## 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)

(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__
nice..

ReplyDeletenice but less examples

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

ReplyDeleteFor Q3. The final voltage for the loop ABCDA is not 6 but -40.

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