Electric Circuits Discussion

Chapter 2Chapter 2

By:Eng. Yousef M.Yazji

Q: 1

If the interconnection in figure below is valid find the total power developed in the circuit ?

Solution :

a

The interconnect is valid since it does not violate Kirchhoff’s laws

a

15+5+I(50v=0 so I(50v)= - 20A.50+10-40+V(5A)=0 so V(5A)= - 20V

p5A = −(5)(20) = −100W p15A = −(15)(50) = −750W p10V = (5)(10) = 50W p50V = (20)(50) = 1000W

p40V = −(5)(40) = −200W

( ) ( )P d e v P ab s=∑ ∑

Q:2Given the circuit shown bellow find

a ) the value of iab ) the value of ibc ) the value of vo

d ) the power dissipation in each resistore ) the power delivered by the 200v source e ) the power delivered by the 200v source

Solution:a)

Write a KVL equation around the right loop−va + vb = −0 so va = vbBy Ohm’s lawva = 300ia and vb = 75ib

Substituting,300ia = 75ib so ib = 4ia

Write a KCL equation, summing the currents leaving:−ig + ia + ib = 0 so ig = ia + ib = ia + 4ia = 5ia−ig + ia + ib = 0 so ig = ia + ib = ia + 4ia = 5ia

Write a KVL equation, −200V + v40 + va = 0

By Ohm’s law,v40 = 40ig and va = 300ia

Substituting,−200V + 40ig + 300ia = 0

Substituting for ig:−200V + 40(5ia) + 300ia = −200V + 200ia + 300ia = −200V + 500ia = 0

500ia = 200V so ia = 200

500

V

b) From part (a), ib = 4ia = 4(0.4A) = 1.6 A.

(c) From the circuit, vo = 75Ω(ib) = 75Ω(1.6A) = 120 V.

(d) Do it with your self

(e)Using the passive sign convention,P source = −(200V)ig = −(200V)(5ia) = −(200V)[5(0.4A)]

= −(200V)(2A) = −400W

Q:3

The current Ia and Ib in the circuit shown bellow are 4 A and -2 A respectively a) Find igb) Find the power dissipated in each resistor c) Find vg

ib←

ia→

Solution

V2=160VV1=100VI1=5AI3=3AVg=190VI4=6AIg=9A

Try to find the power dissipated in each resistor

Q:4

2i→+2i→

+

Find v1 and vg in the circuit shown bellow when vg=5v

−

v1 vg

+

−

Solution:

Q:5

The voltage across the 22.5 resistor in the circuit is 90 V positive at the upper terminal

a) Find the power dissipated in each resistor b) Find the power supplied by the 240V ideal voltage source

Ω

+22.54

1520

90v

+

−

Solution :

Thank You

HW 2.3 ,2.7 , 2.11 , 2.18, 2.25 2.29Due to 12/10/2011