1
\$\begingroup\$

I'm currently working through Cogdell's Foundations of Electrical Circuits and came across problem 4.39 d) as shown below:

enter image description here

Earlier in the question I found the frequency-to-be-used as 43.3x10^3 rad/s. Now, in order to make the circuit appear purely resistive the reactive part needs to sum to 0. In part c) of the question the capacitance was being added in series, so I simply used wL-(1/wC)=0. For d) however, the capacitance is to be in parallel with the inductor, leading to the (seemingly) unsolvable equation:

enter image description here

The book says the answer is C=0.1uF, and that it should lead to a real impedance of 400ohms. The only approach to this question I haven't taken is using the formula for the impedance angle for an R(L||C) circuit, but I'm not actually sure what that formula is. Any insight or hints are appreciated! Thanks.

\$\endgroup\$
4
\$\begingroup\$

No, you should consider the capacitor in parallel to the input ports, like below:

enter image description here

Now it's easier to first find the input admittance and then put the imaginary parts equal to zero to find the right capacitance. Reverse the real part to get the real impedance.

\$\endgroup\$
  • 1
    \$\begingroup\$ Ah, that makes sense! I recalculated the impedance expression for C||(R+L) instead of R+(L||C) and got the answer in the book. Funny how sometimes the hardest part of a question is just setting it up right :P Thanks for the help. \$\endgroup\$ – Eric Aug 9 '18 at 21:23
1
\$\begingroup\$

however, the capacitance is to be in parallel with the inductor

No it isn't because that leads to the whole impedance being infinity; a resistor in series with L||C at resonance --> infinity.

The capacitor is applied across the two terminals on the left and the composite impedance is therefore: -

$$\dfrac{R+j\omega L}{1-\omega^2LC +j\omega RC}$$

If you then take the complex conjugate of the denominator and multiply top and bottom of the equation, the denominator becomes purely real but the numerator is complex. Numerator: -

$$R-RLC\omega^2 -j\omega R^2C+j\omega L-j\omega^3L^2C+\omega^2RLC$$

And the solution needed is for when the imaginery parts are zero i.e.: -

$$=R^2C+L-\omega^2L^2C = 0$$

If you whittle this down a bit, you'll find that the frequency where the impedance is real is: -

$$\omega = \sqrt{\dfrac{1}{LC}-\dfrac{R^2}{L^2}}$$

You know what omega, R and L are so plug those into the formula above and rearrange to find C.

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

The easiest way is \$\omega_o=1/\sqrt{LC}\$ then solve for C for series.

Here the reactance of \$X_C\$ cancels \$X_L\$. \$Z=R-\dfrac{1}{\omega C}+\omega L =R+0\$

For shunt use the sum of admittance , Y.

\$Y=1/R|| (\dfrac{1}{\omega L}-\omega C)=1/R || 0=0\$ thus Z is infinite.

In both cases, the solution is the same easy 1st formula.

\$\endgroup\$
  • \$\begingroup\$ No this is wrong Tony. \$\endgroup\$ – Andy aka Aug 9 '18 at 20:23
  • \$\begingroup\$ Ok Andy. .. pls advise \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 9 '18 at 22:29
  • \$\begingroup\$ I think your assumption that the C is parallel to the L (like the OP assumes) is wrong. C is across the input terminals. \$\endgroup\$ – Andy aka Aug 10 '18 at 7:06
  • \$\begingroup\$ Oops my eyes must be getting worse \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 10 '18 at 11:18

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.