# Question #9f759

##### 3 Answers

Circuit Simplification.

#### Explanation:

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

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

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

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

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

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.

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

For part **(b)**

#### Explanation:

So, your circuit looks like this

You know that the power dissipated by the

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

You know that **equal to**

That happens because both currents pass through

This means that

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

The equivalent resistance between points *parralel*.

This means that the equivalent resistance between points *series*.

This means that

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

The equivalent resistance of the circuit is

#V = I * R#

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