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Simple Power in a Circuit Problem

  • Engineering
  • Thread starter ravenprp
  • Start date
  • #1
74
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1. Homework Statement

Determine I_0 in the network.

2. Homework Equations

P = VI
Passive Sign Convention Rule

3. The Attempt at a Solution

Here's what I did. I calculated the power absorbed/supplied by the bottom left network.

Element 3: P = VI = (6V)(I_0 A) = 6I_0 W
Element 5: P = VI = (6V)(4A) = 24 W
Element 6: P = VI = (8V)(6A) = 48 W
V-Dependent Source: P = VI = (4 * 2 A)(-6A) = -48 W (I did this because of the passive sign convention.

6I_0 + 24 + 48 - 48 = 0
I_0 = (-24) / 6
I_0 = -4 A

It says something is wrong, but I'm not sure why. I'm 99% sure I followed the rules for PSC. I've attached an image of the circuit. I've done it only using this loop, but also, if you check down below, I've done it for every element (I get differing answers).

Thanks :smile: :smile:

(Circuit picture taken from WileyPLUS homework, please do not redistribute/copy).


-------

If I had to calculate each element, in the WHOLE entire network, would this work be correct:

E1: 64W
E2: 60W
E3: I_0 * 6 W
Independent 24V Source: -192W
E4: 32W
E5: 24W
E6: 48W
Dependent 4I_x V Source: -48W

so... (Im going to leave out the W unit in the equation for now, so it doesn't confuse anyone)

64 + 60 + 6I_0 + (-192) + 32 + 24 + 48 - 48 = 0

6I_0 + (-12) = 0
6I_0 = 12

I_0 = 2A
 

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Answers and Replies

  • #2
Well using KCL at that node, I'd say the 2nd answer is right.

What does this tell you about the power around a loop (with a circuit with more than one loop)?
 
  • #3
74
0
Power Supplied = Power Absorbed

So 2Amps would be correct, then right?
 
Last edited:
  • #4
Look at this example I made:

powerexample.jpg
 
  • #5
74
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Ah, I see. Makes sense.
 

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