Question

A box $P$ and a coil $Q$ are connected in series with an ac source of variable frequency. The emf of source is constant at $10V$. Box $P$ contains a capacitance of $1\mu F$ in series with a resistance of $32\Omega$. Coil $Q$ has a self inductance $4.9mH$ and a resistance $68\Omega$ in series. The frequency is adjusted so that the maximum current flows in $P$ and $Q$. At this frequency the voltage across $P$ and $Q$ respectively

• A. $9.76V$, $8.92V$
• B. $6.29V$, $7.96V$
• C. $7.70V$, $10.92V$
• D. $7.70V$, $9.76V$

Answer 1

Answer: (D)

As this circuitis a series $LCR$ circuit, current will be maximum at resonance,

$\therefore$ Resonance frequency
$\omega =\frac{1}{\sqrt{LC}}$
$=\frac{1}{\sqrt{\left(4.9\times 10^{-3}\right)\left(10^{-6}\right)}}$
$=\frac{10^5}{7}rad{s}^{-1}$
and current $I_{max}=\frac{{\epsilon }_0}{R}$
$=\frac{10}{\left(32+68\right)}$
$=\frac{1}{10}A$
So the impedance for box $P$
$Z_p={\left[R_1^2+{\left(\frac{1}{\omega C}\right)}^2\right]}^{1/2}={\left[{\left(32\right)}^2+{\left(\frac{7}{10^5\times 10^{-6}}\right)}^2\right]}^{1/2}$
$=\sqrt{5924}=77\Omega$
and for coil $Q$, $Z_Q-{\left[R_2^2+{\left(\omega L\right)}^2\right]}^{1/2}$
$={\left[{\left(68\right)}^2+{\left(\frac{10^5}{7}\times 4.9\times 10^{-3}\right)}^2\right]}^{1/2}=\sqrt{9524}=97.6\Omega$
and hence, $V_p=I_{max}{Z}_p=\frac{1}{10}\times \left(77\right)=7.70V$
(potential drop across $P$)
and $V_Q=I_{max}{Z}_Q=\frac{1}{10}\times \left(97.6\right)=9.76V$
(potential drop across $Q$)