dmitriylm
Apr22-11, 02:56 AM
1. The problem statement, all variables and given/known data
I'm doing a physics lab and have a question about RC circuits. I'm given four formulas:
1a) dQ/dt = -Q/RC
1b) Q = (Qo)e^(-t/RC)
1c) E - RI - Q/C = 0 ---> R(dQ/dt) + Q/C - E = 0
1d) Q(t) = CE[1 - e^(-t/RC)]
I am told that 1b is the solution to 1a and 1d is the solution to 1c.
a) Equation 1b is called "the solution" to the differential question, 1a because when you plug in Q(t) (from 1b) into equation 1a, the resulting left side of the equation is equal to the right side. Prove that 1b is the solution by plugging it into equation 1a.
b) Demonstrate that RC has the dimensions of time.
c) Plug 1d into 1c and show that it works, as you did in question a.
d) Differentiate Q(t) to obtain an expression for the current.
2. Relevant equations
1a) dQ/dt = -Q/RC
1b) Q = (Qo)e^(-t/RC)
1c) E - RI - Q/C = 0 ---> R(dQ/dt) + Q/C - E = 0
1d) Q(t) = CE[1 - e^(-t/RC)]
3. The attempt at a solution
Not quite sure what should be done to make this work. Can anyone please explain this for me?
I'm doing a physics lab and have a question about RC circuits. I'm given four formulas:
1a) dQ/dt = -Q/RC
1b) Q = (Qo)e^(-t/RC)
1c) E - RI - Q/C = 0 ---> R(dQ/dt) + Q/C - E = 0
1d) Q(t) = CE[1 - e^(-t/RC)]
I am told that 1b is the solution to 1a and 1d is the solution to 1c.
a) Equation 1b is called "the solution" to the differential question, 1a because when you plug in Q(t) (from 1b) into equation 1a, the resulting left side of the equation is equal to the right side. Prove that 1b is the solution by plugging it into equation 1a.
b) Demonstrate that RC has the dimensions of time.
c) Plug 1d into 1c and show that it works, as you did in question a.
d) Differentiate Q(t) to obtain an expression for the current.
2. Relevant equations
1a) dQ/dt = -Q/RC
1b) Q = (Qo)e^(-t/RC)
1c) E - RI - Q/C = 0 ---> R(dQ/dt) + Q/C - E = 0
1d) Q(t) = CE[1 - e^(-t/RC)]
3. The attempt at a solution
Not quite sure what should be done to make this work. Can anyone please explain this for me?