0
\$\begingroup\$

enter image description here

Across the resistor we have

$$ \begin{array}{c} V_{R}=\frac{V_{i n}}{\left(R+X_{C}\right)} \cdot R \\ V_{R}=\frac{V_{i n} \cdot R}{R+\frac{1}{j \omega C}} \\ V_{R}=\frac{V_{m} R j \omega C}{j R \omega C+1} \end{array} $$ Which gives $$\mathrm{V}_{R}=\frac{\omega R C V_{i n}}{\left(1+R^{2} \omega^{2} C^{2}\right)}(\omega R C+j) $$ Which then gives us the phase of voltage across the resistor. Similarly we can do for the capacitor and we have respectively ( assuming phase of input voltage to be zero) : $$ \phi_{R}=\tan \left(\frac{1}{\omega R C}\right) $$ and $$ \phi_{c}=\tan ^{-1}(-\omega R C) $$ And $$ \begin{array}{l} \omega \rightarrow 0 \\ \phi_{c} \rightarrow 0 \\ \phi_{R} \rightarrow \pi/{2} \end{array} $$ so we see that the voltage across the capacitor is in phase with the input voltage. However we have $$ v_{R}=v_{i}-v_{c} $$ and from this equation we see that if the capacitor voltage is in phase with the input voltage, then the voltage across the resistor is in phase with the input voltage. But this contradicts the above equation which says that $$ \phi_{R} \rightarrow \frac{\pi}{2} $$ Where am i mistaken. Please help.

\$\endgroup\$
  • 1
    \$\begingroup\$ Show how you derive the equation for \$\phi _R\$ \$\endgroup\$ – Chu Jun 27 at 6:20
  • \$\begingroup\$ I think that'll make the post unnecessarily long. The same result is given exactly in this post:en.wikipedia.org/wiki/RC_circuit \$\endgroup\$ – Yasir Sadiq Jun 27 at 6:36
  • 1
    \$\begingroup\$ @chu is on to something. look closely. \$\endgroup\$ – P2000 Jun 27 at 6:41
  • 1
    \$\begingroup\$ The expression for phase angle in the wiki article is wrong; should be \$ \phi = 90 -atan (\omega RC)\$ \$\endgroup\$ – Chu Jun 27 at 6:49
  • 1
    \$\begingroup\$ My expression, above, should have been \$\phi_R=0-arctan(\omega RC)\$ \$\endgroup\$ – Chu Jun 27 at 8:06
1
\$\begingroup\$

You write

then the voltage across the resistor is in phase with the input voltage

But this is not correct. Simply because the right-hand-side vi-vc is real does not imply that the left-hand-side vr is real if vi-vc is zero.

For w approaching 0 the capacitor voltage is in phase with the input voltage, and VR approaches 0 with angle approaching +90deg. This is calculated from the voltage divider equation.

Below are simplified plots for Z=-j/w (C=1,R=1), and H(w) for the resistor.

enter image description here

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thank you it cleared it up dear. One more humble request, chu said$$\phi_R=0-arctan(\omega RC)$$ ,this is incorrect then. \$\endgroup\$ – Yasir Sadiq Jun 27 at 9:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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