TGIF! why? cause i don't have classes on Fridays.. XD.. but i do have classes on Saturdays.. DX.. for two weeks now, our broadband net is not working, so i do blogging at our office LOL.. and i don't get much initiative to do a post when i'm here.. DX
anyway, it has been an amazing but stressful week.. my brain's overdoing it again.. DX.. one of our professors is really confusing me, and what he teaches violates my principles.. i might as well share it then? this is circuit analysis btw..
Problem:
The power absorbed by the element is P = 24e^-8t and the voltage across the element is V = 8e^-4t. Calculate:
a. Current(I) at t = 2.5s
b. Charge (q) at 0.5s <= t <= 2s
Given are:
Power(P) = 24e^-8t
Voltage(V) = 8e^-4t
seems easy.. here is how my professor solved it..
a.)
I = P/V d/dt(I) = d/dt(P/V)
I = d/dt(24e^-8t/8e^-4t)
I = d/dt(3e^-4t) I = -12e^-4t
I = -12e^-4(2.5) I = 5.448 x 10^-4 A
b.)
note: ~ is an integrand
q = It
q = ~It
q = ~(Pt/V) q = ~[(24e^-8t * t)/8e^-4t] |0.5 to 2
q = whatever the answer, i don't remember..
got no problem with the b.), but a.) did confused me.. after differentiation, the unit would be A/s ayt? and the unit of current is only A.. so i thought that if it should be differentiated, then it should be done parametrically.. in that case it would be:
I = d/dt(P) / d/dt (V)
dP/dt = d/dt(24e^-8t)
dP/dt = -192e^-8t
dV/dt = d/dt(8e^-4t)
dV/dt = -32e^-4t
I = dP/dt * dt/dV
I = -192e^-8t/-32e^-4t
I = -192e^-8(2.5)/-32e^-4(2.5)
I = 2.724 x 10^-4A
this time, i solved it through parametric differentiation, so that the dt would be cancelled, thus, no extra unit.. only A, and not A/s.. one of my classmate gave me the idea, and we didn't get the chance to ask our professor about it.. don't know which one to do when the exam comes.. DX.. if any of us is wrong, or if neither of us is correct, i'd appreciate it if you'd tell me..
btw, i got you a cool site you can have fun with.. it is..
AKINATOR!
image by Akinator
click the image for the site.. have fun wasting your time!.. =']
and yeah, thanks to Lloyd for the new banner..
anyway, it has been an amazing but stressful week.. my brain's overdoing it again.. DX.. one of our professors is really confusing me, and what he teaches violates my principles.. i might as well share it then? this is circuit analysis btw..
Problem:
The power absorbed by the element is P = 24e^-8t and the voltage across the element is V = 8e^-4t. Calculate:
a. Current(I) at t = 2.5s
b. Charge (q) at 0.5s <= t <= 2s
Given are:
Power(P) = 24e^-8t
Voltage(V) = 8e^-4t
seems easy.. here is how my professor solved it..
a.)
I = P/V d/dt(I) = d/dt(P/V)
I = d/dt(24e^-8t/8e^-4t)
I = d/dt(3e^-4t) I = -12e^-4t
I = -12e^-4(2.5) I = 5.448 x 10^-4 A
b.)
note: ~ is an integrand
q = It
q = ~It
q = ~(Pt/V) q = ~[(24e^-8t * t)/8e^-4t] |0.5 to 2
q = whatever the answer, i don't remember..
got no problem with the b.), but a.) did confused me.. after differentiation, the unit would be A/s ayt? and the unit of current is only A.. so i thought that if it should be differentiated, then it should be done parametrically.. in that case it would be:
I = d/dt(P) / d/dt (V)
dP/dt = d/dt(24e^-8t)
dP/dt = -192e^-8t
dV/dt = d/dt(8e^-4t)
dV/dt = -32e^-4t
I = dP/dt * dt/dV
I = -192e^-8t/-32e^-4t
I = -192e^-8(2.5)/-32e^-4(2.5)
I = 2.724 x 10^-4A
this time, i solved it through parametric differentiation, so that the dt would be cancelled, thus, no extra unit.. only A, and not A/s.. one of my classmate gave me the idea, and we didn't get the chance to ask our professor about it.. don't know which one to do when the exam comes.. DX.. if any of us is wrong, or if neither of us is correct, i'd appreciate it if you'd tell me..
btw, i got you a cool site you can have fun with.. it is..
AKINATOR!
image by Akinator
click the image for the site.. have fun wasting your time!.. =']
and yeah, thanks to Lloyd for the new banner..
2 comments:
NOSE BLEED. Hehehehe! Do u know something about statistics? Maybe u can help me. Hahahaha!!
@ ralph
waaaah... di ko alam un eh, HS pa ung huli qng stat..
Post a Comment