Question

Question #9f759

Answer 1

Circuit Simplification.

Explanation:

Start from the right side of the circuit. 20, 10, 20-ohm resistors are series connected, so their eq. resistance is

`R_"eq 1" =20+10+20=50Omega`.

Then this eq.resistance is connected in parallel with another 50-ohm resistor, so their eq.resistance is

`R_"eq 2" = 50/2 = 25 Omega`

This 25-ohm eq. resistor is again connected in series with a 25 ohm, so the eq. resistor once again becomes

`R_"eq 3" = 25+25= 50Omega`

But then another 50-ohm is connected with this eq. resistor in parallel, so eq. resistance becomes

`R_"eq 4" = 50/2=25Omega`

At last, this 25-ohm eq. resistor is connected with the rest

`R_"eq 5" = 10+5 + 25 = 40Omega`

The 2nd answer is a bit lengthy. Hopefully you don't get frustrated. I'm typing the solution. Until I post the 2nd answer, digest this 1st question.

Answer 2

Oops! Misread it...
Here is the second one.
Good work on part one
https://drive.google.com/a/methuen.k12.ma.us/file/d/0B3FEehgBkvyhTVJxbXJxdHZkTkk/view?usp=docslist_api

Answer 3

For part (b) `V= "160 V"`

Explanation:

So, your circuit looks like this

You know that the power dissipated by the `10Omega` resistor located in the lower right of the circuit is `"10 W"`.

What you have to do is use that power to calculate the intensity of the current that passes through that resistor

`color(blue)(P = I^2 * R implies I = sqrt(P/R))`

In your case, you would get

`I = sqrt( (10color(red)(cancel(color(black)("W"))))/(10color(red)(cancel(color(black)("W")))/"A"^2)) = "1 A"`

Next, you need to work backward from this current to determine the current that works its way from the voltage source.

So, current `I_1` is coming out of the source and going into point `color(blue)(A)`, where it splits into currents `I_2` and `I_3`. Current `I_3` goes into point `color(blue)(B)` and plits into currents `I_4` and `I_5`.

You know that `I_4 = "1 A"`, since that's the current that goes through the `"10-W"` resistor. Now, because the equivalent resistance of those three resistors in series is equal to `50Omega`, `I_5` will be equal to `I_4`.

That happens because both currents pass through `50Omega` resistors.

This means that

`I_3 = I_4 + I_5 = 2 * I_4 = 2 * "1 A" = "2 A"`

The equivalent resistance between points `color(blue)(B)` and `color(blue)(C)` is equal to `25Omega`, since you have two `50Omega` resistors in parralel.

This means that the equivalent resistance between points `color(blue)(A)` and `color(blue)(C)` will be `50Omega`, since you have two `25Omega` resistors in series.

This means that `I_3` is equal to `I_2`, since they both pass through `50Omega` resistors. The current that's coming from the source, `I_1`, will thus be

`I_1 = I_2 + I_3 = 2 * I_3 = 2 * "2 A" = "4 A"`

The equivalent resistance of the circuit is `40Omega`, which means that the voltage rise across the source will be

`V = I * R`

`V = 4color(red)(cancel(color(black)("A"))) * 40"V"/color(red)(cancel(color(black)("A"))) = color(green)("160 V")`