Tuesday, August 28, 2012

SOLVED PROBLEMS Kirchhoff law ( KCL )

SOLVED PROBLEMS ( KCL )

Q1)  Determine the value current in 40 Ohms resistance. Refer figure 8.1



Answer:
  • First we have to apply KCL 1 to the network. See figure 8.2   


Note that there are three unknown variables ( I , I1 , I2 ) and therefore we need to solve three equations simultaneously  to find this unknown variables.

Apply KCL to Circuit ABDA

∑ IR   =   ∑ e.m.f 

( I - I1 )*10 + I2*40 - 20I = 0 

10*I - 30*I1 +40*I2 = 0 -------------------------(1)


Apply KCL to Circuit BCDB

(I - I1 - I2)*30 -  (I1+I2)*15  - I2*40 = 0 

30*I - 45*I1 - 85*I2 = 0 -------------------------(2)


Apply KCL to Circuit ADCEA

I1*20 + (I1+I2)*15 = 2

35*I1 + 15*I2 = 2     -----------------------------(3)

Now solve these three equations

I   = 87/785 A
I1 = 41/785 A
I2 = 9/785 A

So the current in 40 Ohms resistance is  9/785 A  From  B to D.


Q2) Determine the current flowing in each of the batteries and the voltage difference between 10 Ohms resistance. Refer the following figure.



Answer :
  • First we have to apply KCL 1 to the network. See figure 8.3



Applying KCL to loop CEDAC

(I1+I2)*10 + I1*2 = 12 

12*I1 + 10*I2 = 12 ----------------------------(1)

Applying KCL to loop CEDBC

(I1+I2)*10 + I2*1 = 8 

10*I1 + 11*I2 = 8------------------------------(2)


By solving (1) &(2) 

I1 = 1.625A
I2 = -0.750A


Current through 10 Ohms resistance is = I1 + I2 = 0.875A

Voltage across 10 Ohms resistance is = 0.875A x 10 Ohms = 8.750V




Q3) Determine the current in 4 Ohms resistance. Refer Figure8.4


Note: This Network has a current source. So this question will teach you something new. To study about Current sources, see my previous post about "voltage and current sources".


Answer:

  • First we have to apply KCL 1 to the network. See figure 8.5



Applying KCL to loop EFADE


I1*2 - I2*10 - (I - I1 - 6)*1 = 0 

I - 3*I1 + 10*I2 = 6 ------------------------------(1)

Applying KCL to loop ABCDA

(I1+I2+6)*2 - 3*(I - I1 - I2 - 6)  + I2*10 = -10

3*I - 5*I1 +15*I2 = 6 ---------------------------(2) 

Applying KCL to loop EDCGE

(I-I1-6)*1 + (I-I1-I2-6)*3 + I*4 = 24

8*I - 4*I1 3*I2 = 48 ----------------------------(3)

By solving these three equations

I = 4.1A
So the current in 4Ohms resistance = 4.1A 

  







5 comments:

  1. nice... but, make it clear of your loop ...i will follow you

    ReplyDelete
  2. For Q3. The final voltage for the loop ABCDA is not 6 but -40.

    Please check again. The final answer of I=4.1 is correct.

    ReplyDelete

A.C Fundamentals (Solved Examples 2) 1) Find the R.M.S of, V (t) =     Sin 2πt +  Sin 6πt Answer First of all, We have to find t...