Physics-Voltage across each resistor in a circuit with both Parallel and Series Connections

Question

Hello:

I know this is probably simpler or somehow different than I am seeing it, but I do not have complete confidence in how to answer to the following question:

Please refer to this link for the diagram for this problem http://forum.allaboutcircuits.com/attachment.php?attachmentid=3993&d=1217562101

Using the information given in the figure, find (a) the equivalent resistance and the current through the independent voltage source, (b) the voltage across each resistor, (c) the current through each resistor, (d) the energy per second wasted by each resistor, and (e) the energy per second wasted by the equivalent resistance. Note that VAB =50 volts.

Now, I have figured out part (a) [I found Req =5 ohms and Ieq =10 amperes]and I feel confident about how I arrived at my answers. However, part (b) I feel like I do not know how to answer confidently. This may have been an incorrect assumption, but since I could not figure out part (b) I have not attempted any of the subsequent parts yet.

Please help me if you would: I just feel incredibly uncertain about how to figure out this sort of thing. Its seems there are no similar examples in my Physics textbook (or anywhere else I could find).

Answer 1

Number the nodes from A to B going clockwise 1, 2, 3, 4, 5, 6, & 7.

(Note: this makes VAB ≡ V17.  Nodes 4 & 5 are at the same potential, as are nodes 6, 7 & B.)

(a)

R35 = (10)(10)/(10 + 10) = 5 Ω

R25 = (10)(5 + 5)/(10 + 5 + 5) = 5 Ω

R26 = (10)(5 + 5)/(10 + 5 + 5) = 5 Ω

R17 = (10)(5 + 5)/(10 + 5 + 5) = 5 Ω = Req

Ieq = IA1 = 50 V/5 Ω = 10 A

Now for some jumping back & forth between (b) & (c):

(b)

V17 = 50 V

V12 = (5 A)(5 Ω) = 25 V

V26 = 50 V - 25 V = 25 V

V25 = (5 A - 2.5 A)(5 Ω) = 12.5 V

V23 = (1.25 A)(5 Ω) = 6.25 V

V35 = (1.25 A)(5 Ω) = 6.25 V

V56 = (2.5 A)(5 Ω) = 12.5 V

(c)

I17 = 50 V/10 Ω = 5 A

I12 = 10 A - 5 A = 5 A

I26 = (25 V)/(10 Ω) = 2.5 A  

I25 = (12.5 V)/(10 Ω) = 1.25 A

I23 = 5 A - 2.5 A - 1.25A = 1.25 A

I35 = (6.25 V)/(10 Ω) = 0.625 A

I34 = (6.25 V)/(10 Ω) = 0.625 A

I67 = 10 A - 5 A = 5 A

I56 = 5 A - 2.5 A = 2.5 A

(d)

P17 = (10)(5^2) = 250 W

P12 = (5)(5^2) = 125 W

P26 = (10)(2.5^2) = 62.5 W

P25 = (10)(1.25^2) = 15.625 W

P23 = (5)(1.25^2) = 7.8125 W

P34 = (10)(0.625^2) = 3.90625 W

P35 = (10)(0.625^2) = 3.90625 W

P56 = (5)(2.5^2) = 31.25 W

(e)

P = (50)(10) = 500 W, which checks with the sum of (d)