Resistance and DC circuit question

[SmokeyMTNJim]

Here is the problem: I have already lost the ability to get the correct answer for AI

Homework Statement

(a) What is the maximum potential difference that can be applied to the terminals a and b?
33.541

Which resistor will have the largest current? V

(b) For the voltage determined in part (a), what is the power delivered to each resistor?
resistor on the left W
resistor at the top of the loop W
resistor at the bottom of the loop W

(c) What is the total power delivered to the combination of resistors?
W

Homework Equations

:
V=IR, P=IV, P=I^2R
Resistors in series = R+R+R...
Resistors in parallel (1/R)+(1/R)+(1/R)...[/B]

The Attempt at a Solution

My thought process was: If the max power to anyone resistor could be 22.5 watts then use P=IV, substitute the V for V=IR.
My equation is now P=I^2R or 22.5W = I^2 (100Ω) => I = .4743 A for current at the resistor on side A. V=IR => V=.4743(100) => V=47.43

I tried a similar method for the other side after consolidating the Resistors 1/100 + 1/100 = .2 => 1/.2 = 50. Basically a------100Ω--------50Ω--------b

No dang just thought to add these to together and get a Reqiv. for the whole thing. So i just used the same method of P=I^2R to get a I = .3873.
V=IR => 150Ω(.3873) = 58.095 V << Is this what you would get for part a? I don't really want to plug in answers yet for the problem if its wrong

 

[SammyS]

Here is the problem: I have already lost the ability to get the correct answer for AI

Homework Statement

(a) What is the maximum potential difference that can be applied to the terminals a and b?
33.541

Which resistor will have the largest current? V

(b) For the voltage determined in part (a), what is the power delivered to each resistor?
resistor on the left W
resistor at the top of the loop W
resistor at the bottom of the loop W

(c) What is the total power delivered to the combination of resistors?
W

Homework Equations

:
V=IR, P=IV, P=I^2R
Resistors in series = R+R+R...
Resistors in parallel (1/R)+(1/R)+(1/R)...[/B]

The Attempt at a Solution

My thought process was: If the max power to anyone resistor could be 22.5 watts then use P=IV, substitute the V for V=IR.
My equation is now P=I^2R or 22.5W = I^2 (100Ω) => I = .4743 A for current at the resistor on side A. V=IR => V=.4743(100) => V=47.43

I tried a similar method for the other side after consolidating the Resistors 1/100 + 1/100 = .2 => 1/.2 = 50. Basically a------100Ω--------50Ω--------b

No **** just thought to add these to together and get a Reqiv. for the whole thing. So i just used the same method of P=I^2R to get a I = .3873.
V=IR => 150Ω(.3873) = 58.095 V << Is this what you would get for part a? I don't really want to plug in answers yet for the problem if its wrong

Please state the complete problem.

 

[SmokeyMTNJim]

Three 100 Ω resistors are connected as shown in the figure. The maximum power that can safely be delivered to anyone resistor is 22.5W.

(a) What is the maximum potential difference that can be applied to the terminals a and b?
33.541

Which resistor will have the largest current? V

(b) For the voltage determined in part (a), what is the power delivered to each resistor?
resistor on the left

W
resistor at the top of the loop W
resistor at the bottom of the loop W

(c) What is the total power delivered to the combination of resistors?
W

 

[SmokeyMTNJim]

Three 100 Ω resistors are connected as shown in the figure. The maximum power that can safely be delivered to anyone resistor is 22.5W.

(a) What is the maximum potential difference that can be applied to the terminals a and b?
33.541

Which resistor will have the largest current? V

(b) For the voltage determined in part (a), what is the power delivered to each resistor?
resistor on the left

W
resistor at the top of the loop W
resistor at the bottom of the loop W

(c) What is the total power delivered to the combination of resistors?
W

Please state the complete problem.

I apologize for not including that top part.

 

[SammyS]

The Attempt at a Solution

My thought process was: If the max power to anyone resistor could be 22.5 watts then use P=IV, substitute the V for V=IR.
My equation is now P=I^2R or 22.5W = I^2 (100Ω) => I = .4743 A for current at the resistor on side A. V=IR => V=.4743(100) => V=47.43

I tried a similar method for the other side after consolidating the Resistors 1/100 + 1/100 = .2 => 1/.2 = 50. Basically a------100Ω--------50Ω--------b

No **** just thought to add these to together and get a Reqiv. for the whole thing. So i just used the same method of P=I^2R to get a I = .3873.
V=IR => 150Ω(.3873) = 58.095 V << Is this what you would get for part a? I don't really want to plug in answers yet for the problem if its wrong

The resistor closest to point a can have a maximum current of approx. 0.4743 A.

If the equivalent resistance is 150 Ω, what voltage is required to have a current of 0.4743 A ?

 

[SmokeyMTNJim]

The resistor closest to point a can have a maximum current of approx. 0.4743 A.

If the equivalent resistance is 150 Ω, what voltage is required to have a current of 0.4743 A ?

Why do we use the equivalent resistance, but not the equivalent current that I found, was that an erroneous calculation?

 

[SammyS]

Why do we use the equivalent resistance, but not the equivalent current that I found, was that an erroneous calculation?

Look at it another way.

You can have at most 0.4743 A through the resistor on the left. Then how much current will flow through each of the two resistors that are in parallel?

Assuming you get that right, what is the voltage drop across a & b ?

 

[RaulTheUCSCSlug]

It helps to write down the certain rules for parallel and series circuits. When Voltage (potential difference) is the same, when does it change, when does the current change, when it is the same. When I did these questions back in high school, what helped was working the problems out with easier numbers and figuring out V,I,R, at different points in circuits.