ECE 202 Chapter 2 Homework Solutions

Problems 2,3,5,11,12,14,15

(I haven't had a chance to draw diagrams, so refer to the original problem)

  2.2) a) 8 b) 6 c) 4

d) va,R1 ; vb,R2

e) 6 (Assuming the current in the 2 series pairs is the same)

f) (1) va-R1-R4-R2

(2) vb-R2-R5-R3

(3) R4-R6-R5

(4) va-R1-R6-R3-vb

(5) va-R1-R5-R5-R2

(6)va-R2-R4-R6-R3

(7) vb-R2-R4-R6-R3

      2.3) In this problem, we are given the current in the branch with the 25 W resistor, and asked to find the current in the source and the other branch.

First, we'll use Ohm's law to find the voltage in the branch we know: v=ir= 14x25=350V

Next we'll use KVL to find that 350V= v₁₆+v₅₀+v₄

Using Ohm's law:

350V = i_a(16)+i_a(50)+i_a(4) = 70i_a

i_a = 5A

Via KCL , i_g =14A+5A=19A.

Via KVL, the voltage across the source is 350V.

p=vi=(19)(350)=6,650W

    2.5) The interconnection is invalid because KCL is violated. Look at the junction of the 3 current sources: According to KCL, 10-20-5=0, which obviously isn't true.

    2.11) Use KVL around the path of the 40 and 60 ohm resistors, then substitute in Ohm's Law:

40i_a+60i_b = 0        i_a = 1.5i_b

Then, use KCL at the top Node:

i_g-i_a-i_b = 0         i_g = 2.5i_b

Finally, KVL around the outer loop:

300 = 6i_g+60i_b = 75i_b

i_b = 4A which is the anser to part (b), and

i_a = 6A (a)

(c) v_o = 60i_b = 240V

(d) The power in the resistors is simple - since we now know all the currents,

we use the fact that p = vi = (ir)i = i²r at each resistor:

6 ohm: 600W

40 ohm: 1440W

60 ohm: 960W

(e) Total power absorbed is the sum of the 3 resistor values, 3000W

Total power delivered is the product of the voltage and current in the source, (300V)(10A)=3000W

    2.12) (a) Using Ohm's law with the given current, the voltage across the 25 ohm resistor is 100,

so using KVL around the outer loop:

v_8ohm = 180-100 = 80V

The current thru the 8 ohm res. is therefore 80/8=10A

Using KCL , the current thru the 10 ohm resistor is 10A-4A =6A

And the voltage across that resistor is 60V.

Using KCL around the lower right loop:

v_70ohm = 80+60 = 140V

and the current i₁ = 140/70=2A

(b) Power: (use i²R ):

5 ohm: 320W

25 ohm: 400W

70 ohm: 280W

10 ohm: 360W

8 ohm: 800W

    2.14) Use KVL and Ohm's law thru the 3 resistors:

80i_a = 30i_b+90i_b

i_a = 1.5i_b

From KCL at the top node:

4A = i_a+i_b = 2.5i_b

i_b = 4/2.5 = 1.6A

i_a = 2.4A

(b) v_g = 80i_a = 192V

    2.15)