Kirchhoff's Laws

Kirchhoff's Laws

Kirchhoff's Current Law
At any instant the sum of all the currents flowing into any circuit node is equal to the sum of all the currents flowing out of that node:
SI_in = SI_out

Similarly, at any instant the algebraic sum of all the currents at any circuit node is zero:
SI = 0

So, for the node of picture here, For example

I₁+I₂+I₃ = I₄+I₅

Kirchhoff's Voltage Law
At any instant the sum of all the voltage sources in any closed circuit is equal to the sum of all the voltage drops in that circuit:
SV= SIR

Similarly, at any instant the algebraic sum of all the voltages around any closed circuit is zero:
SV - SIR = 0  

So, for the  picture here, For example

                                       Here, V₀  = V₁+V₂+V₃

Example: Given circuit below, find V₂, V₀, I₂, R₁, R₂

I₂ = 3/2  =  1.5A

According to Ohm's law, we have .

Apply KVL to the loop on the right to get:

                                    -V₂+5-3 = 0,        V₂ = 2v

                                              And

                                                         R₂= V₂/I₂    = 2V/1.5A  = 1.33 ohm

According to Ohm's law, we have .

Apply KCL to the middle node on top to get:

2- I₁ – I₂  = 0,

2- I₁ -1.5 =0

I₁ = 0.5 A

And

                                                    R₁ = 5V / 0.5A

                                                     R₁ = 5V / 0.5A

                                                      R₁ = 10 ohm

Again by Ohm's law, we get .

Apply KVL to the loop on the left to get:

-3×2 – 5+ V₀  = 0,   V₀ = 11v                                                                                   

                                                                                          

                                                                                    

Example:   Find all branch current of the circuit,

Solution:

 Apply KVL in Loop ABDA, we get

    5 – x – z + y = 0

Or, x – y + z = 5  ..............(i)

Apply KVL in Loop BCDB, we get

      z – x + 5 + y +z +z =0

or x – y – 3z = 5  ...............(ii)

Apply KVL in Loop ADCEA, we get

.     –y –y – z +10 – x – y = o

Or, x + 3y +z = 10 ..................(iii)

From equation (i) and (ii), we get

    2z = 0   or z = 0

From equation (i) and (iii), we get

  x-y = 5 ...........(iv)

and x+3y = 10 .....(v)

now from equation (iv) and (v), we get

-4y = -5   , or y = 1.25

So,   x = 5+1.25= 6.25

So,  Current of brunch AB = Current of brunch BC= 6.25 Amps

      Current of brunch AD = Current of brunch DC =1.25 Amps

     Current of brunch BD = 0

    Current of brunch CEA = 6.25 + 1.25 = 7.5 Amps

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Example: Find V_AB,  V_BC,   V_CA  from the circuit using Kirshoff`s law

Solution:

We get from Loop ABGEA,   -I₁ R₁ – 10 + 12 = 10,

                                          Or, I₁R_{1 =}  2V   or   V_AB =  2V  (Ans)

Apply KVL in Loop BCFGB, we get

-I₂R₂ – 12 + 10 = 0

I₂R₂ = -2V   or V_BC = -2V (Ans)

Apply KVL in Loop AEFCA, we get

                                              -12 + 12 – I₃R₃ = 0

                                    Or, I₃R₃ = 0V    or, V_CA = OV (Ans)