sorry, I am not sure how to explain in correct physics terms
I just want to understand how it is possible that a usb charger with an output capability of 2 amps can charge a device with an input capability of 1 amp significantly faster than a charger with an output capability of 1 amp? This is...
Ok, so that means that those websites are wrong then? It doesn't matter how high the output current is if the maximum current input stays the same?
But how come people say their devices do charge faster if they use a charger with a higher output current?
For example, I read people say their...
Hi, I'm not sure if this is a correct place to ask this, but i have a question about the charging speed for the usb chargers for phones and tablets.
from what I have gathered, the device being charged is the one that pulls the current from the charger, so if a 1 amp input device is charged with...
uhh...it doesn't seem like you are even trying.. doesn't physics require you to have some knowledge of algebra at least...?
you already have the equation right there: V = I x R
all the problems give you 2 of the unknown variables.. why can't you solve for the one remaining variable?
it is very simple..
the current equals the voltage divided by the resistance...
and in first problem, the voltage equals 9 V (9V battery) and the resistance is 18 ohms...
so try to figure it out...
i learned that ω is the angular frequency.. so am i supposed to use ω = 2 pi f for this problem? i am confused since my teacher keeps calling ω the frequency and not the angular frequency...
Homework Statement
I am given a RLC circuit with a AC power source (e.g. signal generator), and i want to find the reactance of the capacitor.
Given: R = 2200 Ohms, L = .025 H, C = (1 x 10-9) F, V0 = 5 V
f = 3000 Hz
Homework Equations
edit: Sorry i wrote wrong formula.. here is...
oops, i calculated wt and A wrong...
now i get B = 1.99x10-6 and
Q(t) = (1 x 10-6) cos(wt) + (1.99 x 10-6) sin(wt)
Q(1) = 2.37 x 10-7 C
Is this the correct answer for the problem?
i'm not sure what you mean by complementary solution...
but for my problem, i solved for A = 1 x 10-6 C, and B = -6.37 x 1011
and then I did
Q(t) = 1 x 10-6 cos(wt) + (-6.37 x 1011) sin(wt)
= 2.23 x 1011 C
is this answer correct? am i supposed to get an enormous charge at 1 second?
ok, I am still a little confused, since in the lecture, I did not learn these two equations:
Q = Q0cos(ωt + θ)
Q = Q0cos(ω[t - t0]),
I was told that Q(t) = Acos(wt) + Bsin(wt), and that A and B depends on initial conditions
Then, for initial conditions at t = 0, A = Q0, and B = 0...
ok, but i am not sure how to do this..
i thought Q0 was the initial condition charge, and it says at t = 0, the charge is 1.0 microC.. Isn't this the initial condition?
I have no idea what to do with the other charge at the other time..
Homework Statement
An LC circuit constructed of a 1 microF capacitor and a 1 microH inductor
is set in oscillation so that the charge on the capacitor is 1.0 microC at t = 0
and 2.0 microC at t = 1.57 x 10^-6 s. What is the charge on the capacitor at
t = 1.0 s?
Homework Equations
Q(t) =...
I used Io equal to E / R, voltage over resistance from Ohm's Law, because for "long, long time", the initial current is equal to that...
so is my answer correct?