- #1
Color_of_Cyan
- 386
- 0
use "node-voltage" method to find current?
http://imageshack.us/a/img213/8582/homeworktest2prob1.jpg
a. Use the node-voltage method of circuit analysis to find the branch currents Ia, Ib, and Ic in the circuit
b. Find power associated with each source and state whether the source is delivering or absorbing power.
V = IR
Nodal analysis ?
P = IV
I assume it means nodal analysis, so:
http://imageshack.us/a/img594/8576/homeworktest2prob1edit.jpg
my equations starting are:
(Va - Vb)/5Ω = Ia
Vb/10Ω = Ib
Vc/40Ω = Ic
3A + Ia = Ib + Ic
Va = 50V and Vb = Vc right at the beginning
So:
3A + (Va - Vb)/5Ω = Vb/10Ω + Vb/40Ω
3A + (Va - Vb)/5Ω = (4Vb + Vb)/40Ω;
(then added (-Va + Vb)/5Ω to both sides and multiplied for common denominator:)
3A = (-8Va + 8Vb + 4Vb + Vb) / 40Ω
(substitute Va = 50V and combine like terms)
3A = (-400V + 13Vb)/40Ω
120V = -400V + 13Vb;
Vb = 520V/13
Vb = Vc = 40V
Ib = (40/10)A
Ib = 4A
Ic = 40/40A
Ic = 1A
Ia = -3A + 4A + 1A
Ia = 2A
P = IV
P from 50V = 2A*50V
= 100W supplied
P from 3A = 3A * 40V
= 120W supplied
Looks good or no?
and the power absorbed by the resistors adds up to the same amount
Homework Statement
http://imageshack.us/a/img213/8582/homeworktest2prob1.jpg
a. Use the node-voltage method of circuit analysis to find the branch currents Ia, Ib, and Ic in the circuit
b. Find power associated with each source and state whether the source is delivering or absorbing power.
Homework Equations
V = IR
Nodal analysis ?
P = IV
The Attempt at a Solution
I assume it means nodal analysis, so:
http://imageshack.us/a/img594/8576/homeworktest2prob1edit.jpg
my equations starting are:
(Va - Vb)/5Ω = Ia
Vb/10Ω = Ib
Vc/40Ω = Ic
3A + Ia = Ib + Ic
Va = 50V and Vb = Vc right at the beginning
So:
3A + (Va - Vb)/5Ω = Vb/10Ω + Vb/40Ω
3A + (Va - Vb)/5Ω = (4Vb + Vb)/40Ω;
(then added (-Va + Vb)/5Ω to both sides and multiplied for common denominator:)
3A = (-8Va + 8Vb + 4Vb + Vb) / 40Ω
(substitute Va = 50V and combine like terms)
3A = (-400V + 13Vb)/40Ω
120V = -400V + 13Vb;
Vb = 520V/13
Vb = Vc = 40V
Ib = (40/10)A
Ib = 4A
Ic = 40/40A
Ic = 1A
Ia = -3A + 4A + 1A
Ia = 2A
P = IV
P from 50V = 2A*50V
= 100W supplied
P from 3A = 3A * 40V
= 120W supplied
Looks good or no?
and the power absorbed by the resistors adds up to the same amount
Last edited by a moderator: