Question

A Box $P$ and a coil $Q$ are connected in series with an $10VAC$ source of variable frequency. Box $P$ contains a capacitance of $1\mu \text{F}$ in series with a resistance of $32\Omega$. Coil $Q$ has a self inductance $4.9\text{mH}$ and a resistance of $68\Omega$. The frequency is adjusted so that the maximum current flows in $P$ and $Q$. Find the impedance of $P$ and $Q$ at this frequency.

• A. $76\Omega ,98\Omega$
• B. $45\Omega ,72\Omega$
• C. $85\Omega ,20\Omega$
• D. $99\Omega ,10\Omega$

Answer 1

The correct option is A $76\Omega ,98\Omega$

The current is maximum at resonance, when

$|X_L-X_C|=0$

$\Rightarrow (\omega L-\frac{1}{\omega C})=0$

$\Rightarrow \omega L=\frac{1}{\omega C}$

$\Rightarrow {\omega }^2=\frac{1}{LC}$

$\Rightarrow \omega =\frac{1}{\sqrt{(4.9\times {10}^{-3})\left({10}^{-6}\right)}}$

$\therefore \omega =\frac{{10}^5}{7}\text{rad/s}$

The impedance of $P$ is given as

$Z_P=\sqrt{\left(R_1^2\right)+X_C^2}$

$=\sqrt{R_1^2+{\left(\frac{1}{\omega C}\right)}^2}=\sqrt{(32{)}^2+{(\frac{7}{{10}^5}\times \frac{1}{{10}^{-6}})}^2}$

$=\sqrt{(32{)}^2+(70{)}^2}$

$=\sqrt{1024+4900}=\sqrt{5924}$

$\Rightarrow Z_P\approx 76\Omega$

The impedance of $Q$ is given as,

$Z_Q=\sqrt{R_2^2+{\omega }^2{L}^2}$

$=\sqrt{(68{)}^2+{(\frac{{10}^5}{7}\times 4.9\times {10}^{-3})}^2}$

$=\sqrt{(68{)}^2+(70{)}^2}$

$=\sqrt{9524}\approx 98\Omega$

Hence, option $\left(A\right)$ is correct.