Question

The impedance of $Q$ at this frequency is:

• A. $200\Omega$
• B. $\sqrt{1350}\Omega$
• C. $55\Omega$
• D. $\sqrt{9524}\Omega$

Answer 1

Answer: (D)

$\sqrt{9524}\Omega$
has $R_1=32\Omega$ $C_1=1\mu F$
has $R_2=68\Omega$ $L_2=4.9mH$
Total impedance of the series combination of and :
$Z=R_1+R_2+jL\omega -j\frac{1}{C\omega }$
The frequency is adjusted so that maximum current flows in and . that is, frequency is adjusted to have minimum total series impedance. Total series impedance will be minimum when the reactance of L and C cancel each other. Therefore,

$\omega =\frac{1}{\sqrt{LC}}=\frac{100000}{7}rad/s$
Therefore at maximum current, impedance of $Q=R_2+jL\omega =68+70j=\sqrt{9524}\Omega$