Question

rrent in tive at the top. 3.28 For the circuit in Fig. P3.28, find i, and then use cur- rent division to find io. Figure P3.28 15 12 w 212 Sect ²60 120 3.3 125 V 352 2201 3132

Answer 1

Step 1:

Let us consider a current I₁ in fig 1 as label below

Find the current I_g in the circuit below:

39 m nan 15n an YI in (6) 125V § 12n 3 on { 13n y lo snş zoon Fig.1

The resistance 15Ω,12 Ω and 13 Ω are in series so the equivalent resistance of 15Ω,12 Ω and 13 Ω is

= 15 + 12 + 13 = 40Ω

The resistance 5 Ω and 20 Ω are in parallel then the equivalent resistance of 5 Ω and 20 Ω is:

5 x 20 25 -= 422

So that the equivalent circuit diagram is:

2n VII son s uon 125V Чл F192

Step 2:

In the fig.2:

The resistance 6Ω and 4 Ω are in series so the equivalent resistance of 6Ω and 4 Ω is

= 6 +4 = 10Ω

Now, The resistance 10 Ω and 40 Ω are in parallel, then the equivalent resistance of 10Ω and 40 Ω is:

10 x 40 . - = 8Ω 50

So that the equivalent circuit diagram is:

2n son € 1251 Fig.3

Step 3:

From fig.3, Now the find the current Ig is:

125 = 12.5.

From fig.2, calculate the value of current I₁ using the current divider rule:

1,-1, x 40 40 + 10

1. – 12.5 X 40 – 104 50

From fig.1 calculate current I_O using the current divider rule:

l, x 5 1o 5 + 20

10 x 5 -= 1. = 2 A 25

So that the current Ig and Io are:

19 = 12.5 A

1. = 2 A