Finding the Instantaneous Current at t=1.2s

  • Thread starter Thread starter unteng10
  • Start date Start date
  • Tags Tags
    Current
AI Thread Summary
To find the instantaneous current at t=1.2s, the charge equation q=q1t^3 + q2t + q3 is differentiated to obtain I=dq/dt, resulting in I=12C/s^3t^2 + 4.5C/s. Substituting t=1.2s into the equation gives I=12(1.2)^2 + 4.5. After calculating, the correct current is found to be 25.236 A. The discussion highlights the importance of careful substitution and unit conversion in solving physics problems. Simple mistakes can lead to incorrect answers, emphasizing the need for thorough checking.
unteng10
Messages
10
Reaction score
0

Homework Statement


The quantity of a charge passing through a surface of area 2.67 cm^2 varies with time as q=q1t^3 + q2t + q3, where q1=4 C/s^3, q2=4.5 C/s, q3=9.5 C and t is in seconds. What is the instantaneous current through the surface at t=1.2s? Answer in units of A



Homework Equations


I=dq/dt



The Attempt at a Solution


I found that dq/dt = 12C/s^3t^2 + 4.5C/s

To find I, I need to plug in t. Which gives me (12C/s^3)(1.2s) + 4.5C/s, however I am unsure of how to do the conversion to come to the right answer. I know that 1A = 1C/s. I plugged in 25.236 A in but it came out wrong. I know there is some conversions I am missing, any pointers? Thanks
 
Physics news on Phys.org
(12C/s^3)(1.2s) + 4.5C/s
Check the substitution.
 
I got it. Thanks for pointing that out, its amazing how the simple mistakes always get you.
 
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}...
Back
Top