+0.03=-0.03A$.

Here negative sign also means that the direction of $i_{0}$ we assumed also flow to the wrong direction. Right direction will be the opposite.

## Voltage across Resistors

So, we have the current flow through the each resistor. Now we can determine the voltage across each of them using  Ohm’s Law.

Therefore, voltage across 50Ω resistor, $V_{50Ω}= 50*i_{2}=1.5Volts$

Similarly voltage across 25Ω resistor, $V_{25Ω}=25*i_{1}=-1.5Volts$

## Power dissipated by these resistors

Now can easily determine power dissipated by each resistor from the formula P=IV. So, power dissipated by 50Ω resistor, $P_{50Ω}= i_{2}*(1.5)=0.045$ Watts.

Similarly, power dissipated by 25Ω resistor, $P_{25Ω}=i_{1}*(-1.5)=0.09$ Watts.

//This is the end of discussion on circuit analysis technique implementing KCL, and KVL. Next we will talk about circuit simplification using Thevenin and norton equivalent, here in EE section>>

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