- #1
pforpashya
- 12
- 0
please help me solving 2 electrical network problems
file:///C:/Documents%20and%20Settings/user/My%20Documents/My%20Pictures/From%20RSPC_Pashya/Camera%20roll/WP_000127.jpg
In first problem I have to find current through 5Ω resistor using Thevinins eq.
So, in first step i removed 5Ω resistor with open ckt. hence Ix=0 hence the dependent source becomes zero. 4A+1A that is 5A going up. then kcl at above node yields me node voltage 14.28v
so my vth comes 14.28v
now, as Ix becoming zero and dependent source becoming zero we can calculate Rth=1.1428Ω
so to find current thru 5Ω
I will be 14.28/6.1428... hence i got answer as 2.34A.
But answer given to me is 4A.
please help.
In second problem also we need to find current through 24Ω using Thevinins eq.
I applied kvl at input side & kcl at output side got two equations
8Ix-Vx=16
130Ix+Vx=0
solving i got Vx=15v == Vth
then solved for Isc which i got 0.48A
Current throgh 24 ohm i got 0.271A
am i correct?
please help
file:///C:/Documents%20and%20Settings/user/My%20Documents/My%20Pictures/From%20RSPC_Pashya/Camera%20roll/WP_000127.jpg
In first problem I have to find current through 5Ω resistor using Thevinins eq.
So, in first step i removed 5Ω resistor with open ckt. hence Ix=0 hence the dependent source becomes zero. 4A+1A that is 5A going up. then kcl at above node yields me node voltage 14.28v
so my vth comes 14.28v
now, as Ix becoming zero and dependent source becoming zero we can calculate Rth=1.1428Ω
so to find current thru 5Ω
I will be 14.28/6.1428... hence i got answer as 2.34A.
But answer given to me is 4A.
please help.
In second problem also we need to find current through 24Ω using Thevinins eq.
I applied kvl at input side & kcl at output side got two equations
8Ix-Vx=16
130Ix+Vx=0
solving i got Vx=15v == Vth
then solved for Isc which i got 0.48A
Current throgh 24 ohm i got 0.271A
am i correct?
please help