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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.

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    \$\begingroup\$ Show how you derive the equation for \$\phi _R\$ \$\endgroup\$
    – Chu
    Jun 27 '20 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\$
    – Kashmiri
    Jun 27 '20 at 6:36
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    \$\begingroup\$ @chu is on to something. look closely. \$\endgroup\$
    – P2000
    Jun 27 '20 at 6:41
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    \$\begingroup\$ The expression for phase angle in the wiki article is wrong; should be \$ \phi = 90 -atan (\omega RC)\$ \$\endgroup\$
    – Chu
    Jun 27 '20 at 6:49
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    \$\begingroup\$ My expression, above, should have been \$\phi_R=0-arctan(\omega RC)\$ \$\endgroup\$
    – Chu
    Jun 27 '20 at 8:06
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\$\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

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  • \$\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\$
    – Kashmiri
    Jun 27 '20 at 9:33

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