HOw much power is drawn from a one volt battery?

[jersey]

Homework Statement

each bulb in the circuit has a resistence of one ohm. how much power is drawn from the one volt battery?

to see the circuit diagram, click on this link:

it is problem 16 on page 123

To see the circuit diagram, go here:
http://www.lightandmatter.com/bk4c.pdf


GO to the link, scroll to page 123, and the diagram is problem 16.

Homework Equations

Ohms Law
Loop rule
Junction rule

The Attempt at a Solution

I have
D=B+E
A+B=C

and using Loop rule three times:
Va+Vc=1
Vd+Ve=1
Va=Vd+Vb

but i can't go on to solve the the five equatios cause i get lost. Can you help me?

 

[Redbelly98]

Homework Equations

Ohms Law
Loop rule
Junction rule

The Attempt at a Solution

I have
D=B+E
A+B=C

Since you know the resistances, you can rewrite these in terms of the voltages.

For example, D = Vd/Rd, etc.

. . . and using Loop rule three times:
Va+Vc=1
Vd+Ve=1
Va=Vd+Vb

Shouldn't that middle equation be

Vd + Ve + Vf = 1​

?

 

[jersey]

regarding the middle equation, i assumed e anf F to be one single bulb with a resistance of 2 ohms.

I'm not sure how to re wrie each in relation to volatge like you say? Can you help?

 

[jersey]

P=V^2/R for each.

 

[Redbelly98]

regarding the middle equation, i assumed e anf F to be one single bulb with a resistance of 2 ohms.

Okay.

I'm not sure how to re wrie each in relation to volatge like you say? Can you help?

Sure.
From Ohm's law, we know the current through bulb D is

D = Vd / Rd​

So we can replace D in the equations with Vd/Rd instead.
I.e., instead of

D = B + E​

we get

Vd/Rd = B + E​

Then do the same for B and E.

 

[jersey]

Ok, so my new equations are:
Vd/Rd=B+E
Which becomes
Vd/Rd=Bd/Rb+Ve/Re

 

[Redbelly98]

Yes, good. Here are two more things to do:

1. Since you know what Rd, Rb, and Re are, you might as well put those numbers into the equation.

2. Do the same thing for the A+B=C equation.

After doing those, try to solve for the 5 unknowns Va, Vb, etc.

I'm logging off for the night, so good luck.

 

[jersey]

Substituting Rd, Rb and Re = 1 into equation gives:
Vd/1=Bd/1+Ve/1


A+B=C
becomes ?

sorry, I'm trying!

 

[Redbelly98]

Substituting Rd, Rb and Re = 1 into equation gives:
Vd/1=Bd/1+Ve/1

Okay, let's work some more with that equation before moving on.

Doesn't Re=2, since you have combined e and f into one single bulb with a resistance of 2 ohms?

What is "Bd"? I think you meant to write Vb there.

Also, a simpler way to write "Vd/1" would be ____?

 

[jersey]

Vd^-1

 

[Redbelly98]

... a simpler way to write "Vd/1" would be ____?

Vd^-1

What? No. Look at these examples:

\frac{1}{1} = 1

\frac{2}{1} = 2

\frac{3}{1} = 3​

You're just dividing each number by 1, so you get the same number.

\frac{Vd}{1} \ = \ ?​

 

[jersey]

sorry Vd/1 is the same as Vd

 

[jersey]

Vd=Vb+Ve

 

[Redbelly98]

Vd=Vb+Ve

Almost. Just one little error to address:

Doesn't Re=2, since you have combined e and f into one single bulb with a resistance of 2 ohms?

 

[jersey]

Vd=Vb+2Ve

 

[Redbelly98]

No, try again. Look at what you wrote in post #6.

I'm logging off for the night but will be back online tomorrow.