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 $10\, V$. Box $P$ contains a capacitance of $1\, \mu F$ in series with a resistance of $32\,\Omega$. Coil $Q$ has a self inductance $4.9\, mH$ 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. $9.76\,V$, $8.92\,V$
• B. $6.29\, V$, $7.96\,V$
• C. $7.70\,V$, $10.92\,V$
• D. $7.70\,V$, $9.76\,V$

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 $I_{max}=\frac{\varepsilon_{0}}{R}$
$=\frac{10}{\left(32 + 68\right)}$
$=\frac{1}{10}\,A$
So the impedance for box $P$
$Z_{p}=\left[R^{2}_{1}+\left(\frac{1}{\omega C}\right)^{2}\right]^{1/2}=\left[\left(32\right)^{2}+\left(\frac{7}{10^{5} \times 10^{-6}}\right)^{2}\right]^{1/2}$
$=\sqrt{5924}=77\,\Omega$
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, $V_{p}=I_{max}\,Z_{p}=\frac{1}{10} \times \left(77\right)=7.70\,V$
(potential drop across $P$)
and $V_{Q}=I_{max}\,Z_{Q}=\frac{1}{10} \times \left(97.6\right)=9.76\,V$
(potential drop across $Q$)