[Ohm's law] Why does the answer key use the time after 5 seconds?

AI Thread Summary
The discussion revolves around confusion regarding an answer key that states energy dissipation after 5 seconds, while the calculations seem to reference only 3 seconds within that interval. Participants clarify that "after 5 seconds" implies the time has elapsed, but the current only flows for 3 seconds at 2 A during that period. The importance of understanding the question before consulting the solution is emphasized, as it aids in grasping the underlying concepts. The solution confirms that the current is non-zero for only part of the 5-second interval, which is crucial for accurate calculations. Clarity on these points is essential for mastering the application of Ohm's law in this context.
bluesteels
Messages
28
Reaction score
1
Homework Statement
A 10 Ω resistor carries a current that varies as a function of time as
shown. How much energy has been dissipated by the resistor after 5 s?
Relevant Equations
E= I^2 R delta t
Screen Shot 2022-03-18 at 12.25.31 AM.png

This the answer key::

Screen Shot 2022-03-18 at 12.25.19 AM.png


im confused because it say "how much energy has been dissipated after 5s". So shouldn't you be looking at the ground after 5 sec.

but apparently the solution look at it before 5 sec, am I missing anything, can someone explain
 
Last edited:
Physics news on Phys.org
Yes, "after 5 seconds" means that 5 seconds has passed. Think of it as asking, "after 5 seconds has passed". We just get lazy and don't add the "has passed" part.
 
Drakkith said:
Yes, "after 5 seconds" means that 5 seconds has passed. Think of it as asking, "after 5 seconds has passed". We just get lazy and don't add the "has passed" part.
but on the answer key they got 3 sec before 5 second has passed. Do you see what I mean. The first pic is the question 2nd is the answer key.I'm confused on why are getting the value (3s) within 5 second interval when the question ask for after 5 second
 
In the 5 second interval, the current is only non-zero during 3 of those and when it is non-zero it is 2 A.
 
bluesteels said:
but on the answer key they got 3 sec before 5 second has passed.
For how many seconds out of those 5 is there current flowing through the resistor?
 
bluesteels said:
I'm confused on why are getting the value (3s) within 5 second interval when the question ask for after 5 second
So, you are confused by the solution, not by the question? But you are also confused by the question, which prompted you to look at the solution. Try to understand the question and come up with a solution before you look at the solution. Otherwise, you will not be prepared for test questions and won't do well in the course.

The solution states clearly that during the first 5 seconds the current is 2 A for only 3 of those 5 seconds. Can you verify this by looking at the graph, or is it just something you assume must be true because it's stated in the solution?
 
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}...
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Back
Top