Need help finding voltage across a capacitor in an RC circuit?

sl02ggp
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1. Homework Statement

Find the voltage across a capacitor in an RC Circuit, using [V]_{}[/c] (0) = 1, [V]_{}[/i] n(t)=t.

2. Homework Equations
dV/dt = (1/RC)(V)=(1/RC)([V]_{}[/in])


3. The Attempt at a Solution
New to this site: I honestly don't know where to start. Done well in Calculus. Feel DifEq is like a breed of its own..aha

I know you have to use this type of an equation: VC = V0(1− e−t /τ ) where τ = RC and V0 is the initial voltage
 
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anyone?
 
I'm a little confused by your equations..

In general the voltage across a series RC circuit will decay exponentially, on a time constant determined by the value of RC.

The voltage across a resistor is:

V_R = IR = R\frac{dQ}{dt}

The voltage across a capacitor is:

V_C = Q/C

Knowing this, you should be able to set up a simple differential equation which will provide the answer you desire.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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