E, Test One-Additional Prob × https://blackboard olemiss x ( 9 https://blackboar

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E, Test One-Additional Prob × https://blackboard olemiss x ( 9 https://blackboard olemiss x ( G Google Image Result for h/X C fi a https://blackboard.olemiss.edu/webapps/blackboard/execute/content file?cmd=view&content;_id 1602570_1&course;, id=-120566-1 Saif Zahran Abdullah Zahran Al-Rumhi 24 Blackb mar Blackboard Home Page My UM Courses Content Test One - Additional Problem for replacement points Test One - Additional Problem for replacement points Engr 360 Electric Circuit Theory Section 1 2015- 2016 FALL Home Page Content Discussions Groups Tools Help If this item does not open automatically you can open Test One -Additional Problem oftor replacement points here Test 1 - Additional Problem: Name: Solve for Vo in the following circuit using both Loop and Nodal analysis: 1 k12 2 mA 6 V 4 mA 4 V 1 k2

Explanation / Answer

in loop analysis , kirchoff's voltage law (KVL) will be used which states that

" in a loop, net votlage drop is zero."

in nodal analysis, kirchoff's current law (KCL) will be used wich states that

"at any node, net current is zero."

loop analysis:

let the meshes and mesh currents be defined as below:

mesh 1:

upper left mesh

consists of 2 k ohms, 6 volts and 2 mA.

let mesh current be i1 and in clockwise direction.

mesh 2:

upper right mesh

consists of 2 mA, 1 k ohms, and 4 volts.

let mesh current be i2 and in clockwise direction.

mesh 3:

lower left mesh

consists of 6 volts, 4 mA and 1 kilo ohms.

let mesh current be i3 and in clockwise direction

mesh 4:

lower right mesh

consists of 4 volts,4 mA and 2 kilo ohms.

let mesh current be i4 and in clockwise direction

[ Note:

all currents are in mA so that their multiplication with resistances which are in kilo ohms will result in volt

all mesh currents are in clockwise direction and if their magntiude comes out to be negative, then they will be assigned anticlockwsie direction.]

between mesh 1 and mesh 2:

i2-i1=2....(1)

writing KVl for the outer boundary of mesh 1 and emsh 2:

4-6-2*i1-1*i2=0

==>2*i1+i2=-2 ...(2)

solving equation 1 and equation 2:

i2=2/3 mA

and i1=-4/3 mA

between mesh 3 and mesh 4:
i3-i4=4 ...(3)

writing KVL for the boundary of mesh 3 and mesh 4:

6-4-2*i4-1*i3=0

==>i3+2*i4=2...(4)

solving equation 3 and 4 simultaneously:

i4=-2/3 mA

i3=10/3 mA

hence V0=1*i2+2*i4=1*(2/3)+2*(-2/3)=-2/3 volts

Node analysis:

there are in total 5 nodes. lets first name all of them

let the lowest node (junction of 1 kilo ohms, 2 kilo ohm and 4 mA) be named as A and its potential =0 volts

from the lower side, in clockwise direction,

let the nodes are:

B (left side node): junction of 2 kilo ohms, 1 kilo ohms, 6 volts and 4 mA

c (top node) : junction of 1 kilo ohms, 2 kilo ohms and 2 mA

D (right node) : junction of 1 kilo ohms, 2 kilo ohms, and 4 volts

E (centre node) : junction of 6 volts, 2 mA ,4 mA and 4 volts

let voltage all the nodes are :

node A: 0 volt

node B: Vb

node C: Vc

node D: Vd

node E : Ve

now, between node B and node E, 6 volt source is there.

then Ve-Vb=6

==> Vb=Ve-6....(1)

between node E and node D, there is a 4 volt source present

==> Ve-Vd=4

==>Vd=Ve-4...(2)

writing KCL at node A:

(Vb-Va)/1 + 4 + (Vd-Va)/2 =0

noting that Va=0 ,

Vb+4+0.5*Vd=0

using equation 1 and 2:

Ve-6+4+0.5*Ve-2=0

=>1.5*Ve=6+2-4=4

==>Ve=8/3 volts

==>Vb=Ve-6=-10/3 volts

Vd=Ve-4=-4/3 volts

now, writing KCL at node C (top node):

(Vb-Vc)/2 + 2 + (Vd-Vc)/1-0

==>0.5*Vb-0.5*Vc+2+Vd-Vc=0

==> 1.5*Vc=0.5*Vb+2+Vd

using Vb=-10/3 volts and Vd=-4/3 volts

we get

1.5*Vc=-1

==>Vc=-2/3 volts

then required voltage V0=Vc-Va=(-2/3) volts

as can be observed, we are getting same result from both node analysis and loop analysis.

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