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. Bx P contains a capacitance of 1 $\mu$F is series with a resistance of 32$\Omega$. Coil Q has 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 the 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

The correct option is D 7.70 V, 9.76 V

Maximum current will be flow in resonating condition:

$\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{(4.9\times {10}^{-3})({10}^{-6}}}=\frac{{10}^5}{7}\frac{rad}{s}$

with $I_{max}=\frac{E_0}{R}=\frac{10V}{(32+68)}=\frac{1}{10}A$

So the impedance for box $P$:

$Z_p={[R_1^2+{\left(\frac{1}{\omega C}\right)}^2]}^{1/2}={\left[\right(32{)}^2+{\left(\frac{7}{{10}^5\times {10}^{-6}}\right)}^2]}^{1/2}=\sqrt{5924}=77\Omega$

And for coil $Q$:

$Z_Q=[R^2+(\omega L{)}^2{]}^{1/2}=\left[\right(68{)}^2+(4.9\times {10}^{-3}\times {10}^5/7^2{]}^{1/2}=\sqrt{9524}=97.6\Omega$

Hence

$V_p=I{Z}_p=\frac{1}{10}\times \left(77\right)=7.7V$ ( potential drop across $P$)

and $V_Q=I{Z}_Q=\frac{1}{10}\times \left(97.6\right)=9.76V$ ( potential drop across $Q$)