Capacitor current in function of time

AI Thread Summary
To calculate the current (i) in a circuit with a generator (U), resistor (R), and capacitor (C) as a function of time, one must consider the relationship between voltage (V), charge (Q), and current. The voltage across the capacitor is given by V = Q/C, while the voltage drop across the resistor can be expressed as V = U - R.i. To find the current, integrating the equation I = C(dV/dt) is necessary, but it requires understanding how voltage changes over time. The energy dissipated by the resistor during the capacitor's charging can be calculated using the integral of the power (P = i^2R) over the charging time. Properly combining these equations will yield the desired results.
jaumzaum
Messages
433
Reaction score
33
In a circuit of 1 generator (U) 1 resistor (R) and one capacitor (C), how do I calculate the current i in function of time,
as well as the energy dissipated by the resistor during the time the capacitor is charged?
 
Physics news on Phys.org
Sorry, you will need to show us some effort before you get any help.
 
I have no idea on how to solve this.


I would try to integrate I

\int \frac{C.dV}{dt}

But V changes during time

I think V = Q/C
V=U-R.i and
Q = \int I.dt

But how do I put all together?
 
Last edited:
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