0
\$\begingroup\$

In an RC high pass circuit, when you apply a step input, the capacitor doesn't have any drop initially. My professor's explanation for this was that the capacitor doesn't change voltages instantaneously. He presented the formula \$ V= \frac1C\int_0^T idt\$.

Could someone explain to me why there is no drop across the capacitor? I didn't understand the explanation given by my professor.

\$\endgroup\$
1
  • \$\begingroup\$ how about accepting any of these answer? \$\endgroup\$ Dec 20, 2014 at 11:34

3 Answers 3

2
\$\begingroup\$

From the professor's explanation you know that the voltage across the capacitor is proportional to the integral of current wrt time, so for the voltage to change instantly, the current would have to be infinite.

Unless you have one if those imaginary circuits with no resistance or inductance in a loop, that can't happen. In your case you have the R of the RC, so the current is limited to input voltage divided by R.

\$\endgroup\$
1
\$\begingroup\$

Can someone explain to me why there is no drop across the Capacitor

For an ideal capacitor, the current through is proportional to the time rate of change of voltage across

$$i_C = C \frac{dv_C}{dt}$$

The faster the voltage changes, the larger the magnitude of the current. A voltage 'jump', an instantaneous change in voltage, would imply an infinitely large current. Thus, the voltage across the capacitor must be continuous (no 'jumps') for the current through to have a value.

If the input voltage to the RC circuit is an ideal step, the voltage across the capacitor, the instant before the input voltage changes, is zero. Since the voltage across the capacitor must be continuous, the instant after the input voltage changes, the voltage across the capacitor must be zero (no jumps).

However, the current through the capacitor can instantaneously change. Before the input voltage changes, there is zero current. The instant after the input voltage changes, the capacitor current is maximum and then decays exponentially to zero.

\$\endgroup\$
0
\$\begingroup\$

Since an instantaneous step contains all frequencies, and the reactance of a capacitor decreases as frequency increases, at the instant of the step the reactance of the cap will be zero and it'll look like a short circuit.

\$\endgroup\$
3
  • \$\begingroup\$ I see your answer can help a lot getting the idea, but I'd say that only the "infinite frequency" component should pass unharmed... I mean, I get your example but it might be misleading for others in my opinion. \$\endgroup\$ Aug 3, 2014 at 8:07
  • \$\begingroup\$ @Vladimir: So far so good... :-) \$\endgroup\$
    – EM Fields
    Aug 3, 2014 at 8:17
  • \$\begingroup\$ Yep I really appreciate simple examples that help to understand how things work but I really can't think of a way of improving yours. I guess it's good as is ;) \$\endgroup\$ Aug 3, 2014 at 8:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.