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
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
This means that the equivalent resistance between points
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")#