# Maximal voltages across cap. and res. in serial RC circuit with pulse voltage source

1. Jan 31, 2013

### GalMichaeli

1. The problem statement, all variables and given/known data
For the circuit in the picture below, with $V_{c}(t=0) = 0$ and a voltage source with period T described by
$$V_s(t) = \sum_{-\infty}^\infty (-1)^{n}g(t-nT)$$
where
$$g(t) = 7[u(t)-u(t-T)]\quad [Volt]$$
and $u(t)$ is a step function described by
$$u(t) = \begin{cases} 1 & \text{if } t \geq 0 \\ 0 & \text{if } t < 0 \end{cases}$$
What is the maximal voltages across the capacitor?
What is the maximal voltages across the resistor?

2. Relevant equations

3. The attempt at a solution
The voltage source is a pulse train with amplitude $\pm 14 \quad [Volts]$ and since at time $t = 0$ we may consider the cap. as a short-circuit, we have $V_{R}(t=0) = 14 \quad [Volts].$
I'm having trouble figuring out what the maximal voltage across the cap. is.
Should I apply transient analysis?

Thanx.

#### Attached Files:

• ###### circuit.bmp
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2. Jan 31, 2013

### rude man

Re: Maximal voltages across cap. and res. in serial RC circuit with pulse voltage sou

First, never mind the voltage anywhere at t=0. You don't know what t=0 is. This voltage wavetrain has been running since t = -∞.

Now my gut reaction was Fourier series. But that's the hard way. Instead:

1. realize that v(t), the voltage across C, will vary symmetrically about zero volts since that is the average value of your input, from Vmin to Vmax = |Vmin|.

2.Then realize that the most negative C voltage is just before the input goes from -E to +E (why?). Then realize by symmetry that the max C voltage will occur just before the input transitions from +E to -E (again, be able to justify this statement).

3. Write the KVL: current thru R = current into C starting with t=0 at the -E to +E input transition. This will be a differential equation, easy to solve, in capacitor voltage v(t). Solve with the initial condition v(0+) = Vmin, then solve for v(T) = Vmax. The rest should "follow immediately" as the textbooks say.