# Circuit questions. i dont know how to answer

#### medontknowye

Joined Nov 9, 2011
12

its quite messy but only take notice of the Solid coloured numbers.
Question 1: evaluate current I

Question 2:Calculate the power associated with the dependant power source

Question 3:Calculate the power Delivered by the independant current source

#### medontknowye

Joined Nov 9, 2011
12
bump lkokololjhgjgfdxcvvbbnmjjhgd

#### Georacer

Joined Nov 25, 2009
5,182
Your thread has been moved to the Homework Help section, where it is more appropriate.

The Homework Help section asks that you post up the work you have done so far, so we can see it and find where you went wrong, or suggest better approaches. We will not do all your work for you.

Thank you.

#### medontknowye

Joined Nov 9, 2011
12
someone help me please i dont know where to start,

#### medontknowye

Joined Nov 9, 2011
12
hey bro, please teach me how to answer some of this question i really ont know where to start.

#### t_n_k

Joined Mar 6, 2009
5,455
Sometimes all the 'bumping' and pleading for help won't get much of a response.

There is a reasonable requirement on Homework forum posters to show what attempt(s) they have made to solve the problem. Otherwise there is often a reluctance on the part of other forum members to help.

Have you any working you can post to progress this?

#### medontknowye

Joined Nov 9, 2011
12
the thing is bro i dont even know where to start. can u guys give me tips atleast
of what to do where to start..that would be helpful..or if they can show me how to answer it..i promise i know nothing on this.. it mostly because i got enrolled last week to my uni and we have a test the next week. missed quite alot...so i kinda need help. im pleading not because im lazy its because im genuinly lost

#### t_n_k

Joined Mar 6, 2009
5,455
Surprising as it may seem this is a fairly simple problem.

The voltage at the center node is 5*i volts. Let's call it Vx, where Vx=5*i

The voltage drop [call it Vd] across the 2Ω resistor is Vd=(21-Vx) volts.

The current in the 2Ω is therefore i=Vd/2=(21-Vx)/2=(21-5*i)/2

So we have

i=(21-5*i)/2

One now simply re-arranges this equation to find the actual value of i - at task which I'll leave to you.

#### medontknowye

Joined Nov 9, 2011
12
wow thank you so the answer is 21/7 which is 3...so the current on i(A) is 3 amps. thank you very much....ur are good...how would u do the second one

#### Kermit2

Joined Feb 5, 2010
4,162
Perhaps you might consider a career as a welder, or one in HVAC?

#### medontknowye

Joined Nov 9, 2011
12
please help me out its jst this once....its jst im really behind..a way to do it but it seems very wrong ok..here it is anyway

-5I+21+2I=0
after i re arrange it i get 21/3 which is 7
so I is 7
then i use P=IV but im soo confuse help please

#### medontknowye

Joined Nov 9, 2011
12
wow harsh but i guess its true...maybe i should... i sort of know the answer so..here it is anyway...but please try and help...
-5i+2i+21=0
re-arrange and i get 7amps

i place that one the equation P=IV
7*21 will equal to 142
or should i use the 3amps i got earlier