Finding Resistance Between A & B: Solved

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The discussion revolves around calculating the resistance between points A and B using resistors R1 (3.8 Ω), R2 (80.0 Ω), and R3 (20.0 Ω). Participants clarify that the task is to find the equivalent resistance of the network at points A and B. The solution involves combining resistors in series and parallel, ultimately leading to an equivalent resistance of 32.0 Ω. There is a reminder that the focus should be on guiding problem-solving rather than providing direct answers. The conversation emphasizes understanding the principles of series and parallel resistor combinations.
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[SOLVED] Finding Resistance

Homework Statement


A number of resistors of values R1 = 3.8 Ω, R2 = 80.0 Ω, and R3 = 20.0 Ω are connected as shown in the figure. What is the resistance between points A and B?


Homework Equations





The Attempt at a Solution



Im confused because all I've been calculating is the equivalent resistance, what is the difference between the two?
 

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n77ler said:

Homework Statement


A number of resistors of values R1 = 3.8 Ω, R2 = 80.0 Ω, and R3 = 20.0 Ω are connected as shown in the figure. What is the resistance between points A and B?


Homework Equations





The Attempt at a Solution



Im confused because all I've been calculating is the equivalent resistance, what is the difference between the two?


They are just asking for the equivalent resistance of that network, as seen at the entry points A & B. Just use your series & parallel combination rules to reduce the network to a single (equivalent) resistor between A & B, and you will have your answer.
 
ok thank-you
 
The Four R1 and R3 are in series thus those five resistors to the right can be combined by adding the value...(4*3.8+20)=35.2Ω
That resistance is in parallel with the R2 thus the combination of the five and this resistor is equal to 1/(1/35.2+1/80) = 24.4Ω (3sf)
Now the total becomes this plus 2*R1

Total = 24.4+3.8*2 = 32.0Ω (3s.f.)

Just incase you were still stuck, that and I like to go back to some easy electronics now and again to remind me that my head hasn't exploded :)
 
You shouldn't give out the answers. We're not here to give solutions, but rather how to figure out a problem.
 
My Apologies, I was only going from what I saw in the rest of the forum.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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