Question

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

The impedance of Q at this frequency is

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

Answer 1

The correct option is D $\sqrt{9524}\Omega$

For impedence in coil Q
$Z=\sqrt{R_2^2+(X_L{)}^2}$ ....(i)
Now if the maximum current flows through P and Q, then circuit is at resonance. Also at resonance $\omega =\frac{1}{\sqrt{LC}}$
$\omega =\frac{1}{\sqrt{4.9\times {10}^{-3}\times {10}^{-6}}}\omega =\frac{{10}^5}{7}$
So we can find $X_L$ as
$X_L=\omega L=\frac{{10}^5}{7}\times 4.9\times {10}^{-3}=70\Omega$
Putting values in eq(i)
$Z=\sqrt{{68}^2+{70}^2}=\sqrt{9524}\Omega$