# Use node-voltage method to find current?

use "node-voltage" method to find current?

## Homework Statement

http://imageshack.us/a/img213/8582/homeworktest2prob1.jpg [Broken]

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

## The Attempt at a Solution

I assume it means nodal analysis, so:

http://imageshack.us/a/img594/8576/homeworktest2prob1edit.jpg [Broken]

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

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phinds
Gold Member
2019 Award

I did a quick check and got Ia = 2 so I figure you've got the rest right as well, but I didn't actually check any of them

berkeman
Mentor

It looks funny to have both Vb and Vc, since they are the same node. I'd just label it Vb.